Research ArticleCombinatorial Interpretation of General Eulerian Numbers

Tingyao Xiong1 Jonathan I Hall2 and Hung-Ping Tsao3

1 Department of Mathematics and Statistics Radford University Radford VA 24141 USA2Department of Mathematics Michigan State University East Lansing MI 48824 USA3Department of Decision Sciences San Francisco State University Novato CA 94132 USA

Correspondence should be addressed to Tingyao Xiong txiongradfordedu

Received 29 August 2013 Accepted 10 October 2013 Published 2 January 2014

Academic Editor Pantelimon Stanica

Copyright copy 2014 Tingyao Xiong et alThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Since the 1950s mathematicians have successfully interpreted the traditional Eulerian numbers and 119902-Eulerian numberscombinatorially In this paper the authors give a combinatorial interpretation to the general Eulerian numbers defined on generalarithmetic progressions 119886 119886 + 119889 119886 + 2119889

1 Introduction

Definition 1 Given a positive integer 119899 define Ω119899as the set

of all permutations of [119899] = 1 2 3 119899 For a permutation120587 = 119901

111990121199013 119901119899isin Ω119899 119894 is called an ascent of 120587 if 119901

119894lt 119901119894+1

119894 is called a weak exceedance of 120587 if 119901

119894ge 119894

It is well known that a traditional Eulerian number 119860119899119896

is the number of permutations 120587 isin Ω119899that have 119896 weak

exceedances [1 page 215] And 119860119899119896

satisfies the recurrence1198601198991

= 1 (119899 ge 1) 119860119899119896

= 0 (119896 gt 119899)

119860119899119896

= 119896119860119899minus1119896

+ (119899 + 1 minus 119896)119860119899minus1119896minus1

(1 le 119896 le 119899) (1)

Besides the recursive formula (1) 119860119899119896

can be calculateddirectly by the following analytic formula [2 page 8]

119860119899119896

=

119896minus1

sum

119894=0

(minus1)119894

(119896 minus 119894)119899

(119899 + 1

119894) (1 le 119896 le 119899) (2)

Definition 2 Given a permutation 120587 = 119901111990121199013 119901119899isin Ω119899

define functions

maj120587 = sum

119901119895gt119901119895+1

119895

119886 (119899 119896 119894) = 120587 | maj 120587 = 119894 amp 120587 has 119896 ascents (3)

Since the 1950s Carlitz [3 4] and his successors havegeneralized Eulerrsquos results to 119902-sequences 1 119902 119902

2

1199023

UnderCarlitzrsquos definition the 119902-Eulerian numbers119860119899119896(119902) are

given by

119860119899119896(119902) = 119902

(119898minus119896+1)(119898minus119896)2

119896(119899minus119896minus1)

sum

119894=0

119886 (119899 119899 minus 119896 119894) 119902119894

(4)

where functions 119886(119899 119896 119894) are as defined in Definition 2In [5] instead of studying 119902-sequences the authors have

generalized Eulerian numbers to any general arithmeticprogression

119886 119886 + 119889 119886 + 2119889 119886 + 3119889 (5)Under the new definition and given an arithmetic pro-

gression as defined in (5) the general Eulerian numbers119860119899119896(119886 119889) can be calculated directly by the following equation

[5 Lemma 26]

119860119899119896(119886 119889) =

119896

sum

119894=0

(minus1)119894

[(119896 + 1 minus 119894) 119889 minus 119886]119899

(119899 + 1

119894) (6)

Interested readers can find more results about the generalEulerian numbers and even general Eulerian polynomials in[5]

2 Combinatorial Interpretation ofGeneral Eulerian Numbers

The following concepts and properties will be heavily used inthis section

Hindawi Publishing CorporationJournal of Discrete MathematicsVolume 2014 Article ID 870596 6 pageshttpdxdoiorg1011552014870596

2 Journal of Discrete Mathematics

Definition 3 Let 119882119899119896

be the set of 119899-permutations with 119896

weak exceedances Then |119882119899119896| = 119860

119899119896 Furthermore given

a permutation 120587 = 119901111990121199013 119901119899 let 119876

119899(120587) = 119894 where 119901

119894= 119899

Given a permutation 120587 isin Ω119899 it is known that 120587 can be

written as a one-line form like 120587 = 119901111990121199013 119901119899 or 120587 can be

written in a disjoint union of distinct cycles For 120587 written ina cycle form we can use a standard representation by writing(a) each cycle starting with its largest element and (b) thecycles in increasing order of their largest element Moreovergiven a permutation 120587 written in a standard representationcycle form define a function 119891 as 119891(120587) to be the permutationobtained from 120587 by erasing the parenthesesThen119891 is knownas the fundamental bijection from Ω

119899to itself [6 page 30]

Indeed the inverse map 119891minus1 of the fundamental bijection

function 119891 is also famous in illustrating the relation betweenthe ascents and weak exceedances as follows [2 page 98]

Proposition 4 The function 119891minus1 gives a bijection between theset of permutations on [119899] with 119896 ascents and the set119882

119899119896+1

Example 5 The standard representation of permutation 120587 =

5243716 is (2)(43)(7615) isin Ω7 and119891(120587) = 2437615119876

7(120587) =

5 120587 = 5243716 has 3 ascents while 119891minus1(120587) = (5243)(716) =

6453271 isin 11988274

has 3 + 1 = 4 weak excedances because 1199011=

6 gt 1 1199012= 4 gt 2 119901

3= 5 gt 3 and 119901

6= 7 gt 6

Now suppose we want to construct a sequence consistingof 119896 vertical bars and the first 119899 positive integers Then the 119896vertical bars divide these 119899 numbers into 119896+1 compartmentsIn each compartment there is either no number or all thenumbers are listed in a decreasing order The followingdefinition is analogous to the definition of [2 page 8]

Definition 6 A bar in the above construction is calledextraneous if either

(a) it is immediately followed by another bar or

(b) each of the rest compartment is either empty orconsists of integers in a decreasing order if this baris removed

Example 7 Suppose 119899 = 7 119896 = 4 then in the followingarrangement

32 |1| |7654| (7)

the 1st 2nd and 4th bars are extraneous

Now we are ready to give combinatorial interpretationsto the general Eulerian numbers 119860

119899119896(119886 119889) First note that

(6) implies that 119860119899119896(119886 119889) is a homogeneous polynomial of

degree 119899 with respect to 119886 and 119889 Indeed

119860119899119896(119886 119889)

=

119896

sum

119894=0

(minus1)119894

[(119896 + 1 minus 119894) 119889 minus 119886]119899

(119899 + 1

119894)

=

119896

sum

119894=0

(minus1)119894

[(119896 + 1 minus 119894) (119889 minus 119886) + (119896 minus 119894) 119886]119899

(119899 + 1

119894)

=

119899

sum

119895=0

[

119896

sum

119894=0

(minus1)119894

(119896 + 1 minus 119894)119899minus119895

(119896 minus 119894)119895

(119899 + 1

119894)]

times (119899

119895) (119889 minus 119886)

119899minus119895

119886119895

=

119899

sum

119895=0

119888119899119896(119895) (

119899

119895) (119889 minus 119886)

119899minus119895

119886119895

(8)

where

119888119899119896(119895) =

119896

sum

119894=0

(minus1)119894

(119896 + 1 minus 119894)119899minus119895

(119896 minus 119894)119895

(119899 + 1

119894)

0 le 119895 le 119899

(9)

The following theorem gives combinatorial interpreta-tions to the coefficients 119888

119899119896(119895) 0 le 119895 le 119899

Theorem 8 Let the general Eulerian numbers 119860119899119896(119886 119889) be

written as in (8) Then

119888119899119896(119895) = 120587 isin 119882

119899119896+1 119895 lt 119876

119899(120587) le 119899

+ 120587 isin 119882119899119896

1 le 119876119899(120587) le 119895

(10)

Proof We can check the result in (10) for two special values119895 = 0 and 119895 = 119899 quickly By (2)

when 119895 = 0 119888119899119896(0) = sum

119896

119894=0(minus1)119894

(119896 + 1 minus 119894)119899

( 119899+1119894) =

119860119899119896+1

when 119895 = 119899 119888119899119896(119899) = sum

119896

119894=0(minus1)119894

(119896 minus 119894)119899

( 119899+1119894) = 119860

119899119896

Therefore (10) is true for 119895 = 0 and 119895 = 119899

Generally for 1 le 119895 le 119899 minus 1 we write down 119896 bars with119896 + 1 compartments in between Place each element of [119899] ina compartment If none of the 119896 bars is extraneous then thearrangement corresponds to a permutation with 119896 ascentsLet 119861 be the set of arrangements with at most one extraneousbar at the end and none of integers 1 2 119895 locating in thelast compartment We will show that 119888

119899119896(119895) = |119861|

To achieve that goal we use the Principle of Inclusion andExclusion There are (119896 + 1)119899minus119895119896119895 ways to put 119899 numbers into119896 + 1 compartments with elements 1 2 119895 avoiding thelast compartments

Let 119861119894be the number of arrangements with the following

features

(1) none of 1 2 119895 sits in the last compartment(2) each arrangement in 119861

119894has at least 119894 extraneous

bars(3) in each arrangement in 119861

119894 any two extraneous

bars are not located right next to each other

Journal of Discrete Mathematics 3

Then the Principle of Inclusion and Exclusion shows that

|119861| = (119896 + 1)119899minus119895

119896119895

minus 1198611+ 1198612+ sdot sdot sdot + (minus1)

119896

119861119896 (11)

Nowwe consider the value of119861119894 where 1 le 119894 le 119896 Suppose

that we have 119896+1minus119894 compartments with 119896minus119894 bars in betweenThere are (119896 + 1 minus 119894)119899minus119895(119896 minus 119894)119895 ways to insert 119899 numbers intothese 119896 + 1 minus 119894 compartments with first 119895 integers avoidingthe last compartment and list integers in each component ina decreasing order Then insert 119894 separating extraneous barsinto 119899 + 1 positions So we get

119861119894= (119896 + 1 minus 119894)

119899minus119895

(119896 minus 119894)119895

(119899 + 1

119894) (12)

Plug formula (12) into (11) we have 119888119899119896(119895) = |119861|

Given an arrangement 120587 isin 119861 if we remove the bars thenwe obtain a permutation 120587 isin Ω

119899 So without confusion we

just use the samenotation120587 to represent both an arrangementin set 119861 and a permutation on [119899] Now for each 120587 isin 119861 120587either

(case 1) has no extraneous bar and none of 1 2 119895locates in the last compartment or(case 2) has only one extraneous bar at the end

If 120587 is in case 1 then 120587 has 119896 ascents since each bar isnon-extraneous And the last compartment of 120587 is nonemptyTherefore the last cycle of 119891minus1(120587) has to be (119899 119901

119892) In other

words119876119899(119891minus1

(120587)) = 119901119892gt 119895 since none of 1 2 119895 locates

in the last compartment And by Proposition 4 119891minus1(120587) isin

119882119899119896+1

If 120587 is in case 2 then 120587 has 119896minus1 ascents since only the last

bar is extraneous Note that in this case the arrangement withno elements of 1 2 119895 in the compartment second to thelast or the last nonempty compartment has been removedby the Principle of Inclusion and Exclusion Equivalently atleast one number of 1 2 119895 has to be in the compartmentsecond to the last So the last cycle of 119891minus1(120587) has to be(119899 119901

119897) and 119876

119899(119891minus1

(120587)) = 119901119897le 119895 Also by Proposition 4

119891minus1

(120587) isin 119882119899119896

Combing all the results above statement (10) is correct

The next Theorem describes some interesting propertiesof the coefficients 119888

119899119896

Theorem9 Let the coefficients 119888119899119896

be as described inTheorem8 Then

(1) sum119899119896=0

119888119899119896(119895) = 119899 for any 0 le 119895 le 119899

(2) 119888119899119896(119895) = 119888

119899119899minus119896(119899 minus 119895) for all 0 le 119895 119896 le 119899

Before we can prove Theorem 9 we need the followinglemma which is also interesting by itself

Lemma 10 Given a positive integer 119899 then

120587 isin 119882119899119896

amp 119876119899(120587) = 119895

= 120587 isin 119882119899119899+1minus119896

amp 119876119899(120587) = 119899 + 1 minus 119895

(13)

for any 1 le 119896 119895 le 119899

Proof First of all given a positive integer 119899 we define afunction 119892 Ω

119899rarr Ω119899as follows

for 120587 = 11990111199012 119901119899isin Ω119899

119892 (120587) = (119899 + 1 minus 1199011) (119899 + 1 minus 119901

2) (119899 + 1 minus 119901

119899)

(14)

For instance for 120587 = 53214 isin Ω5 119892(120587) = 13452 119892 is

obviously a bijection ofΩ119899to itself

Now for some fixed 1 le 119896 119895 le 119899 suppose 119878119896119895

= 120587 isin

119882119899119896

amp 119876119899(120587) = 119895 and 119879

119896119895= 120587 isin 119882

119899119899+1minus119896amp 119876119899(120587) =

119899 + 1 minus 119895 For any 120587 isin 119878119896119895 we write 120587 in the standard

representation cycle form So 120587 = (119901119906 ) (119899 119895) and

119891(120587) = 119901119906 119899 119895 has 119896 minus 1 ascents by Proposition 4

Now we compose 119891(120587) with the bijection function 119892 as justdefined Then 119892(119891(120587)) = 119899 + 1 minus 119901

119906 1 119899 + 1 minus 119895 has

119899 minus 119896 ascents which implies that 119891minus1(119892(119891(120587))) has 119899 + 1 minus 119896weak excedances So 119891

minus1

(119892(119891(120587))) isin 119882119899119899+1minus119896

Note thatthe last cycle of 119891minus1(119892(119891(120587))) has to be (119899 119899 + 1 minus 119895)Therefore 119891minus1(119892(119891(120587))) isin 119879

119896119895 Since both 119891 and 119892 are

bijection functions 119891minus1119892119891 gives a bijection between 119878119896119895

and119879119896119895

Now we are ready to proveTheorem 9

Proof of Theorem 9 For part 1 by Theorem 8119899

sum

119896=0

119888119899119896(119895) =

119899

sum

119896=0

120587 isin 119882119899119896+1

119895 lt 119876119899(120587) le 119899

+

119899

sum

119896=0

120587 isin 119882119899119896

1 le 119876119899(120587) le 119895

=

119899

sum

119896=0

120587 isin 119882119899119896 =

1003816100381610038161003816Ω1198991003816100381610038161003816 = 119899

(15)

For part 2 also byTheorem 8

119888119899119896(119895) =

119899

sum

119894=119895+1

120587 isin 119882119899119896+1

119876119899(120587) = 119894

+

119895

sum

119898=1

120587 isin 119882119899119896 119876119899(120587) = 119898

=

119899

sum

119894=119895+1

120587 isin 119882119899119899minus119896

119876119899(120587) = 119899 + 1 minus 119894

+

119895

sum

119898=1

120587 isin 119882119899119899+1minus119896

119876119899(120587) = 119899 + 1 minus 119898 by Lemma 10

= 120587 isin 119882119899119896

1 le 119876119899(120587) le 119899 minus 119895

+ 120587 isin 119882119899119899+1minus119896

119899 minus 119895 lt 119876119899(120587) le 119899

= 119888119899119899minus119896

(119899 minus 119895)

(16)

4 Journal of Discrete Mathematics

Remark 11 Using the analytic formula of 119888119899119896(119895) as in (9) part

2 of Theorem 9 implies the following identity119896

sum

119894=0

(minus1)119894

(119896 + 1 minus 119894)119899minus119895

(119896 minus 119894)119895

(119899 + 1

119894)

=

119899minus119896

sum

119897=0

(minus1)119897

(119899 + 1 minus 119896 minus 119897)119895

(119899 minus 119896 minus 119897)119899minus119895

(119899 + 1

119897)

(17)

where 119899 is a positive integer and 0 le 119895 119896 le 119899

3 Another Combinatorial Interpretation of119888119899119896(1) and 119888

119899119896(119899minus1)

In pursuing the combinatorial meanings of the coefficients119888119899119896 the authors have found some other interesting properties

about permutations The results in this section will revealclose connections between the traditional Eulerian numbers119860119899119896

and 119888119899119896(119895) where 119895 = 1 or 119895 = 119899 minus 1

One fundamental concept of permutation combinatoricsis inversion A pair (119901

119894 119901119895) is called an inversion of the

permutation 120587 = 11990111199012 119901119899if 119894 lt 119895 and 119901

119894gt 119901119895[6 page

36] The following definition provides the main concepts ofthis section

Definition 12 For a fixed positive integer 119899 let 119860119882119899119896

= 120587 =

119901111990121199013 119901119899| 120587 isin 119882

119899119896and 119901

1lt 119901119899 (or (119901

1 119901119899) is not

an inversion) and 119861119882119899119896

= 119882119899119896

119860119882119899119896

(or (1199011 119901119899) is an

inversion)It is obvious that |119860119882

119899119896| + |119861119882

119899119896| = 119860119899119896 The following

theorem interprets coefficients 119888119899119896(1) and 119888

119899119896(119899 minus 1) in terms

of 119860119882119899119896

and 119861119882119899119896

Theorem 13 Let the coefficients 119888119899119896

of the general Euleriannumbers be written as in (9) 119860119882

119899119896and 119861119882

119899119896are as defined

in Definition 12 Then(1) 119888119899119896(1) = 2|119860119882

119899119896+1|

(2) 119888119899119896(119899 minus 1) = 2|119861119882

119899119896|

Proof For part (1) by Theorem 8 119888119899119896(1) = |119878

1| + |1198782| where

1198781= 120587 = 119901

11199012 119901119899| 120587 isin 119882

119899119896+1amp 1199011

= 119899 1198782= 120587 =

11990111199012 119901119899| 120587 isin 119882

119899119896amp 1199011= 119899 Given a permutation

120587 = 11990111199012 119901119899isin 1198781and 119901

119899= 119899 then both 119901

11199012 119901119899and

1199011198991199012 1199011belong to 119878

1 so one of them has to be in 119860119882

119899119896+1

If 120587 = 11990111199012 119901119899isin 1198781and 119901

119899= 119899 then 120587 isin 119860119882

119899119896+1 but

1199011198991199012 1199011isin 1198782 Therefore (12)119888

119899119896(1) = |119860119882

119899119896+1|

Part (2) can be proved using exactly the same method Sowe leave it to the readers as an exercise

|119860119882119899119896| and |119861119882

119899119896| are interesting combinatorial

concepts by themselves Note that generally speaking|119860119882119899119896| = |119861119882

119899119896| Indeed |119860119882

119899119896| = |119861119882

119899119899+1minus119896|

Theorem 14 For any positive integer 119899 ge 2 the sets119860119882119899119896

and119861119882119899119896

are defined in Definition 12Then |119860119882119899119896| = |119861119882

119899119899+1minus119896|

for 1 le 119896 le 119899

Proof It is an obvious result of part 2 of Theorems 9 and 13

Our last result of this paper is the following theoremwhich reveals that both |119860119882

119899119896| and |119861119882

119899119896| take exactly the

same recursive formula as the traditional Eulerian numbers119860119899119896

as shown in (1)

Theorem 15 For a fixed positive integer 119899 let 119860119882119899119896

and119861119882119899119896

be as defined in Definition 12 then

1198961003816100381610038161003816119860119882119899minus1119896

1003816100381610038161003816 + (119899 + 1 minus 119896)1003816100381610038161003816119860119882119899minus1119896minus1

1003816100381610038161003816 =1003816100381610038161003816119860119882119899119896

1003816100381610038161003816 (18)

1198961003816100381610038161003816119861119882119899minus1119896

1003816100381610038161003816 + (119899 + 1 minus 119896)1003816100381610038161003816119861119882119899minus1119896minus1

1003816100381610038161003816 =1003816100381610038161003816119861119882119899119896

1003816100381610038161003816 (19)

Proof A computational proof can be obtained straightfor-ward by using (9) and Theorem 13 But here we provide aproof in a flavor of combinatorics

Idea of the Proof For (18) given a permutation 1198601=

119901111990121199013 119901119899minus1

isin 119860119882119899minus1119896

for each position 119894 with 119901119894ge 119894

we insert 119899 into a certain place of 1198601 such that the new

permutation 1198601015840

1is in 119860119882

119899119896 There are 119896 such positions so

we can get 119896 new permutations in 119860119882119899119896 Similarly if 119860

2=

119901111990121199013 119901119899minus1

isin 119860119882119899minus1119896minus1

for each position 119894 with 119901119894lt 119894

and the position at the end of 1198602 we insert 119899 into a specific

position of 1198602and the resulting new permutation 119860

1015840

2is in

119860119882119899119896 There are 119899 + 1 minus 119896 such positions so we can get

119899 + 1 minus 119896 new permutations in 119860119882119899119896 We will show that

all the permutations obtained from the above constructionsare distinct and they have exhausted all the permutations in119860119882119899119896

For any fixed 1198601015840 = 120587112058721205873 120587119899isin 119860119882

119899119896 then 120587

1lt 120587119899

We classify 1198601015840 into the following disjoint cases

Case a Consider that 120587119894

= 119899 with 119894 lt 119899 So 1198601015840

=

12058711205872 120587119894minus1119899120587119894+1

120587119899minus1

120587119899

(a1) 1205871lt 120587119899minus1

and 120587119899ge 119894

(a2) 1205871lt 120587119899minus1

and 120587119899lt 119894

(a3) 1205871gt 120587119899minus1

120587119899lt 119899 minus 1 and 120587

119899ge 119894

(a4) 1205871gt 120587119899minus1

120587119899lt 119899 minus 1 and 120587

119899lt 119894

(a5) 1205871gt 120587119899minus1

and 120587119899= 119899 minus 1

Case b Consider that 120587119899= 119899 So 120587

119894= 119899minus1 for some 119894 lt 119899 and

1198601015840

= 12058711205872 120587119894minus1119899 minus 1 120587

119899minus1119899

(b1) 1205871lt 120587119899minus1

(b2) 120587

119899minus1lt 1205871lt 119899 minus 1 and 120587

119899minus1ge 119894

(b3) 120587119899minus1

lt 1205871lt 119899 minus 1 and 120587

119899minus1lt 119894

(b4) 1205871= 119899 minus 1

Based on the classifications listed above we can construct amap 119891 119860119882

119899minus1119896 119860119882119899minus1119896minus1

rarr 119860119882119899119896

by applying the ideaof the proof we have illustrated at the beginning of the proofTo save space the map 119891 is demonstrated in Table 1 FromTable 1 we can see that in each case the positions of inserting119899 are all different So all the images obtained in a certain caseare different Since all the cases are disjoint all the images1198601015840

isin 119860119882119899119896

are distinct

Journal of Discrete Mathematics 5

Table 1 The map 119891 119860119882119899minus1119896

119860119882119899minus1119896minus1

rarr 119860119882119899119896

119860 = 11990111199012 119901119899minus1

Position 119894 Condition 1198601015840

isin 119860119882119899119896

119860 isin 119860119882119899minus1119896

1 lt 119894 le 119899 minus 1

and119901119894ge 119894

119901119894gt 1199011

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011lt 119901119899minus1

1199011lt 119901119894

Case (a1)

119901119894lt 1199011and

119901119899minus1

lt 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus2119901119894119901119899minus1

With 119901119894lt 1199011lt 119901119899minus1

lt 119899 minus 1

Case (a3)

119901119894lt 1199011and

119901119899minus1

= 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899 minus 1119901

119894+1 119901119899minus2119901119894119899

With 119901119894lt 1199011119901119894ge 119894

Case (b2)

119894 = 1

119901119895= 119899 minus 1

and 119895 lt 119899 minus 1

1198601015840

= 119901119899minus11199012 119901119895minus1119899119901119895+1

119901119899minus2

1199011119899 minus 1

With 1199011lt 119901119899minus1

Case (a5)

119901119899minus1

= 119899 minus 11198601015840

= 119899 minus 11199012 119901119899minus21199011119899

Case (b4)

119860 isin 119860119882119899minus1119896minus1

1 lt 119894 le 119899 minus 1

and119901119894lt 119894

119901119894gt 1199011

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011lt 119901119899minus1

1199011lt 119901119894

Case (a2)

119901119894lt 1199011and

119901119899minus1

lt 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus2119901119894119901119899minus1

With 119901119894lt 1199011lt 119901119899minus1

lt 119899 minus 1

Case (a4)

119901119894lt 1199011and

119901119899minus1

= 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899 minus 1119901

119894+1 119901119899minus2119901119894119899

With 119901119894lt 1199011119901119894lt 119894

Case (b3)

119894 = 1198991198601015840

= 11990111199012 119901119899minus1119899

Case (b1)

Table 2 The map 119892 119861119882119899minus1119896

119861119882119899minus1119896minus1

rarr 119861119882119899119896

119861 = 11990111199012 119901119899minus1

Position 119894 Condition 1198611015840

isin 119861119882119899119896

119861 isin 119882119861119899minus1119896

1 lt 119894 lt 119899 minus 1

and119901119894ge 119894

1199011gt 119901119894

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011gt 119901119899minus1

and 119901119894ge 119894

Case (c1)

1199011lt 119901119894lt 119899 minus 1

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119894119901119899minus1

With 1199011lt 119901119894lt 119899 minus 1 119901

119894ge 119894

Case (c3)

1199011lt 119901119894= 119899 minus 1

1198611015840

= 1198991199012 119901119894minus1119899 minus 1119901

119894+1 1199011119901119899minus1

With 119901119894= 119899 minus 1 119901

1gt 119901119899minus1

Case (d2)

119894 = 1 1199011ge 1

1198611015840

= 1198991199012 119901119899minus11199011

With 119901119899minus1

lt 1199011

Case (d1)

119861 isin 119882119861119899minus1119896minus1

1 lt 119894 lt 119899 minus 1

and119901119894lt 119894

1199011gt 119901119894

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011gt 119901119899minus1

and 119901119894lt 119894

Case (c2)

1199011lt 119901119894

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119894119901119899minus1

With 1199011lt 119901119894lt 119899 minus 1 119901

119894lt 119894

Case (c4)119894 = 119899 minus 1

119901119894lt 119894

1199011gt 119901119894= 119901119899minus1

1198611015840

= 1199011 119901119899minus2119899119901119899minus1

Case (c6)1 le 119894 lt 119899 minus 1

and119901119894= 119899 minus 1

119901119894= 119899 minus 1

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119899minus2119899 minus 1119901

119899minus1

Case (c5)

6 Journal of Discrete Mathematics

Similarly for each 1198611015840 = 120587112058721205873 120587119899isin 119861119882

119899119896 then 120587

1gt

120587119899 We classify 1198611015840 into the following disjoint cases

Case c Consider that 120587119894= 119899 with 1 lt 119894 le 119899 minus 1 So 1198611015840 =

12058711205872 120587119894minus1119899120587119894+1

120587119899minus1

120587119899

(c1) 1205871gt 120587119899minus1

and 120587119899ge 119894

(c2) 1205871gt 120587119899minus1

and 120587119899lt 119894

(c3) 1205871lt 120587119899minus1

lt 119899 minus 1 120587119899minus1

ge 119894(c4) 120587

1lt 120587119899minus1

lt 119899 minus 1 120587119899minus1

lt 119894(c5) 120587

119899minus1= 119899 minus 1

(c6) 120587119899minus1

= 119899

Case d Consider that 1205871= 119899 So 1198611015840 = 119899120587

2 120587119899minus2

120587119899minus1

(d1) 120587119899minus2

lt 120587119899minus1

(d2) 120587

119899minus2gt 120587119899minus1

To prove (19) we use a similar idea of proof as shown aboveIf 1198611= 119901111990121199013 119901119899minus1

isin 119861119882119899minus1119896

for each position 119894 with119901119894ge 119894 we insert 119899 into a certain place of 119861

1to get 1198611015840

1isin 119860119882

119899119896

If 1198612= 119901111990121199013 119901119899minus1

isin 119861119882119899minus1119896minus1

for each position 119894 with119901119894lt 119894 and the position 119894 where 119901

119894= 119899 minus 1 we insert 119899 into

a specific position of 1198612to obtain 119861

1015840

2isin 119860119882

119899119896 Such a map

119892 119861119882119899minus1119896

119861119882119899minus1119896minus1

rarr 119861119882119899119896

is illustrated in Table 2And the distinct images under 119892 exhaust all the permutationsin 119861119882

119899119896

Here is a concrete example for the constructions illus-trated in Table 2

Example 16 Suppose 119899 = 4 119896 = 2 We want to obtain11986111988242

= 3142 3412 3421 4132 4213 4312 4321 from11986111988232

= 321 231 and 11986111988231

= 312 For 321 isin 11986111988232 1199011=

3 ge 1 then it corresponds to 1198611015840 = 4213 which is case (d1) inTable 2 119901

2= 2 ge 2 then it corresponds to 1198611015840 = 3412which is

case (c1) in Table 2 Similarly we can construct 4312 4321from 231 isin 119861119882

32and 3421 3142 4132 from 312 isin 119861119882

31

using Table 2

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Riordan An Introduction to Combinatorial Analysis WileyPublications in Mathematical Statistics John Wiley amp SonsNew York NY USA 1958

[2] M Bona Combinatorics of Permutations Discrete Mathematicsand its Applications Chapman amp HallCRC Boca Raton FlaUSA 2004

[3] L Carlitz ldquo119902-Bernoulli and Eulerian numbersrdquo Transactions ofthe American Mathematical Society vol 76 pp 332ndash350 1954

[4] L Carlitz ldquoA combinatorial property of 119902-Eulerian numbersrdquoThe American Mathematical Monthly vol 82 pp 51ndash54 1975

[5] T Xiong H P Tsao and J I Hall ldquoGeneral Eulerian numbersand Eulerian polynomialsrdquo Journal of Mathematics vol 2013Article ID 629132 9 pages 2013

[6] R P Stanley Enumerative Combinatorics vol 1 of CambridgeStudies in Advanced Mathematics Cambridge University PressCambridge UK 1996

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Operations ResearchAdvances in

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Stochastic AnalysisInternational Journal of

2 Journal of Discrete Mathematics

Definition 3 Let 119882119899119896

be the set of 119899-permutations with 119896

weak exceedances Then |119882119899119896| = 119860

119899119896 Furthermore given

a permutation 120587 = 119901111990121199013 119901119899 let 119876

119899(120587) = 119894 where 119901

119894= 119899

Given a permutation 120587 isin Ω119899 it is known that 120587 can be

written as a one-line form like 120587 = 119901111990121199013 119901119899 or 120587 can be

written in a disjoint union of distinct cycles For 120587 written ina cycle form we can use a standard representation by writing(a) each cycle starting with its largest element and (b) thecycles in increasing order of their largest element Moreovergiven a permutation 120587 written in a standard representationcycle form define a function 119891 as 119891(120587) to be the permutationobtained from 120587 by erasing the parenthesesThen119891 is knownas the fundamental bijection from Ω

119899to itself [6 page 30]

Indeed the inverse map 119891minus1 of the fundamental bijection

function 119891 is also famous in illustrating the relation betweenthe ascents and weak exceedances as follows [2 page 98]

Proposition 4 The function 119891minus1 gives a bijection between theset of permutations on [119899] with 119896 ascents and the set119882

119899119896+1

Example 5 The standard representation of permutation 120587 =

5243716 is (2)(43)(7615) isin Ω7 and119891(120587) = 2437615119876

7(120587) =

5 120587 = 5243716 has 3 ascents while 119891minus1(120587) = (5243)(716) =

6453271 isin 11988274

has 3 + 1 = 4 weak excedances because 1199011=

6 gt 1 1199012= 4 gt 2 119901

3= 5 gt 3 and 119901

6= 7 gt 6

Now suppose we want to construct a sequence consistingof 119896 vertical bars and the first 119899 positive integers Then the 119896vertical bars divide these 119899 numbers into 119896+1 compartmentsIn each compartment there is either no number or all thenumbers are listed in a decreasing order The followingdefinition is analogous to the definition of [2 page 8]

Definition 6 A bar in the above construction is calledextraneous if either

(a) it is immediately followed by another bar or

(b) each of the rest compartment is either empty orconsists of integers in a decreasing order if this baris removed

Example 7 Suppose 119899 = 7 119896 = 4 then in the followingarrangement

32 |1| |7654| (7)

the 1st 2nd and 4th bars are extraneous

Now we are ready to give combinatorial interpretationsto the general Eulerian numbers 119860

119899119896(119886 119889) First note that

(6) implies that 119860119899119896(119886 119889) is a homogeneous polynomial of

degree 119899 with respect to 119886 and 119889 Indeed

119860119899119896(119886 119889)

=

119896

sum

119894=0

(minus1)119894

[(119896 + 1 minus 119894) 119889 minus 119886]119899

(119899 + 1

119894)

=

119896

sum

119894=0

(minus1)119894

[(119896 + 1 minus 119894) (119889 minus 119886) + (119896 minus 119894) 119886]119899

(119899 + 1

119894)

=

119899

sum

119895=0

[

119896

sum

119894=0

(minus1)119894

(119896 + 1 minus 119894)119899minus119895

(119896 minus 119894)119895

(119899 + 1

119894)]

times (119899

119895) (119889 minus 119886)

119899minus119895

119886119895

=

119899

sum

119895=0

119888119899119896(119895) (

119899

119895) (119889 minus 119886)

119899minus119895

119886119895

(8)

where

119888119899119896(119895) =

119896

sum

119894=0

(minus1)119894

(119896 + 1 minus 119894)119899minus119895

(119896 minus 119894)119895

(119899 + 1

119894)

0 le 119895 le 119899

(9)

The following theorem gives combinatorial interpreta-tions to the coefficients 119888

119899119896(119895) 0 le 119895 le 119899

Theorem 8 Let the general Eulerian numbers 119860119899119896(119886 119889) be

written as in (8) Then

119888119899119896(119895) = 120587 isin 119882

119899119896+1 119895 lt 119876

119899(120587) le 119899

+ 120587 isin 119882119899119896

1 le 119876119899(120587) le 119895

(10)

Proof We can check the result in (10) for two special values119895 = 0 and 119895 = 119899 quickly By (2)

when 119895 = 0 119888119899119896(0) = sum

119896

119894=0(minus1)119894

(119896 + 1 minus 119894)119899

( 119899+1119894) =

119860119899119896+1

when 119895 = 119899 119888119899119896(119899) = sum

119896

119894=0(minus1)119894

(119896 minus 119894)119899

( 119899+1119894) = 119860

119899119896

Therefore (10) is true for 119895 = 0 and 119895 = 119899

Generally for 1 le 119895 le 119899 minus 1 we write down 119896 bars with119896 + 1 compartments in between Place each element of [119899] ina compartment If none of the 119896 bars is extraneous then thearrangement corresponds to a permutation with 119896 ascentsLet 119861 be the set of arrangements with at most one extraneousbar at the end and none of integers 1 2 119895 locating in thelast compartment We will show that 119888

119899119896(119895) = |119861|

To achieve that goal we use the Principle of Inclusion andExclusion There are (119896 + 1)119899minus119895119896119895 ways to put 119899 numbers into119896 + 1 compartments with elements 1 2 119895 avoiding thelast compartments

Let 119861119894be the number of arrangements with the following

features

(1) none of 1 2 119895 sits in the last compartment(2) each arrangement in 119861

119894has at least 119894 extraneous

bars(3) in each arrangement in 119861

119894 any two extraneous

bars are not located right next to each other

Journal of Discrete Mathematics 3

Then the Principle of Inclusion and Exclusion shows that

|119861| = (119896 + 1)119899minus119895

119896119895

minus 1198611+ 1198612+ sdot sdot sdot + (minus1)

119896

119861119896 (11)

Nowwe consider the value of119861119894 where 1 le 119894 le 119896 Suppose

that we have 119896+1minus119894 compartments with 119896minus119894 bars in betweenThere are (119896 + 1 minus 119894)119899minus119895(119896 minus 119894)119895 ways to insert 119899 numbers intothese 119896 + 1 minus 119894 compartments with first 119895 integers avoidingthe last compartment and list integers in each component ina decreasing order Then insert 119894 separating extraneous barsinto 119899 + 1 positions So we get

119861119894= (119896 + 1 minus 119894)

119899minus119895

(119896 minus 119894)119895

(119899 + 1

119894) (12)

Plug formula (12) into (11) we have 119888119899119896(119895) = |119861|

Given an arrangement 120587 isin 119861 if we remove the bars thenwe obtain a permutation 120587 isin Ω

119899 So without confusion we

just use the samenotation120587 to represent both an arrangementin set 119861 and a permutation on [119899] Now for each 120587 isin 119861 120587either

(case 1) has no extraneous bar and none of 1 2 119895locates in the last compartment or(case 2) has only one extraneous bar at the end

If 120587 is in case 1 then 120587 has 119896 ascents since each bar isnon-extraneous And the last compartment of 120587 is nonemptyTherefore the last cycle of 119891minus1(120587) has to be (119899 119901

119892) In other

words119876119899(119891minus1

(120587)) = 119901119892gt 119895 since none of 1 2 119895 locates

in the last compartment And by Proposition 4 119891minus1(120587) isin

119882119899119896+1

If 120587 is in case 2 then 120587 has 119896minus1 ascents since only the last

bar is extraneous Note that in this case the arrangement withno elements of 1 2 119895 in the compartment second to thelast or the last nonempty compartment has been removedby the Principle of Inclusion and Exclusion Equivalently atleast one number of 1 2 119895 has to be in the compartmentsecond to the last So the last cycle of 119891minus1(120587) has to be(119899 119901

119897) and 119876

119899(119891minus1

(120587)) = 119901119897le 119895 Also by Proposition 4

119891minus1

(120587) isin 119882119899119896

Combing all the results above statement (10) is correct

The next Theorem describes some interesting propertiesof the coefficients 119888

119899119896

Theorem9 Let the coefficients 119888119899119896

be as described inTheorem8 Then

(1) sum119899119896=0

119888119899119896(119895) = 119899 for any 0 le 119895 le 119899

(2) 119888119899119896(119895) = 119888

119899119899minus119896(119899 minus 119895) for all 0 le 119895 119896 le 119899

Before we can prove Theorem 9 we need the followinglemma which is also interesting by itself

Lemma 10 Given a positive integer 119899 then

120587 isin 119882119899119896

amp 119876119899(120587) = 119895

= 120587 isin 119882119899119899+1minus119896

amp 119876119899(120587) = 119899 + 1 minus 119895

(13)

for any 1 le 119896 119895 le 119899

Proof First of all given a positive integer 119899 we define afunction 119892 Ω

119899rarr Ω119899as follows

for 120587 = 11990111199012 119901119899isin Ω119899

119892 (120587) = (119899 + 1 minus 1199011) (119899 + 1 minus 119901

2) (119899 + 1 minus 119901

119899)

(14)

For instance for 120587 = 53214 isin Ω5 119892(120587) = 13452 119892 is

obviously a bijection ofΩ119899to itself

Now for some fixed 1 le 119896 119895 le 119899 suppose 119878119896119895

= 120587 isin

119882119899119896

amp 119876119899(120587) = 119895 and 119879

119896119895= 120587 isin 119882

119899119899+1minus119896amp 119876119899(120587) =

119899 + 1 minus 119895 For any 120587 isin 119878119896119895 we write 120587 in the standard

representation cycle form So 120587 = (119901119906 ) (119899 119895) and

119891(120587) = 119901119906 119899 119895 has 119896 minus 1 ascents by Proposition 4

Now we compose 119891(120587) with the bijection function 119892 as justdefined Then 119892(119891(120587)) = 119899 + 1 minus 119901

119906 1 119899 + 1 minus 119895 has

119899 minus 119896 ascents which implies that 119891minus1(119892(119891(120587))) has 119899 + 1 minus 119896weak excedances So 119891

minus1

(119892(119891(120587))) isin 119882119899119899+1minus119896

Note thatthe last cycle of 119891minus1(119892(119891(120587))) has to be (119899 119899 + 1 minus 119895)Therefore 119891minus1(119892(119891(120587))) isin 119879

119896119895 Since both 119891 and 119892 are

bijection functions 119891minus1119892119891 gives a bijection between 119878119896119895

and119879119896119895

Now we are ready to proveTheorem 9

Proof of Theorem 9 For part 1 by Theorem 8119899

sum

119896=0

119888119899119896(119895) =

119899

sum

119896=0

120587 isin 119882119899119896+1

119895 lt 119876119899(120587) le 119899

+

119899

sum

119896=0

120587 isin 119882119899119896

1 le 119876119899(120587) le 119895

=

119899

sum

119896=0

120587 isin 119882119899119896 =

1003816100381610038161003816Ω1198991003816100381610038161003816 = 119899

(15)

For part 2 also byTheorem 8

119888119899119896(119895) =

119899

sum

119894=119895+1

120587 isin 119882119899119896+1

119876119899(120587) = 119894

+

119895

sum

119898=1

120587 isin 119882119899119896 119876119899(120587) = 119898

=

119899

sum

119894=119895+1

120587 isin 119882119899119899minus119896

119876119899(120587) = 119899 + 1 minus 119894

+

119895

sum

119898=1

120587 isin 119882119899119899+1minus119896

119876119899(120587) = 119899 + 1 minus 119898 by Lemma 10

= 120587 isin 119882119899119896

1 le 119876119899(120587) le 119899 minus 119895

+ 120587 isin 119882119899119899+1minus119896

119899 minus 119895 lt 119876119899(120587) le 119899

= 119888119899119899minus119896

(119899 minus 119895)

(16)

4 Journal of Discrete Mathematics

Remark 11 Using the analytic formula of 119888119899119896(119895) as in (9) part

2 of Theorem 9 implies the following identity119896

sum

119894=0

(minus1)119894

(119896 + 1 minus 119894)119899minus119895

(119896 minus 119894)119895

(119899 + 1

119894)

=

119899minus119896

sum

119897=0

(minus1)119897

(119899 + 1 minus 119896 minus 119897)119895

(119899 minus 119896 minus 119897)119899minus119895

(119899 + 1

119897)

(17)

where 119899 is a positive integer and 0 le 119895 119896 le 119899

3 Another Combinatorial Interpretation of119888119899119896(1) and 119888

119899119896(119899minus1)

In pursuing the combinatorial meanings of the coefficients119888119899119896 the authors have found some other interesting properties

about permutations The results in this section will revealclose connections between the traditional Eulerian numbers119860119899119896

and 119888119899119896(119895) where 119895 = 1 or 119895 = 119899 minus 1

One fundamental concept of permutation combinatoricsis inversion A pair (119901

119894 119901119895) is called an inversion of the

permutation 120587 = 11990111199012 119901119899if 119894 lt 119895 and 119901

119894gt 119901119895[6 page

36] The following definition provides the main concepts ofthis section

Definition 12 For a fixed positive integer 119899 let 119860119882119899119896

= 120587 =

119901111990121199013 119901119899| 120587 isin 119882

119899119896and 119901

1lt 119901119899 (or (119901

1 119901119899) is not

an inversion) and 119861119882119899119896

= 119882119899119896

119860119882119899119896

(or (1199011 119901119899) is an

inversion)It is obvious that |119860119882

119899119896| + |119861119882

119899119896| = 119860119899119896 The following

theorem interprets coefficients 119888119899119896(1) and 119888

119899119896(119899 minus 1) in terms

of 119860119882119899119896

and 119861119882119899119896

Theorem 13 Let the coefficients 119888119899119896

of the general Euleriannumbers be written as in (9) 119860119882

119899119896and 119861119882

119899119896are as defined

in Definition 12 Then(1) 119888119899119896(1) = 2|119860119882

119899119896+1|

(2) 119888119899119896(119899 minus 1) = 2|119861119882

119899119896|

Proof For part (1) by Theorem 8 119888119899119896(1) = |119878

1| + |1198782| where

1198781= 120587 = 119901

11199012 119901119899| 120587 isin 119882

119899119896+1amp 1199011

= 119899 1198782= 120587 =

11990111199012 119901119899| 120587 isin 119882

119899119896amp 1199011= 119899 Given a permutation

120587 = 11990111199012 119901119899isin 1198781and 119901

119899= 119899 then both 119901

11199012 119901119899and

1199011198991199012 1199011belong to 119878

1 so one of them has to be in 119860119882

119899119896+1

If 120587 = 11990111199012 119901119899isin 1198781and 119901

119899= 119899 then 120587 isin 119860119882

119899119896+1 but

1199011198991199012 1199011isin 1198782 Therefore (12)119888

119899119896(1) = |119860119882

119899119896+1|

Part (2) can be proved using exactly the same method Sowe leave it to the readers as an exercise

|119860119882119899119896| and |119861119882

119899119896| are interesting combinatorial

concepts by themselves Note that generally speaking|119860119882119899119896| = |119861119882

119899119896| Indeed |119860119882

119899119896| = |119861119882

119899119899+1minus119896|

Theorem 14 For any positive integer 119899 ge 2 the sets119860119882119899119896

and119861119882119899119896

are defined in Definition 12Then |119860119882119899119896| = |119861119882

119899119899+1minus119896|

for 1 le 119896 le 119899

Proof It is an obvious result of part 2 of Theorems 9 and 13

Our last result of this paper is the following theoremwhich reveals that both |119860119882

119899119896| and |119861119882

119899119896| take exactly the

same recursive formula as the traditional Eulerian numbers119860119899119896

as shown in (1)

Theorem 15 For a fixed positive integer 119899 let 119860119882119899119896

and119861119882119899119896

be as defined in Definition 12 then

1198961003816100381610038161003816119860119882119899minus1119896

1003816100381610038161003816 + (119899 + 1 minus 119896)1003816100381610038161003816119860119882119899minus1119896minus1

1003816100381610038161003816 =1003816100381610038161003816119860119882119899119896

1003816100381610038161003816 (18)

1198961003816100381610038161003816119861119882119899minus1119896

1003816100381610038161003816 + (119899 + 1 minus 119896)1003816100381610038161003816119861119882119899minus1119896minus1

1003816100381610038161003816 =1003816100381610038161003816119861119882119899119896

1003816100381610038161003816 (19)

Proof A computational proof can be obtained straightfor-ward by using (9) and Theorem 13 But here we provide aproof in a flavor of combinatorics

Idea of the Proof For (18) given a permutation 1198601=

119901111990121199013 119901119899minus1

isin 119860119882119899minus1119896

for each position 119894 with 119901119894ge 119894

we insert 119899 into a certain place of 1198601 such that the new

permutation 1198601015840

1is in 119860119882

119899119896 There are 119896 such positions so

we can get 119896 new permutations in 119860119882119899119896 Similarly if 119860

2=

119901111990121199013 119901119899minus1

isin 119860119882119899minus1119896minus1

for each position 119894 with 119901119894lt 119894

and the position at the end of 1198602 we insert 119899 into a specific

position of 1198602and the resulting new permutation 119860

1015840

2is in

119860119882119899119896 There are 119899 + 1 minus 119896 such positions so we can get

119899 + 1 minus 119896 new permutations in 119860119882119899119896 We will show that

all the permutations obtained from the above constructionsare distinct and they have exhausted all the permutations in119860119882119899119896

For any fixed 1198601015840 = 120587112058721205873 120587119899isin 119860119882

119899119896 then 120587

1lt 120587119899

We classify 1198601015840 into the following disjoint cases

Case a Consider that 120587119894

= 119899 with 119894 lt 119899 So 1198601015840

=

12058711205872 120587119894minus1119899120587119894+1

120587119899minus1

120587119899

(a1) 1205871lt 120587119899minus1

and 120587119899ge 119894

(a2) 1205871lt 120587119899minus1

and 120587119899lt 119894

(a3) 1205871gt 120587119899minus1

120587119899lt 119899 minus 1 and 120587

119899ge 119894

(a4) 1205871gt 120587119899minus1

120587119899lt 119899 minus 1 and 120587

119899lt 119894

(a5) 1205871gt 120587119899minus1

and 120587119899= 119899 minus 1

Case b Consider that 120587119899= 119899 So 120587

119894= 119899minus1 for some 119894 lt 119899 and

1198601015840

= 12058711205872 120587119894minus1119899 minus 1 120587

119899minus1119899

(b1) 1205871lt 120587119899minus1

(b2) 120587

119899minus1lt 1205871lt 119899 minus 1 and 120587

119899minus1ge 119894

(b3) 120587119899minus1

lt 1205871lt 119899 minus 1 and 120587

119899minus1lt 119894

(b4) 1205871= 119899 minus 1

Based on the classifications listed above we can construct amap 119891 119860119882

119899minus1119896 119860119882119899minus1119896minus1

rarr 119860119882119899119896

by applying the ideaof the proof we have illustrated at the beginning of the proofTo save space the map 119891 is demonstrated in Table 1 FromTable 1 we can see that in each case the positions of inserting119899 are all different So all the images obtained in a certain caseare different Since all the cases are disjoint all the images1198601015840

isin 119860119882119899119896

are distinct

Journal of Discrete Mathematics 5

Table 1 The map 119891 119860119882119899minus1119896

119860119882119899minus1119896minus1

rarr 119860119882119899119896

119860 = 11990111199012 119901119899minus1

Position 119894 Condition 1198601015840

isin 119860119882119899119896

119860 isin 119860119882119899minus1119896

1 lt 119894 le 119899 minus 1

and119901119894ge 119894

119901119894gt 1199011

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011lt 119901119899minus1

1199011lt 119901119894

Case (a1)

119901119894lt 1199011and

119901119899minus1

lt 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus2119901119894119901119899minus1

With 119901119894lt 1199011lt 119901119899minus1

lt 119899 minus 1

Case (a3)

119901119894lt 1199011and

119901119899minus1

= 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899 minus 1119901

119894+1 119901119899minus2119901119894119899

With 119901119894lt 1199011119901119894ge 119894

Case (b2)

119894 = 1

119901119895= 119899 minus 1

and 119895 lt 119899 minus 1

1198601015840

= 119901119899minus11199012 119901119895minus1119899119901119895+1

119901119899minus2

1199011119899 minus 1

With 1199011lt 119901119899minus1

Case (a5)

119901119899minus1

= 119899 minus 11198601015840

= 119899 minus 11199012 119901119899minus21199011119899

Case (b4)

119860 isin 119860119882119899minus1119896minus1

1 lt 119894 le 119899 minus 1

and119901119894lt 119894

119901119894gt 1199011

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011lt 119901119899minus1

1199011lt 119901119894

Case (a2)

119901119894lt 1199011and

119901119899minus1

lt 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus2119901119894119901119899minus1

With 119901119894lt 1199011lt 119901119899minus1

lt 119899 minus 1

Case (a4)

119901119894lt 1199011and

119901119899minus1

= 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899 minus 1119901

119894+1 119901119899minus2119901119894119899

With 119901119894lt 1199011119901119894lt 119894

Case (b3)

119894 = 1198991198601015840

= 11990111199012 119901119899minus1119899

Case (b1)

Table 2 The map 119892 119861119882119899minus1119896

119861119882119899minus1119896minus1

rarr 119861119882119899119896

119861 = 11990111199012 119901119899minus1

Position 119894 Condition 1198611015840

isin 119861119882119899119896

119861 isin 119882119861119899minus1119896

1 lt 119894 lt 119899 minus 1

and119901119894ge 119894

1199011gt 119901119894

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011gt 119901119899minus1

and 119901119894ge 119894

Case (c1)

1199011lt 119901119894lt 119899 minus 1

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119894119901119899minus1

With 1199011lt 119901119894lt 119899 minus 1 119901

119894ge 119894

Case (c3)

1199011lt 119901119894= 119899 minus 1

1198611015840

= 1198991199012 119901119894minus1119899 minus 1119901

119894+1 1199011119901119899minus1

With 119901119894= 119899 minus 1 119901

1gt 119901119899minus1

Case (d2)

119894 = 1 1199011ge 1

1198611015840

= 1198991199012 119901119899minus11199011

With 119901119899minus1

lt 1199011

Case (d1)

119861 isin 119882119861119899minus1119896minus1

1 lt 119894 lt 119899 minus 1

and119901119894lt 119894

1199011gt 119901119894

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011gt 119901119899minus1

and 119901119894lt 119894

Case (c2)

1199011lt 119901119894

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119894119901119899minus1

With 1199011lt 119901119894lt 119899 minus 1 119901

119894lt 119894

Case (c4)119894 = 119899 minus 1

119901119894lt 119894

1199011gt 119901119894= 119901119899minus1

1198611015840

= 1199011 119901119899minus2119899119901119899minus1

Case (c6)1 le 119894 lt 119899 minus 1

and119901119894= 119899 minus 1

119901119894= 119899 minus 1

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119899minus2119899 minus 1119901

119899minus1

Case (c5)

6 Journal of Discrete Mathematics

Similarly for each 1198611015840 = 120587112058721205873 120587119899isin 119861119882

119899119896 then 120587

1gt

120587119899 We classify 1198611015840 into the following disjoint cases

Case c Consider that 120587119894= 119899 with 1 lt 119894 le 119899 minus 1 So 1198611015840 =

12058711205872 120587119894minus1119899120587119894+1

120587119899minus1

120587119899

(c1) 1205871gt 120587119899minus1

and 120587119899ge 119894

(c2) 1205871gt 120587119899minus1

and 120587119899lt 119894

(c3) 1205871lt 120587119899minus1

lt 119899 minus 1 120587119899minus1

ge 119894(c4) 120587

1lt 120587119899minus1

lt 119899 minus 1 120587119899minus1

lt 119894(c5) 120587

119899minus1= 119899 minus 1

(c6) 120587119899minus1

= 119899

Case d Consider that 1205871= 119899 So 1198611015840 = 119899120587

2 120587119899minus2

120587119899minus1

(d1) 120587119899minus2

lt 120587119899minus1

(d2) 120587

119899minus2gt 120587119899minus1

To prove (19) we use a similar idea of proof as shown aboveIf 1198611= 119901111990121199013 119901119899minus1

isin 119861119882119899minus1119896

for each position 119894 with119901119894ge 119894 we insert 119899 into a certain place of 119861

1to get 1198611015840

1isin 119860119882

119899119896

If 1198612= 119901111990121199013 119901119899minus1

isin 119861119882119899minus1119896minus1

for each position 119894 with119901119894lt 119894 and the position 119894 where 119901

119894= 119899 minus 1 we insert 119899 into

a specific position of 1198612to obtain 119861

1015840

2isin 119860119882

119899119896 Such a map

119892 119861119882119899minus1119896

119861119882119899minus1119896minus1

rarr 119861119882119899119896

is illustrated in Table 2And the distinct images under 119892 exhaust all the permutationsin 119861119882

119899119896

Here is a concrete example for the constructions illus-trated in Table 2

Example 16 Suppose 119899 = 4 119896 = 2 We want to obtain11986111988242

= 3142 3412 3421 4132 4213 4312 4321 from11986111988232

= 321 231 and 11986111988231

= 312 For 321 isin 11986111988232 1199011=

3 ge 1 then it corresponds to 1198611015840 = 4213 which is case (d1) inTable 2 119901

2= 2 ge 2 then it corresponds to 1198611015840 = 3412which is

case (c1) in Table 2 Similarly we can construct 4312 4321from 231 isin 119861119882

32and 3421 3142 4132 from 312 isin 119861119882

31

using Table 2

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Riordan An Introduction to Combinatorial Analysis WileyPublications in Mathematical Statistics John Wiley amp SonsNew York NY USA 1958

[2] M Bona Combinatorics of Permutations Discrete Mathematicsand its Applications Chapman amp HallCRC Boca Raton FlaUSA 2004

[3] L Carlitz ldquo119902-Bernoulli and Eulerian numbersrdquo Transactions ofthe American Mathematical Society vol 76 pp 332ndash350 1954

[4] L Carlitz ldquoA combinatorial property of 119902-Eulerian numbersrdquoThe American Mathematical Monthly vol 82 pp 51ndash54 1975

[5] T Xiong H P Tsao and J I Hall ldquoGeneral Eulerian numbersand Eulerian polynomialsrdquo Journal of Mathematics vol 2013Article ID 629132 9 pages 2013

[6] R P Stanley Enumerative Combinatorics vol 1 of CambridgeStudies in Advanced Mathematics Cambridge University PressCambridge UK 1996

Submit your manuscripts athttpwwwhindawicom

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Differential EquationsInternational Journal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Stochastic AnalysisInternational Journal of

Journal of Discrete Mathematics 3

Then the Principle of Inclusion and Exclusion shows that

|119861| = (119896 + 1)119899minus119895

119896119895

minus 1198611+ 1198612+ sdot sdot sdot + (minus1)

119896

119861119896 (11)

Nowwe consider the value of119861119894 where 1 le 119894 le 119896 Suppose

that we have 119896+1minus119894 compartments with 119896minus119894 bars in betweenThere are (119896 + 1 minus 119894)119899minus119895(119896 minus 119894)119895 ways to insert 119899 numbers intothese 119896 + 1 minus 119894 compartments with first 119895 integers avoidingthe last compartment and list integers in each component ina decreasing order Then insert 119894 separating extraneous barsinto 119899 + 1 positions So we get

119861119894= (119896 + 1 minus 119894)

119899minus119895

(119896 minus 119894)119895

(119899 + 1

119894) (12)

Plug formula (12) into (11) we have 119888119899119896(119895) = |119861|

Given an arrangement 120587 isin 119861 if we remove the bars thenwe obtain a permutation 120587 isin Ω

119899 So without confusion we

just use the samenotation120587 to represent both an arrangementin set 119861 and a permutation on [119899] Now for each 120587 isin 119861 120587either

(case 1) has no extraneous bar and none of 1 2 119895locates in the last compartment or(case 2) has only one extraneous bar at the end

If 120587 is in case 1 then 120587 has 119896 ascents since each bar isnon-extraneous And the last compartment of 120587 is nonemptyTherefore the last cycle of 119891minus1(120587) has to be (119899 119901

119892) In other

words119876119899(119891minus1

(120587)) = 119901119892gt 119895 since none of 1 2 119895 locates

in the last compartment And by Proposition 4 119891minus1(120587) isin

119882119899119896+1

If 120587 is in case 2 then 120587 has 119896minus1 ascents since only the last

bar is extraneous Note that in this case the arrangement withno elements of 1 2 119895 in the compartment second to thelast or the last nonempty compartment has been removedby the Principle of Inclusion and Exclusion Equivalently atleast one number of 1 2 119895 has to be in the compartmentsecond to the last So the last cycle of 119891minus1(120587) has to be(119899 119901

119897) and 119876

119899(119891minus1

(120587)) = 119901119897le 119895 Also by Proposition 4

119891minus1

(120587) isin 119882119899119896

Combing all the results above statement (10) is correct

The next Theorem describes some interesting propertiesof the coefficients 119888

119899119896

Theorem9 Let the coefficients 119888119899119896

be as described inTheorem8 Then

(1) sum119899119896=0

119888119899119896(119895) = 119899 for any 0 le 119895 le 119899

(2) 119888119899119896(119895) = 119888

119899119899minus119896(119899 minus 119895) for all 0 le 119895 119896 le 119899

Before we can prove Theorem 9 we need the followinglemma which is also interesting by itself

Lemma 10 Given a positive integer 119899 then

120587 isin 119882119899119896

amp 119876119899(120587) = 119895

= 120587 isin 119882119899119899+1minus119896

amp 119876119899(120587) = 119899 + 1 minus 119895

(13)

for any 1 le 119896 119895 le 119899

Proof First of all given a positive integer 119899 we define afunction 119892 Ω

119899rarr Ω119899as follows

for 120587 = 11990111199012 119901119899isin Ω119899

119892 (120587) = (119899 + 1 minus 1199011) (119899 + 1 minus 119901

2) (119899 + 1 minus 119901

119899)

(14)

For instance for 120587 = 53214 isin Ω5 119892(120587) = 13452 119892 is

obviously a bijection ofΩ119899to itself

Now for some fixed 1 le 119896 119895 le 119899 suppose 119878119896119895

= 120587 isin

119882119899119896

amp 119876119899(120587) = 119895 and 119879

119896119895= 120587 isin 119882

119899119899+1minus119896amp 119876119899(120587) =

119899 + 1 minus 119895 For any 120587 isin 119878119896119895 we write 120587 in the standard

representation cycle form So 120587 = (119901119906 ) (119899 119895) and

119891(120587) = 119901119906 119899 119895 has 119896 minus 1 ascents by Proposition 4

Now we compose 119891(120587) with the bijection function 119892 as justdefined Then 119892(119891(120587)) = 119899 + 1 minus 119901

119906 1 119899 + 1 minus 119895 has

119899 minus 119896 ascents which implies that 119891minus1(119892(119891(120587))) has 119899 + 1 minus 119896weak excedances So 119891

minus1

(119892(119891(120587))) isin 119882119899119899+1minus119896

Note thatthe last cycle of 119891minus1(119892(119891(120587))) has to be (119899 119899 + 1 minus 119895)Therefore 119891minus1(119892(119891(120587))) isin 119879

119896119895 Since both 119891 and 119892 are

bijection functions 119891minus1119892119891 gives a bijection between 119878119896119895

and119879119896119895

Now we are ready to proveTheorem 9

Proof of Theorem 9 For part 1 by Theorem 8119899

sum

119896=0

119888119899119896(119895) =

119899

sum

119896=0

120587 isin 119882119899119896+1

119895 lt 119876119899(120587) le 119899

+

119899

sum

119896=0

120587 isin 119882119899119896

1 le 119876119899(120587) le 119895

=

119899

sum

119896=0

120587 isin 119882119899119896 =

1003816100381610038161003816Ω1198991003816100381610038161003816 = 119899

(15)

For part 2 also byTheorem 8

119888119899119896(119895) =

119899

sum

119894=119895+1

120587 isin 119882119899119896+1

119876119899(120587) = 119894

+

119895

sum

119898=1

120587 isin 119882119899119896 119876119899(120587) = 119898

=

119899

sum

119894=119895+1

120587 isin 119882119899119899minus119896

119876119899(120587) = 119899 + 1 minus 119894

+

119895

sum

119898=1

120587 isin 119882119899119899+1minus119896

119876119899(120587) = 119899 + 1 minus 119898 by Lemma 10

= 120587 isin 119882119899119896

1 le 119876119899(120587) le 119899 minus 119895

+ 120587 isin 119882119899119899+1minus119896

119899 minus 119895 lt 119876119899(120587) le 119899

= 119888119899119899minus119896

(119899 minus 119895)

(16)

4 Journal of Discrete Mathematics

Remark 11 Using the analytic formula of 119888119899119896(119895) as in (9) part

2 of Theorem 9 implies the following identity119896

sum

119894=0

(minus1)119894

(119896 + 1 minus 119894)119899minus119895

(119896 minus 119894)119895

(119899 + 1

119894)

=

119899minus119896

sum

119897=0

(minus1)119897

(119899 + 1 minus 119896 minus 119897)119895

(119899 minus 119896 minus 119897)119899minus119895

(119899 + 1

119897)

(17)

where 119899 is a positive integer and 0 le 119895 119896 le 119899

3 Another Combinatorial Interpretation of119888119899119896(1) and 119888

119899119896(119899minus1)

In pursuing the combinatorial meanings of the coefficients119888119899119896 the authors have found some other interesting properties

about permutations The results in this section will revealclose connections between the traditional Eulerian numbers119860119899119896

and 119888119899119896(119895) where 119895 = 1 or 119895 = 119899 minus 1

One fundamental concept of permutation combinatoricsis inversion A pair (119901

119894 119901119895) is called an inversion of the

permutation 120587 = 11990111199012 119901119899if 119894 lt 119895 and 119901

119894gt 119901119895[6 page

36] The following definition provides the main concepts ofthis section

Definition 12 For a fixed positive integer 119899 let 119860119882119899119896

= 120587 =

119901111990121199013 119901119899| 120587 isin 119882

119899119896and 119901

1lt 119901119899 (or (119901

1 119901119899) is not

an inversion) and 119861119882119899119896

= 119882119899119896

119860119882119899119896

(or (1199011 119901119899) is an

inversion)It is obvious that |119860119882

119899119896| + |119861119882

119899119896| = 119860119899119896 The following

theorem interprets coefficients 119888119899119896(1) and 119888

119899119896(119899 minus 1) in terms

of 119860119882119899119896

and 119861119882119899119896

Theorem 13 Let the coefficients 119888119899119896

of the general Euleriannumbers be written as in (9) 119860119882

119899119896and 119861119882

119899119896are as defined

in Definition 12 Then(1) 119888119899119896(1) = 2|119860119882

119899119896+1|

(2) 119888119899119896(119899 minus 1) = 2|119861119882

119899119896|

Proof For part (1) by Theorem 8 119888119899119896(1) = |119878

1| + |1198782| where

1198781= 120587 = 119901

11199012 119901119899| 120587 isin 119882

119899119896+1amp 1199011

= 119899 1198782= 120587 =

11990111199012 119901119899| 120587 isin 119882

119899119896amp 1199011= 119899 Given a permutation

120587 = 11990111199012 119901119899isin 1198781and 119901

119899= 119899 then both 119901

11199012 119901119899and

1199011198991199012 1199011belong to 119878

1 so one of them has to be in 119860119882

119899119896+1

If 120587 = 11990111199012 119901119899isin 1198781and 119901

119899= 119899 then 120587 isin 119860119882

119899119896+1 but

1199011198991199012 1199011isin 1198782 Therefore (12)119888

119899119896(1) = |119860119882

119899119896+1|

Part (2) can be proved using exactly the same method Sowe leave it to the readers as an exercise

|119860119882119899119896| and |119861119882

119899119896| are interesting combinatorial

concepts by themselves Note that generally speaking|119860119882119899119896| = |119861119882

119899119896| Indeed |119860119882

119899119896| = |119861119882

119899119899+1minus119896|

Theorem 14 For any positive integer 119899 ge 2 the sets119860119882119899119896

and119861119882119899119896

are defined in Definition 12Then |119860119882119899119896| = |119861119882

119899119899+1minus119896|

for 1 le 119896 le 119899

Proof It is an obvious result of part 2 of Theorems 9 and 13

Our last result of this paper is the following theoremwhich reveals that both |119860119882

119899119896| and |119861119882

119899119896| take exactly the

same recursive formula as the traditional Eulerian numbers119860119899119896

as shown in (1)

Theorem 15 For a fixed positive integer 119899 let 119860119882119899119896

and119861119882119899119896

be as defined in Definition 12 then

1198961003816100381610038161003816119860119882119899minus1119896

1003816100381610038161003816 + (119899 + 1 minus 119896)1003816100381610038161003816119860119882119899minus1119896minus1

1003816100381610038161003816 =1003816100381610038161003816119860119882119899119896

1003816100381610038161003816 (18)

1198961003816100381610038161003816119861119882119899minus1119896

1003816100381610038161003816 + (119899 + 1 minus 119896)1003816100381610038161003816119861119882119899minus1119896minus1

1003816100381610038161003816 =1003816100381610038161003816119861119882119899119896

1003816100381610038161003816 (19)

Proof A computational proof can be obtained straightfor-ward by using (9) and Theorem 13 But here we provide aproof in a flavor of combinatorics

Idea of the Proof For (18) given a permutation 1198601=

119901111990121199013 119901119899minus1

isin 119860119882119899minus1119896

for each position 119894 with 119901119894ge 119894

we insert 119899 into a certain place of 1198601 such that the new

permutation 1198601015840

1is in 119860119882

119899119896 There are 119896 such positions so

we can get 119896 new permutations in 119860119882119899119896 Similarly if 119860

2=

119901111990121199013 119901119899minus1

isin 119860119882119899minus1119896minus1

for each position 119894 with 119901119894lt 119894

and the position at the end of 1198602 we insert 119899 into a specific

position of 1198602and the resulting new permutation 119860

1015840

2is in

119860119882119899119896 There are 119899 + 1 minus 119896 such positions so we can get

119899 + 1 minus 119896 new permutations in 119860119882119899119896 We will show that

all the permutations obtained from the above constructionsare distinct and they have exhausted all the permutations in119860119882119899119896

For any fixed 1198601015840 = 120587112058721205873 120587119899isin 119860119882

119899119896 then 120587

1lt 120587119899

We classify 1198601015840 into the following disjoint cases

Case a Consider that 120587119894

= 119899 with 119894 lt 119899 So 1198601015840

=

12058711205872 120587119894minus1119899120587119894+1

120587119899minus1

120587119899

(a1) 1205871lt 120587119899minus1

and 120587119899ge 119894

(a2) 1205871lt 120587119899minus1

and 120587119899lt 119894

(a3) 1205871gt 120587119899minus1

120587119899lt 119899 minus 1 and 120587

119899ge 119894

(a4) 1205871gt 120587119899minus1

120587119899lt 119899 minus 1 and 120587

119899lt 119894

(a5) 1205871gt 120587119899minus1

and 120587119899= 119899 minus 1

Case b Consider that 120587119899= 119899 So 120587

119894= 119899minus1 for some 119894 lt 119899 and

1198601015840

= 12058711205872 120587119894minus1119899 minus 1 120587

119899minus1119899

(b1) 1205871lt 120587119899minus1

(b2) 120587

119899minus1lt 1205871lt 119899 minus 1 and 120587

119899minus1ge 119894

(b3) 120587119899minus1

lt 1205871lt 119899 minus 1 and 120587

119899minus1lt 119894

(b4) 1205871= 119899 minus 1

Based on the classifications listed above we can construct amap 119891 119860119882

119899minus1119896 119860119882119899minus1119896minus1

rarr 119860119882119899119896

by applying the ideaof the proof we have illustrated at the beginning of the proofTo save space the map 119891 is demonstrated in Table 1 FromTable 1 we can see that in each case the positions of inserting119899 are all different So all the images obtained in a certain caseare different Since all the cases are disjoint all the images1198601015840

isin 119860119882119899119896

are distinct

Journal of Discrete Mathematics 5

Table 1 The map 119891 119860119882119899minus1119896

119860119882119899minus1119896minus1

rarr 119860119882119899119896

119860 = 11990111199012 119901119899minus1

Position 119894 Condition 1198601015840

isin 119860119882119899119896

119860 isin 119860119882119899minus1119896

1 lt 119894 le 119899 minus 1

and119901119894ge 119894

119901119894gt 1199011

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011lt 119901119899minus1

1199011lt 119901119894

Case (a1)

119901119894lt 1199011and

119901119899minus1

lt 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus2119901119894119901119899minus1

With 119901119894lt 1199011lt 119901119899minus1

lt 119899 minus 1

Case (a3)

119901119894lt 1199011and

119901119899minus1

= 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899 minus 1119901

119894+1 119901119899minus2119901119894119899

With 119901119894lt 1199011119901119894ge 119894

Case (b2)

119894 = 1

119901119895= 119899 minus 1

and 119895 lt 119899 minus 1

1198601015840

= 119901119899minus11199012 119901119895minus1119899119901119895+1

119901119899minus2

1199011119899 minus 1

With 1199011lt 119901119899minus1

Case (a5)

119901119899minus1

= 119899 minus 11198601015840

= 119899 minus 11199012 119901119899minus21199011119899

Case (b4)

119860 isin 119860119882119899minus1119896minus1

1 lt 119894 le 119899 minus 1

and119901119894lt 119894

119901119894gt 1199011

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011lt 119901119899minus1

1199011lt 119901119894

Case (a2)

119901119894lt 1199011and

119901119899minus1

lt 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus2119901119894119901119899minus1

With 119901119894lt 1199011lt 119901119899minus1

lt 119899 minus 1

Case (a4)

119901119894lt 1199011and

119901119899minus1

= 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899 minus 1119901

119894+1 119901119899minus2119901119894119899

With 119901119894lt 1199011119901119894lt 119894

Case (b3)

119894 = 1198991198601015840

= 11990111199012 119901119899minus1119899

Case (b1)

Table 2 The map 119892 119861119882119899minus1119896

119861119882119899minus1119896minus1

rarr 119861119882119899119896

119861 = 11990111199012 119901119899minus1

Position 119894 Condition 1198611015840

isin 119861119882119899119896

119861 isin 119882119861119899minus1119896

1 lt 119894 lt 119899 minus 1

and119901119894ge 119894

1199011gt 119901119894

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011gt 119901119899minus1

and 119901119894ge 119894

Case (c1)

1199011lt 119901119894lt 119899 minus 1

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119894119901119899minus1

With 1199011lt 119901119894lt 119899 minus 1 119901

119894ge 119894

Case (c3)

1199011lt 119901119894= 119899 minus 1

1198611015840

= 1198991199012 119901119894minus1119899 minus 1119901

119894+1 1199011119901119899minus1

With 119901119894= 119899 minus 1 119901

1gt 119901119899minus1

Case (d2)

119894 = 1 1199011ge 1

1198611015840

= 1198991199012 119901119899minus11199011

With 119901119899minus1

lt 1199011

Case (d1)

119861 isin 119882119861119899minus1119896minus1

1 lt 119894 lt 119899 minus 1

and119901119894lt 119894

1199011gt 119901119894

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011gt 119901119899minus1

and 119901119894lt 119894

Case (c2)

1199011lt 119901119894

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119894119901119899minus1

With 1199011lt 119901119894lt 119899 minus 1 119901

119894lt 119894

Case (c4)119894 = 119899 minus 1

119901119894lt 119894

1199011gt 119901119894= 119901119899minus1

1198611015840

= 1199011 119901119899minus2119899119901119899minus1

Case (c6)1 le 119894 lt 119899 minus 1

and119901119894= 119899 minus 1

119901119894= 119899 minus 1

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119899minus2119899 minus 1119901

119899minus1

Case (c5)

6 Journal of Discrete Mathematics

Similarly for each 1198611015840 = 120587112058721205873 120587119899isin 119861119882

119899119896 then 120587

1gt

120587119899 We classify 1198611015840 into the following disjoint cases

Case c Consider that 120587119894= 119899 with 1 lt 119894 le 119899 minus 1 So 1198611015840 =

12058711205872 120587119894minus1119899120587119894+1

120587119899minus1

120587119899

(c1) 1205871gt 120587119899minus1

and 120587119899ge 119894

(c2) 1205871gt 120587119899minus1

and 120587119899lt 119894

(c3) 1205871lt 120587119899minus1

lt 119899 minus 1 120587119899minus1

ge 119894(c4) 120587

1lt 120587119899minus1

lt 119899 minus 1 120587119899minus1

lt 119894(c5) 120587

119899minus1= 119899 minus 1

(c6) 120587119899minus1

= 119899

Case d Consider that 1205871= 119899 So 1198611015840 = 119899120587

2 120587119899minus2

120587119899minus1

(d1) 120587119899minus2

lt 120587119899minus1

(d2) 120587

119899minus2gt 120587119899minus1

To prove (19) we use a similar idea of proof as shown aboveIf 1198611= 119901111990121199013 119901119899minus1

isin 119861119882119899minus1119896

for each position 119894 with119901119894ge 119894 we insert 119899 into a certain place of 119861

1to get 1198611015840

1isin 119860119882

119899119896

If 1198612= 119901111990121199013 119901119899minus1

isin 119861119882119899minus1119896minus1

for each position 119894 with119901119894lt 119894 and the position 119894 where 119901

119894= 119899 minus 1 we insert 119899 into

a specific position of 1198612to obtain 119861

1015840

2isin 119860119882

119899119896 Such a map

119892 119861119882119899minus1119896

119861119882119899minus1119896minus1

rarr 119861119882119899119896

is illustrated in Table 2And the distinct images under 119892 exhaust all the permutationsin 119861119882

119899119896

Here is a concrete example for the constructions illus-trated in Table 2

Example 16 Suppose 119899 = 4 119896 = 2 We want to obtain11986111988242

= 3142 3412 3421 4132 4213 4312 4321 from11986111988232

= 321 231 and 11986111988231

= 312 For 321 isin 11986111988232 1199011=

3 ge 1 then it corresponds to 1198611015840 = 4213 which is case (d1) inTable 2 119901

2= 2 ge 2 then it corresponds to 1198611015840 = 3412which is

case (c1) in Table 2 Similarly we can construct 4312 4321from 231 isin 119861119882

32and 3421 3142 4132 from 312 isin 119861119882

31

using Table 2

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Riordan An Introduction to Combinatorial Analysis WileyPublications in Mathematical Statistics John Wiley amp SonsNew York NY USA 1958

[2] M Bona Combinatorics of Permutations Discrete Mathematicsand its Applications Chapman amp HallCRC Boca Raton FlaUSA 2004

[3] L Carlitz ldquo119902-Bernoulli and Eulerian numbersrdquo Transactions ofthe American Mathematical Society vol 76 pp 332ndash350 1954

[4] L Carlitz ldquoA combinatorial property of 119902-Eulerian numbersrdquoThe American Mathematical Monthly vol 82 pp 51ndash54 1975

[5] T Xiong H P Tsao and J I Hall ldquoGeneral Eulerian numbersand Eulerian polynomialsrdquo Journal of Mathematics vol 2013Article ID 629132 9 pages 2013

[6] R P Stanley Enumerative Combinatorics vol 1 of CambridgeStudies in Advanced Mathematics Cambridge University PressCambridge UK 1996

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Stochastic AnalysisInternational Journal of

4 Journal of Discrete Mathematics

Remark 11 Using the analytic formula of 119888119899119896(119895) as in (9) part

2 of Theorem 9 implies the following identity119896

sum

119894=0

(minus1)119894

(119896 + 1 minus 119894)119899minus119895

(119896 minus 119894)119895

(119899 + 1

119894)

=

119899minus119896

sum

119897=0

(minus1)119897

(119899 + 1 minus 119896 minus 119897)119895

(119899 minus 119896 minus 119897)119899minus119895

(119899 + 1

119897)

(17)

where 119899 is a positive integer and 0 le 119895 119896 le 119899

3 Another Combinatorial Interpretation of119888119899119896(1) and 119888

119899119896(119899minus1)

In pursuing the combinatorial meanings of the coefficients119888119899119896 the authors have found some other interesting properties

about permutations The results in this section will revealclose connections between the traditional Eulerian numbers119860119899119896

and 119888119899119896(119895) where 119895 = 1 or 119895 = 119899 minus 1

One fundamental concept of permutation combinatoricsis inversion A pair (119901

119894 119901119895) is called an inversion of the

permutation 120587 = 11990111199012 119901119899if 119894 lt 119895 and 119901

119894gt 119901119895[6 page

36] The following definition provides the main concepts ofthis section

Definition 12 For a fixed positive integer 119899 let 119860119882119899119896

= 120587 =

119901111990121199013 119901119899| 120587 isin 119882

119899119896and 119901

1lt 119901119899 (or (119901

1 119901119899) is not

an inversion) and 119861119882119899119896

= 119882119899119896

119860119882119899119896

(or (1199011 119901119899) is an

inversion)It is obvious that |119860119882

119899119896| + |119861119882

119899119896| = 119860119899119896 The following

theorem interprets coefficients 119888119899119896(1) and 119888

119899119896(119899 minus 1) in terms

of 119860119882119899119896

and 119861119882119899119896

Theorem 13 Let the coefficients 119888119899119896

of the general Euleriannumbers be written as in (9) 119860119882

119899119896and 119861119882

119899119896are as defined

in Definition 12 Then(1) 119888119899119896(1) = 2|119860119882

119899119896+1|

(2) 119888119899119896(119899 minus 1) = 2|119861119882

119899119896|

Proof For part (1) by Theorem 8 119888119899119896(1) = |119878

1| + |1198782| where

1198781= 120587 = 119901

11199012 119901119899| 120587 isin 119882

119899119896+1amp 1199011

= 119899 1198782= 120587 =

11990111199012 119901119899| 120587 isin 119882

119899119896amp 1199011= 119899 Given a permutation

120587 = 11990111199012 119901119899isin 1198781and 119901

119899= 119899 then both 119901

11199012 119901119899and

1199011198991199012 1199011belong to 119878

1 so one of them has to be in 119860119882

119899119896+1

If 120587 = 11990111199012 119901119899isin 1198781and 119901

119899= 119899 then 120587 isin 119860119882

119899119896+1 but

1199011198991199012 1199011isin 1198782 Therefore (12)119888

119899119896(1) = |119860119882

119899119896+1|

Part (2) can be proved using exactly the same method Sowe leave it to the readers as an exercise

|119860119882119899119896| and |119861119882

119899119896| are interesting combinatorial

concepts by themselves Note that generally speaking|119860119882119899119896| = |119861119882

119899119896| Indeed |119860119882

119899119896| = |119861119882

119899119899+1minus119896|

Theorem 14 For any positive integer 119899 ge 2 the sets119860119882119899119896

and119861119882119899119896

are defined in Definition 12Then |119860119882119899119896| = |119861119882

119899119899+1minus119896|

for 1 le 119896 le 119899

Proof It is an obvious result of part 2 of Theorems 9 and 13

Our last result of this paper is the following theoremwhich reveals that both |119860119882

119899119896| and |119861119882

119899119896| take exactly the

same recursive formula as the traditional Eulerian numbers119860119899119896

as shown in (1)

Theorem 15 For a fixed positive integer 119899 let 119860119882119899119896

and119861119882119899119896

be as defined in Definition 12 then

1198961003816100381610038161003816119860119882119899minus1119896

1003816100381610038161003816 + (119899 + 1 minus 119896)1003816100381610038161003816119860119882119899minus1119896minus1

1003816100381610038161003816 =1003816100381610038161003816119860119882119899119896

1003816100381610038161003816 (18)

1198961003816100381610038161003816119861119882119899minus1119896

1003816100381610038161003816 + (119899 + 1 minus 119896)1003816100381610038161003816119861119882119899minus1119896minus1

1003816100381610038161003816 =1003816100381610038161003816119861119882119899119896

1003816100381610038161003816 (19)

Proof A computational proof can be obtained straightfor-ward by using (9) and Theorem 13 But here we provide aproof in a flavor of combinatorics

Idea of the Proof For (18) given a permutation 1198601=

119901111990121199013 119901119899minus1

isin 119860119882119899minus1119896

for each position 119894 with 119901119894ge 119894

we insert 119899 into a certain place of 1198601 such that the new

permutation 1198601015840

1is in 119860119882

119899119896 There are 119896 such positions so

we can get 119896 new permutations in 119860119882119899119896 Similarly if 119860

2=

119901111990121199013 119901119899minus1

isin 119860119882119899minus1119896minus1

for each position 119894 with 119901119894lt 119894

and the position at the end of 1198602 we insert 119899 into a specific

position of 1198602and the resulting new permutation 119860

1015840

2is in

119860119882119899119896 There are 119899 + 1 minus 119896 such positions so we can get

119899 + 1 minus 119896 new permutations in 119860119882119899119896 We will show that

all the permutations obtained from the above constructionsare distinct and they have exhausted all the permutations in119860119882119899119896

For any fixed 1198601015840 = 120587112058721205873 120587119899isin 119860119882

119899119896 then 120587

1lt 120587119899

We classify 1198601015840 into the following disjoint cases

Case a Consider that 120587119894

= 119899 with 119894 lt 119899 So 1198601015840

=

12058711205872 120587119894minus1119899120587119894+1

120587119899minus1

120587119899

(a1) 1205871lt 120587119899minus1

and 120587119899ge 119894

(a2) 1205871lt 120587119899minus1

and 120587119899lt 119894

(a3) 1205871gt 120587119899minus1

120587119899lt 119899 minus 1 and 120587

119899ge 119894

(a4) 1205871gt 120587119899minus1

120587119899lt 119899 minus 1 and 120587

119899lt 119894

(a5) 1205871gt 120587119899minus1

and 120587119899= 119899 minus 1

Case b Consider that 120587119899= 119899 So 120587

119894= 119899minus1 for some 119894 lt 119899 and

1198601015840

= 12058711205872 120587119894minus1119899 minus 1 120587

119899minus1119899

(b1) 1205871lt 120587119899minus1

(b2) 120587

119899minus1lt 1205871lt 119899 minus 1 and 120587

119899minus1ge 119894

(b3) 120587119899minus1

lt 1205871lt 119899 minus 1 and 120587

119899minus1lt 119894

(b4) 1205871= 119899 minus 1

Based on the classifications listed above we can construct amap 119891 119860119882

119899minus1119896 119860119882119899minus1119896minus1

rarr 119860119882119899119896

by applying the ideaof the proof we have illustrated at the beginning of the proofTo save space the map 119891 is demonstrated in Table 1 FromTable 1 we can see that in each case the positions of inserting119899 are all different So all the images obtained in a certain caseare different Since all the cases are disjoint all the images1198601015840

isin 119860119882119899119896

are distinct

Journal of Discrete Mathematics 5

Table 1 The map 119891 119860119882119899minus1119896

119860119882119899minus1119896minus1

rarr 119860119882119899119896

119860 = 11990111199012 119901119899minus1

Position 119894 Condition 1198601015840

isin 119860119882119899119896

119860 isin 119860119882119899minus1119896

1 lt 119894 le 119899 minus 1

and119901119894ge 119894

119901119894gt 1199011

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011lt 119901119899minus1

1199011lt 119901119894

Case (a1)

119901119894lt 1199011and

119901119899minus1

lt 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus2119901119894119901119899minus1

With 119901119894lt 1199011lt 119901119899minus1

lt 119899 minus 1

Case (a3)

119901119894lt 1199011and

119901119899minus1

= 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899 minus 1119901

119894+1 119901119899minus2119901119894119899

With 119901119894lt 1199011119901119894ge 119894

Case (b2)

119894 = 1

119901119895= 119899 minus 1

and 119895 lt 119899 minus 1

1198601015840

= 119901119899minus11199012 119901119895minus1119899119901119895+1

119901119899minus2

1199011119899 minus 1

With 1199011lt 119901119899minus1

Case (a5)

119901119899minus1

= 119899 minus 11198601015840

= 119899 minus 11199012 119901119899minus21199011119899

Case (b4)

119860 isin 119860119882119899minus1119896minus1

1 lt 119894 le 119899 minus 1

and119901119894lt 119894

119901119894gt 1199011

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011lt 119901119899minus1

1199011lt 119901119894

Case (a2)

119901119894lt 1199011and

119901119899minus1

lt 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus2119901119894119901119899minus1

With 119901119894lt 1199011lt 119901119899minus1

lt 119899 minus 1

Case (a4)

119901119894lt 1199011and

119901119899minus1

= 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899 minus 1119901

119894+1 119901119899minus2119901119894119899

With 119901119894lt 1199011119901119894lt 119894

Case (b3)

119894 = 1198991198601015840

= 11990111199012 119901119899minus1119899

Case (b1)

Table 2 The map 119892 119861119882119899minus1119896

119861119882119899minus1119896minus1

rarr 119861119882119899119896

119861 = 11990111199012 119901119899minus1

Position 119894 Condition 1198611015840

isin 119861119882119899119896

119861 isin 119882119861119899minus1119896

1 lt 119894 lt 119899 minus 1

and119901119894ge 119894

1199011gt 119901119894

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011gt 119901119899minus1

and 119901119894ge 119894

Case (c1)

1199011lt 119901119894lt 119899 minus 1

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119894119901119899minus1

With 1199011lt 119901119894lt 119899 minus 1 119901

119894ge 119894

Case (c3)

1199011lt 119901119894= 119899 minus 1

1198611015840

= 1198991199012 119901119894minus1119899 minus 1119901

119894+1 1199011119901119899minus1

With 119901119894= 119899 minus 1 119901

1gt 119901119899minus1

Case (d2)

119894 = 1 1199011ge 1

1198611015840

= 1198991199012 119901119899minus11199011

With 119901119899minus1

lt 1199011

Case (d1)

119861 isin 119882119861119899minus1119896minus1

1 lt 119894 lt 119899 minus 1

and119901119894lt 119894

1199011gt 119901119894

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011gt 119901119899minus1

and 119901119894lt 119894

Case (c2)

1199011lt 119901119894

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119894119901119899minus1

With 1199011lt 119901119894lt 119899 minus 1 119901

119894lt 119894

Case (c4)119894 = 119899 minus 1

119901119894lt 119894

1199011gt 119901119894= 119901119899minus1

1198611015840

= 1199011 119901119899minus2119899119901119899minus1

Case (c6)1 le 119894 lt 119899 minus 1

and119901119894= 119899 minus 1

119901119894= 119899 minus 1

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119899minus2119899 minus 1119901

119899minus1

Case (c5)

6 Journal of Discrete Mathematics

Similarly for each 1198611015840 = 120587112058721205873 120587119899isin 119861119882

119899119896 then 120587

1gt

120587119899 We classify 1198611015840 into the following disjoint cases

Case c Consider that 120587119894= 119899 with 1 lt 119894 le 119899 minus 1 So 1198611015840 =

12058711205872 120587119894minus1119899120587119894+1

120587119899minus1

120587119899

(c1) 1205871gt 120587119899minus1

and 120587119899ge 119894

(c2) 1205871gt 120587119899minus1

and 120587119899lt 119894

(c3) 1205871lt 120587119899minus1

lt 119899 minus 1 120587119899minus1

ge 119894(c4) 120587

1lt 120587119899minus1

lt 119899 minus 1 120587119899minus1

lt 119894(c5) 120587

119899minus1= 119899 minus 1

(c6) 120587119899minus1

= 119899

Case d Consider that 1205871= 119899 So 1198611015840 = 119899120587

2 120587119899minus2

120587119899minus1

(d1) 120587119899minus2

lt 120587119899minus1

(d2) 120587

119899minus2gt 120587119899minus1

To prove (19) we use a similar idea of proof as shown aboveIf 1198611= 119901111990121199013 119901119899minus1

isin 119861119882119899minus1119896

for each position 119894 with119901119894ge 119894 we insert 119899 into a certain place of 119861

1to get 1198611015840

1isin 119860119882

119899119896

If 1198612= 119901111990121199013 119901119899minus1

isin 119861119882119899minus1119896minus1

for each position 119894 with119901119894lt 119894 and the position 119894 where 119901

119894= 119899 minus 1 we insert 119899 into

a specific position of 1198612to obtain 119861

1015840

2isin 119860119882

119899119896 Such a map

119892 119861119882119899minus1119896

119861119882119899minus1119896minus1

rarr 119861119882119899119896

is illustrated in Table 2And the distinct images under 119892 exhaust all the permutationsin 119861119882

119899119896

Here is a concrete example for the constructions illus-trated in Table 2

Example 16 Suppose 119899 = 4 119896 = 2 We want to obtain11986111988242

= 3142 3412 3421 4132 4213 4312 4321 from11986111988232

= 321 231 and 11986111988231

= 312 For 321 isin 11986111988232 1199011=

3 ge 1 then it corresponds to 1198611015840 = 4213 which is case (d1) inTable 2 119901

2= 2 ge 2 then it corresponds to 1198611015840 = 3412which is

case (c1) in Table 2 Similarly we can construct 4312 4321from 231 isin 119861119882

32and 3421 3142 4132 from 312 isin 119861119882

31

using Table 2

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Riordan An Introduction to Combinatorial Analysis WileyPublications in Mathematical Statistics John Wiley amp SonsNew York NY USA 1958

[2] M Bona Combinatorics of Permutations Discrete Mathematicsand its Applications Chapman amp HallCRC Boca Raton FlaUSA 2004

[3] L Carlitz ldquo119902-Bernoulli and Eulerian numbersrdquo Transactions ofthe American Mathematical Society vol 76 pp 332ndash350 1954

[4] L Carlitz ldquoA combinatorial property of 119902-Eulerian numbersrdquoThe American Mathematical Monthly vol 82 pp 51ndash54 1975

[5] T Xiong H P Tsao and J I Hall ldquoGeneral Eulerian numbersand Eulerian polynomialsrdquo Journal of Mathematics vol 2013Article ID 629132 9 pages 2013

[6] R P Stanley Enumerative Combinatorics vol 1 of CambridgeStudies in Advanced Mathematics Cambridge University PressCambridge UK 1996

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Discrete Mathematics 5

Table 1 The map 119891 119860119882119899minus1119896

119860119882119899minus1119896minus1

rarr 119860119882119899119896

119860 = 11990111199012 119901119899minus1

Position 119894 Condition 1198601015840

isin 119860119882119899119896

119860 isin 119860119882119899minus1119896

1 lt 119894 le 119899 minus 1

and119901119894ge 119894

119901119894gt 1199011

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011lt 119901119899minus1

1199011lt 119901119894

Case (a1)

119901119894lt 1199011and

119901119899minus1

lt 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus2119901119894119901119899minus1

With 119901119894lt 1199011lt 119901119899minus1

lt 119899 minus 1

Case (a3)

119901119894lt 1199011and

119901119899minus1

= 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899 minus 1119901

119894+1 119901119899minus2119901119894119899

With 119901119894lt 1199011119901119894ge 119894

Case (b2)

119894 = 1

119901119895= 119899 minus 1

and 119895 lt 119899 minus 1

1198601015840

= 119901119899minus11199012 119901119895minus1119899119901119895+1

119901119899minus2

1199011119899 minus 1

With 1199011lt 119901119899minus1

Case (a5)

119901119899minus1

= 119899 minus 11198601015840

= 119899 minus 11199012 119901119899minus21199011119899

Case (b4)

119860 isin 119860119882119899minus1119896minus1

1 lt 119894 le 119899 minus 1

and119901119894lt 119894

119901119894gt 1199011

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011lt 119901119899minus1

1199011lt 119901119894

Case (a2)

119901119894lt 1199011and

119901119899minus1

lt 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899119901119894+1

119901119899minus2119901119894119901119899minus1

With 119901119894lt 1199011lt 119901119899minus1

lt 119899 minus 1

Case (a4)

119901119894lt 1199011and

119901119899minus1

= 119899 minus 1

1198601015840

= 11990111199012 119901119894minus1119899 minus 1119901

119894+1 119901119899minus2119901119894119899

With 119901119894lt 1199011119901119894lt 119894

Case (b3)

119894 = 1198991198601015840

= 11990111199012 119901119899minus1119899

Case (b1)

Table 2 The map 119892 119861119882119899minus1119896

119861119882119899minus1119896minus1

rarr 119861119882119899119896

119861 = 11990111199012 119901119899minus1

Position 119894 Condition 1198611015840

isin 119861119882119899119896

119861 isin 119882119861119899minus1119896

1 lt 119894 lt 119899 minus 1

and119901119894ge 119894

1199011gt 119901119894

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011gt 119901119899minus1

and 119901119894ge 119894

Case (c1)

1199011lt 119901119894lt 119899 minus 1

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119894119901119899minus1

With 1199011lt 119901119894lt 119899 minus 1 119901

119894ge 119894

Case (c3)

1199011lt 119901119894= 119899 minus 1

1198611015840

= 1198991199012 119901119894minus1119899 minus 1119901

119894+1 1199011119901119899minus1

With 119901119894= 119899 minus 1 119901

1gt 119901119899minus1

Case (d2)

119894 = 1 1199011ge 1

1198611015840

= 1198991199012 119901119899minus11199011

With 119901119899minus1

lt 1199011

Case (d1)

119861 isin 119882119861119899minus1119896minus1

1 lt 119894 lt 119899 minus 1

and119901119894lt 119894

1199011gt 119901119894

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119899minus1119901119894

With 1199011gt 119901119899minus1

and 119901119894lt 119894

Case (c2)

1199011lt 119901119894

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119894119901119899minus1

With 1199011lt 119901119894lt 119899 minus 1 119901

119894lt 119894

Case (c4)119894 = 119899 minus 1

119901119894lt 119894

1199011gt 119901119894= 119901119899minus1

1198611015840

= 1199011 119901119899minus2119899119901119899minus1

Case (c6)1 le 119894 lt 119899 minus 1

and119901119894= 119899 minus 1

119901119894= 119899 minus 1

1198611015840

= 1199011 119901119894minus1119899119901119894+1

119901119899minus2119899 minus 1119901

119899minus1

Case (c5)

6 Journal of Discrete Mathematics

Similarly for each 1198611015840 = 120587112058721205873 120587119899isin 119861119882

119899119896 then 120587

1gt

120587119899 We classify 1198611015840 into the following disjoint cases

Case c Consider that 120587119894= 119899 with 1 lt 119894 le 119899 minus 1 So 1198611015840 =

12058711205872 120587119894minus1119899120587119894+1

120587119899minus1

120587119899

(c1) 1205871gt 120587119899minus1

and 120587119899ge 119894

(c2) 1205871gt 120587119899minus1

and 120587119899lt 119894

(c3) 1205871lt 120587119899minus1

lt 119899 minus 1 120587119899minus1

ge 119894(c4) 120587

1lt 120587119899minus1

lt 119899 minus 1 120587119899minus1

lt 119894(c5) 120587

119899minus1= 119899 minus 1

(c6) 120587119899minus1

= 119899

Case d Consider that 1205871= 119899 So 1198611015840 = 119899120587

2 120587119899minus2

120587119899minus1

(d1) 120587119899minus2

lt 120587119899minus1

(d2) 120587

119899minus2gt 120587119899minus1

To prove (19) we use a similar idea of proof as shown aboveIf 1198611= 119901111990121199013 119901119899minus1

isin 119861119882119899minus1119896

for each position 119894 with119901119894ge 119894 we insert 119899 into a certain place of 119861

1to get 1198611015840

1isin 119860119882

119899119896

If 1198612= 119901111990121199013 119901119899minus1

isin 119861119882119899minus1119896minus1

for each position 119894 with119901119894lt 119894 and the position 119894 where 119901

119894= 119899 minus 1 we insert 119899 into

a specific position of 1198612to obtain 119861

1015840

2isin 119860119882

119899119896 Such a map

119892 119861119882119899minus1119896

119861119882119899minus1119896minus1

rarr 119861119882119899119896

is illustrated in Table 2And the distinct images under 119892 exhaust all the permutationsin 119861119882

119899119896

Here is a concrete example for the constructions illus-trated in Table 2

Example 16 Suppose 119899 = 4 119896 = 2 We want to obtain11986111988242

= 3142 3412 3421 4132 4213 4312 4321 from11986111988232

= 321 231 and 11986111988231

= 312 For 321 isin 11986111988232 1199011=

3 ge 1 then it corresponds to 1198611015840 = 4213 which is case (d1) inTable 2 119901

2= 2 ge 2 then it corresponds to 1198611015840 = 3412which is

case (c1) in Table 2 Similarly we can construct 4312 4321from 231 isin 119861119882

32and 3421 3142 4132 from 312 isin 119861119882

31

using Table 2

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Riordan An Introduction to Combinatorial Analysis WileyPublications in Mathematical Statistics John Wiley amp SonsNew York NY USA 1958

[2] M Bona Combinatorics of Permutations Discrete Mathematicsand its Applications Chapman amp HallCRC Boca Raton FlaUSA 2004

[3] L Carlitz ldquo119902-Bernoulli and Eulerian numbersrdquo Transactions ofthe American Mathematical Society vol 76 pp 332ndash350 1954

[4] L Carlitz ldquoA combinatorial property of 119902-Eulerian numbersrdquoThe American Mathematical Monthly vol 82 pp 51ndash54 1975

[5] T Xiong H P Tsao and J I Hall ldquoGeneral Eulerian numbersand Eulerian polynomialsrdquo Journal of Mathematics vol 2013Article ID 629132 9 pages 2013

[6] R P Stanley Enumerative Combinatorics vol 1 of CambridgeStudies in Advanced Mathematics Cambridge University PressCambridge UK 1996

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Journal of Discrete Mathematics

Similarly for each 1198611015840 = 120587112058721205873 120587119899isin 119861119882

119899119896 then 120587

1gt

120587119899 We classify 1198611015840 into the following disjoint cases

Case c Consider that 120587119894= 119899 with 1 lt 119894 le 119899 minus 1 So 1198611015840 =

12058711205872 120587119894minus1119899120587119894+1

120587119899minus1

120587119899

(c1) 1205871gt 120587119899minus1

and 120587119899ge 119894

(c2) 1205871gt 120587119899minus1

and 120587119899lt 119894

(c3) 1205871lt 120587119899minus1

lt 119899 minus 1 120587119899minus1

ge 119894(c4) 120587

1lt 120587119899minus1

lt 119899 minus 1 120587119899minus1

lt 119894(c5) 120587

119899minus1= 119899 minus 1

(c6) 120587119899minus1

= 119899

Case d Consider that 1205871= 119899 So 1198611015840 = 119899120587

2 120587119899minus2

120587119899minus1

(d1) 120587119899minus2

lt 120587119899minus1

(d2) 120587

119899minus2gt 120587119899minus1

To prove (19) we use a similar idea of proof as shown aboveIf 1198611= 119901111990121199013 119901119899minus1

isin 119861119882119899minus1119896

for each position 119894 with119901119894ge 119894 we insert 119899 into a certain place of 119861

1to get 1198611015840

1isin 119860119882

119899119896

If 1198612= 119901111990121199013 119901119899minus1

isin 119861119882119899minus1119896minus1

for each position 119894 with119901119894lt 119894 and the position 119894 where 119901

119894= 119899 minus 1 we insert 119899 into

a specific position of 1198612to obtain 119861

1015840

2isin 119860119882

119899119896 Such a map

119892 119861119882119899minus1119896

119861119882119899minus1119896minus1

rarr 119861119882119899119896

is illustrated in Table 2And the distinct images under 119892 exhaust all the permutationsin 119861119882

119899119896

Here is a concrete example for the constructions illus-trated in Table 2

Example 16 Suppose 119899 = 4 119896 = 2 We want to obtain11986111988242

= 3142 3412 3421 4132 4213 4312 4321 from11986111988232

= 321 231 and 11986111988231

= 312 For 321 isin 11986111988232 1199011=

3 ge 1 then it corresponds to 1198611015840 = 4213 which is case (d1) inTable 2 119901

2= 2 ge 2 then it corresponds to 1198611015840 = 3412which is

case (c1) in Table 2 Similarly we can construct 4312 4321from 231 isin 119861119882

32and 3421 3142 4132 from 312 isin 119861119882

31

using Table 2

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Riordan An Introduction to Combinatorial Analysis WileyPublications in Mathematical Statistics John Wiley amp SonsNew York NY USA 1958

[2] M Bona Combinatorics of Permutations Discrete Mathematicsand its Applications Chapman amp HallCRC Boca Raton FlaUSA 2004

[3] L Carlitz ldquo119902-Bernoulli and Eulerian numbersrdquo Transactions ofthe American Mathematical Society vol 76 pp 332ndash350 1954

[4] L Carlitz ldquoA combinatorial property of 119902-Eulerian numbersrdquoThe American Mathematical Monthly vol 82 pp 51ndash54 1975

[5] T Xiong H P Tsao and J I Hall ldquoGeneral Eulerian numbersand Eulerian polynomialsrdquo Journal of Mathematics vol 2013Article ID 629132 9 pages 2013

[6] R P Stanley Enumerative Combinatorics vol 1 of CambridgeStudies in Advanced Mathematics Cambridge University PressCambridge UK 1996

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

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Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of