225

Зборник на трудови од III конгрес на математичарите на Македонија Струга, Македонија, 29.IX-2.X-2005

Стр. 225-238

Proceedings of III congress of mathematicians of Macedonia

Struga, Macedonia, 29.IX-2.X-2005 Pages 225-238

QUASICOMPOSITIONS OF NATURAL NUMBER

Adi Dani1

In this paper number 0 is included in set of natural numbers. By

{ }0,1,..., 1mI m= − is denoted the set of all natural numbers lesser than given natural number m. Definition 1. Each m sequence ( )0 1 1 1

, ,..., m kc c c c − +

= with terms in set

{ }1 0,1,...,kI k+ = with condition 0 1 1... mc c c k−+ + + = is called quasicomposition of natural number k in m parts. Theorem 1. Each m quasicomposition of number k defines a k nondecreasing (or nonincreasing) sequence with terms in mI end vice versa. Proof. Show the biective correspondence below

( )0 11

0 1 1 1, ,..., 0,0,...0,1,1,...,1,..., 1, 1,..., 1

m

m kc c c

m

c c c m m m−

− +

⎛ ⎞⎜ ⎟↔ − − −⎜ ⎟⎝ ⎠

Theorem 2. Each nondecreasing k sequence with terms in Im defines a k subset of 1m kI + − and vice versa.

Proof. Let be ( )0 1 1, ,..., k mv v v v −= a nondecreasing k sequence with terms in mI

following definition we have ( )

( )1

, 0 1 ( 1)

(0 2)k i m i i

i i m k

i I v I v m i v i m i

v i m k v i I + −

∀ ∈ ∈ ⇒ ≤ ≤ − ⇒ ≤ + ≤ + − ⇒

⇒ ≤ + ≤ + − ⇒ + ∈

that mean sequence v define (determine) set ( ) { } 1:i k m kI v v i i I I + −= + ∈ ⊆

clearly ( )I v k=

because( ), , , 0,k i j i j i j i ji j I i j v v j i v v j i v i v j v i v j∀ ∈ < ≤ ⇒ − > < + − ⇒ + < + ⇒ + ≠ +

Conversely: Let be { }0 1 1 1, ,..., k m kB b b b I− + −= ⊆ a k set then we can suppose that

( )0 1 10 ... 2 1 1kb b b m k m k−≤ < < < ≤ + − = + − −

then 1 0 1k i i i mi I i b m i b i m b i I∀ ∈ ⇒ ≤ ≤ + − ⇒ ≤ − ≤ − ⇒ − ∈

A. Dani

226

1 1 1 1i i i i i ib b b i b i b i b i+ + +< ⇒ − < − ⇒ − ≤ − − that mean set B defines a nondecreasing k sequence

( ) ( )( )0 1 10, 1,..., 1k mv B b b b k−= − − − −

who’s terms are in set mI .

Denote by ( )mq k the number of all quasicompositions of number k in m-parts.

According to theorem 1 and theorem 2 that number is equal at the number 1m k

k+ −⎛ ⎞

⎜ ⎟⎝ ⎠

of all k subsets of set 1m kI + − . We can resume

[1] ( )0 1 1 0 1 1 0 1 1... 0 ... 1 0 ... 2

11 1 1

m k k

mc c c k v v v m b b b m k

m kq k

k− − −+ + = ≤ ≤ ≤ ≤ ≤ − ≤ < < < ≤ + −

+ −⎛ ⎞= = = = ⎜ ⎟

⎝ ⎠∑ ∑ ∑

for example.

( )3

3 4 1 6 6 6 54 154 4 2 2 1

q+ −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⋅= = = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⋅⎝ ⎠ ⎝ ⎠ ⎝ ⎠

that mean there exists exactly 15 • Quasicompositions of number 4 in 3 parts • No decreasing 4 sequences with terms in 3I • 4 subsets of set 6I

In extension, we present all those configurations by colons ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

5 3 65 3 65 3 65 3 65 3 6

0,0, 4 2,2,2,2 2,3,4,50,1,3 1,2,2,2 1,3,4,50,2,2 1,1,2,2 1,2,4,50,3,1 1,1,1,2 1,2,3,50,4,0 1,1,1,1 1,2,3,4

↔ ↔↔ ↔↔ ↔↔ ↔↔ ↔

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

5 3 6

5 3 6

5 3 6

5 3 6

5 3 6

1,0,3 0,2,2,2 0,3,4,5

1,1,2 0,1,2,2 0, 2,4,5

1,2,1 0,1,1,2 0,2,3,5

1,3,0 0,1,1,1 0, 2,3,4

2,0, 2 0,0,2,2 0,1,4,5

↔ ↔

↔ ↔

↔ ↔

↔ ↔

↔ ↔

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

5 3 6

5 3 6

5 3 6

5 3 6

5 3 6

2,1,1 0,0,1,2 0,1,3,5

2,2,0 0,0,1,1 0,1,3,4

3,0,1 0,0,0,2 0,1,2,5

3,1,0 0,0,0,1 0,1,2,4

4,0,0 0,0,0,0 0,1,2,3

↔ ↔

↔ ↔

↔ ↔

↔ ↔

↔ ↔

Quasicompositions of natural number

227

Each quasicomposition ( )0 1 1 1, ,..., m k

c c c c − += of number k in m parts explicitly we can

present as array of boxes with common borders with objects placed in boxes in i-th box we put ic objects (dots) mi I∀ ∈

0 1 1

... ... ... ...mc c c −

ii i ii i ii i

That scheme of boxes we call putting of k similar objects in m different boxes. For example

( )41, 2,0,3 ↔ i ii iii

Theorem 3. The number ( )mq k satisfy the following relations

( )( )( ) ( ) ( )

0

1

1. 0

2. 0 1, 0

3. 1 , 0, 0m

m m m

q k

q m

q k q k q k m k−

=

= >

= + − > >

Proof. The first two equations are evident, third equation is true because the set of all quasicompositions of k in m parts can be parted in two disjoint set. One set contain all quasicompositions whose first term (part) is 0 and the other set contain the rest of quasicompositions. Each quasicomposition of first set defines a quasicomposition of k in 1m − parts; in other hand, each quasicomposition of second set defines a quasicomposition of 1k − in m parts. The numbers ( )mq k can be placed in an infinite table with rows m and columns k

• All elements of first row 0m = in accordance with 1. are 0 • First element of first column k = 0 is 0 others elements in accordance with

2. are 1 • Every element of table that is not in first column or in first row in

accordance with recurrence 3. is a sum of his neighbor on left with his neighbor up to him. \ 0 1 2 3 4 5 6 7 8 9 .0 0 0 0 0 0 0 0 0 0 0 .1 1 1 1 1 1 1 1 1 1 1 .2 1 2 3 4 5 6 7 8 9 10 .3 1 3 6 10 15 21 28 36 45 55 .4 1 4 10 20 35 56 84 120 165 220 .5 1 5 15 35 70 126 210 330 495 715 .6 1 6 21 56 126 252 462 792 1287 2002 .7 1 7 28 84 210 462 924 1716 3003 5005 .8 1 8 36 120 330 792 1716 3432 643

m k

5 11440 .9 1 9 45 165 495 1287 3003 6435 12870 24310 .. . . . . . . . . . . .

Table 1

A. Dani

228

Definition. The formal power series ( ) ( )0

km m

kq x q k x

∞

=

=∑

is called generating function of number of quasicompositions of number k in m parts. From

[2] ( )0 1 1...

11

m

mc c c k

m kq k

k−+ + + =

+ −⎛ ⎞= = ⎜ ⎟

⎝ ⎠∑

we have

[3] ( ) ( )0 11

0 1 1 0 1 10 0 ... 0 0 0

... 1m

m m

mc cck km

k k c c c k c c c

q k x x x x x x−

− −

∞ ∞ ∞ ∞ ∞−

= = + + + = = = =

= = = −∑ ∑ ∑ ∑ ∑ ∑

Conclusion: if 1x < then

[4] ( )0

11 m k

k

m kx x

k

∞−

=

+ −⎛ ⎞− = ⋅⎜ ⎟

⎝ ⎠∑

Definition. Let be ( )0 1 1 1, ,..., m k

c c c c − += a quasicomposition of k in m parts then

vector ( ) ( )0 1 1

, ,..., k mg c t t t

+= where 1,i kt i I +∀ ∈ express the number of coordinates of c

those are equal at i, is called trace of quasicomposition c. Obviously the trace satisfy the conditions

1 2

0 1

1 2 ......

k

k

t t kt kt t t m

+ + + =+ + + =

Denote by ( )0 1 1, ,..., k m

P t t t+

the set of all quasicompositions c of number k in m parts

with same trace ( ) ( )0 1 1, ,..., k m

g c t t t+

= and let be ( )0 1 1 1, ,..., k m

c P t t t − +∈ then

,i kt i I∀ ∈ is the number of coordinates of c who’s are equal at i, if we rearrange them in !it ways then all m coordinates of c can be arranged in 0 1! !... !kt t t ways. All of !m

possible arrangements of m coordinates of quasicompositions of ( )0 1 1, ,..., k m

P t t t+

we

can get, if that process repeat for each element of set ( )0 1 1, ,..., k m

P t t t+

that give equation

( )0 1 0 11, ,..., ! !... ! !k km

P t t t t t t m+

= or ( )0 1 10 1

!, ,...,! !... !k m

k

mP t t tt t t+

=

finally if this process apply over all traces of quasicompositions of number k in m parts we get the formula

[5] ( )1 2

0 1

1 2 ... 0 1 ...

!! !... !

kk

mt t kt k kt t t m

mq kt t t+ + + =

+ + + =

= ∑

Quasicompositions of natural number

229

Quasicompositions of natural number with upper homogenous restrictions Definition. Each quasicomposition ( )0 1 1 1

, ,..., m kr r r − +

of k in m parts whose terms

fulfill the condition ,i mr s i I≤ ∀ ∈ is called quasicomposition of number k in m parts no greater than s or simply m quasicomposition of k of order s. Each m quasicomposition of k of order s define a putting of k similar objects in m different boxes and in each box we can put at most s objects Denote by

[[6]] 0 1 1...

,

1m

i m

r r r ksr s i I

mk

−+ + + =≤ ∀ ∈

⎛ ⎞=⎜ ⎟

⎝ ⎠∑

number of all m quasicomposition of k of order s, then we can prove the following equations

[1s] 0

10 s

⎛ ⎞=⎜ ⎟

⎝ ⎠

[2s] 0,s

mk ms

k⎛ ⎞

= >⎜ ⎟⎝ ⎠

[3s] min ,

0

1k s

is s

m mk k i=

−⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠∑

[4s] s s

m mk ms k

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

Proof.

0 1 1... 0 0 ,

1 10

mi m

c c csc s i I

m

−+ + + =≤ ≤ ∀ ∈

⎛ ⎞= =⎜ ⎟

⎝ ⎠∑

0 1 1... 0 ,

1 0,m

i m

c c c ksc s i I

mk ms

k−+ + + =

≤ ≤ ∀ ∈

⎛ ⎞= = >⎜ ⎟

⎝ ⎠∑

0 1 1 1 0 1 2 1 0 1 2 0 1 21 1

... 0 ... 0 ... 1 ... 0 , 0 , 0 , 0 ,

1 1 1 1

1

m m m m m mi m i m i m i m

s k s

c c c k c c c c k c i c c c k i i k c c c k isc s i I c s i I c s i I c s i I

mk

mk i

− − − − − −− −

+ + + = = + + + = − = + + + = − = + + + + = −≤ ≤ ∀ ∈ ≤ ≤ ∀ ∈ ≤ ≤ ∀ ∈ ≤ ≤ ∀ ∈

⎛ ⎞= = = + =⎜ ⎟

⎝ ⎠

−⎛=

−⎝

∑ ∑ ∑ ∑ ∑ ∑ ∑

0,0

k

i s

k s=

⎞< <⎜ ⎟

⎠∑

A. Dani

230

0 1 1 1 0 1 2 1 0 1 21

... 0 ... 0 ... 0 , 0 , 0 ,

0

1 1 1

1 ,

m m m m mi m i m i m

s s

c c c k c c c c k c i c c c k isc s i I c s i I c s i I

s

i s

mk

ms k ms

k i

− − − − −−

+ + + = = + + + = − = + + + = −≤ ≤ ∀ ∈ ≤ ≤ ∀ ∈ ≤ ≤ ∀ ∈

=

⎛ ⎞= = = =⎜ ⎟

⎝ ⎠

−⎛ ⎞= < ≤⎜ ⎟−⎝ ⎠

∑ ∑ ∑ ∑ ∑

∑

0 1 1 0 1 1 0 1 1... ( ) ( ) ... ( ) ... 0 , 0 , 0 ,

1 1 1

m m mi m i m i m

c c c k s c s c s c ms k t t t ms ksc s i I s c s i I t s i I

s

mk

mms k

− − −+ + + = − + − + + − = − + + + = −≤ ≤ ∀ ∈ ≤ − ≤ ∀ ∈ ≤ ≤ ∀ ∈

⎛ ⎞= = = =⎜ ⎟

⎝ ⎠

⎛ ⎞= ⎜ ⎟−⎝ ⎠

∑ ∑ ∑

Definition. Formal potential series

( ),0 0

msk k

m sk ks s

m mq x x x

k k

∞

= =

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠∑ ∑

is generating function of numbers of m quasicompositions of order s considering [6] we conclude

[7] ( )0 11

0 1 1 0 1 10 0 ... 0 0 0 ,

... 1 ...m

m mi m

ms ms s s s mc cck k s

k k c c c k c c csc s i I

mx x x x x x x

k−

− −= = + + + = = = =≤ ∀ ∈

⎛ ⎞= = = + + +⎜ ⎟

⎝ ⎠∑ ∑ ∑ ∑ ∑ ∑

[7a] 1

0

1 , 11

msmsk

k s

m xx xk x

+

=

⎛ ⎞⎛ ⎞ −= ≠⎜ ⎟⎜ ⎟ −⎝ ⎠ ⎝ ⎠∑

that generating function is polynomial, the numbers s

mk

⎛ ⎞⎜ ⎟⎝ ⎠

are called polynomial

coefficients and [7] is called polynomial formula. If in [7] we put 1x = that give

[8] ( )0

1ms

m

k s

ms

k=

⎛ ⎞= +⎜ ⎟

⎝ ⎠∑

for 1x = − from [7a] we have

[8a] ( ) ( )( ) ( )

1

0

1 01 1

1 1 11 1 02

msmsk s

k s

mmk m

+

=

=⎧⎛ ⎞− −⎛ ⎞ ⎪− = =⎜ ⎟ ⎨⎜ ⎟ + −⎜ ⎟− − >⎝ ⎠ ⎪⎝ ⎠ ⎩

∑

Polynomial Theorem

If in [5] we put 1 2 ... 0s s kt t t+ += = = = then we get an other formula for counting the quasicompositions of order s

Quasicompositions of natural number

231

[9] ( )1 2

0 1

1 2 ... 0 1...

!! !... !

ss

sm

t t st k sst t t m

m mq kk t t t+ + + =

+ + + =

⎛ ⎞= =⎜ ⎟

⎝ ⎠∑

that fact lead to

[10]

( ) ( ) ( )

1 2

1 2 0 10 1

0 1

0 1

2 ...

0 0 2 ... ...0 1 0 1...

... 0 1

! !! ! ... ! ! ! ... !

! 1 ...! ! ... !

s

s ss

s

s

ms msr r srk k

k k r r sr k r r r ms ssr r r m

rr r s

r r r m s

m m mx x xk r r r r r r

m x xr r r

+ + +

= = + + + = + + + =+ + + =

+ + + =

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

=

∑ ∑ ∑ ∑

∑

from [7] and [10] we get

[11] ( ) ( ) ( ) ( )0 1

0 1 ... 0 1

!1 ... 1 ...! ! ... !

s

s

m rr rs s

r r r m s

mx x x xr r r+ + + =

+ + + = ∑

if in [11] we put 10

,i is

ax i Ia += ∈ then we get the list of equations

0 1

0 1

01 1

...0 0 0 1 0 0 0

!1 ... ...! !... !

s

s

m r r r

s s

r r r m s

a a aa ama a r r r a a a+ + + =

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠∑

0 1

0 10 1

0 1 0 1...

...0 0 1 0

... ...!! !... !

s

ss

m r rrs s

r r rr r r m s

a a a a a ama r r r a + + +

+ + + =

⎛ ⎞+ + + =⎜ ⎟⎝ ⎠

∑

( ) 0 1

0 1

0 1 0 1

...0 0 1 0

... ...!! !... !

s

s

m r rrs s

m mr r r m s

a a a a a ama r r r a+ + + =

+ + += ∑

[12] ( ) 0 1

0 1

0 1 0 1... 0 1

!... ...! !... !

s

s

m r rrs s

r r r m s

ma a a a a ar r r+ + + =

+ + + = ∑

the last equation is called polynomial theorem If in [9] we put 1s = then

( ) ( )1 10 1

1 0 1 1 11

! ! !! ! ! ! ! !t k t k

t t m

m mm m mk kt t m t t m k k= =

+ =

⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠∑ ∑

If in formula, [7] put 1s = that give Newton binomial formula

[13] ( )0

1m

mk

k

mx x

k=

⎛ ⎞= +⎜ ⎟

⎝ ⎠∑

Now in [13] we suppose that axb

= then we get binomial theorem

[13a] ( )0

mm k m k

k

ma b a b

k−

=

⎛ ⎞+ = ⎜ ⎟

⎝ ⎠∑

A. Dani

232

Generalized Fibonacci sequence Let be s ∈ a fixed natural number then infinite sequence

( )(0), (1),..., ( ),...s s s sF F F F m=

who fulfill the conditions ( )0 0,F m m= ∈ for 0s >

( )

( ) ( )

( ) ( )

1

0

1

0

0 1

1 ,

1 ,

s

m

s si

s

s si

F

F m F m i m s

F m F m i m s

−

=

−

=

=

= − − <

= − − ≥

∑

∑

is called Fibonacci sequence of order s. This definition implies that the Fibonacci sequences of order 0 and 1 are constant all those terms are 0 respectively 1.

( ) ( )0 10, 1,F m F m m N= = ∀ ∈

Fibonacci sequence of order 2s = we denote by 2F F= is defined by

( ) ( ) ( ) ( ) ( )0 1 1 ; 1 2 , 2F F F m F m F m m= = = − + − ≥ this is well known usual Fibonacci sequence. In table below are listed the ten first terms of Fibonacci sequences of order no greater than 4

0

1

2

3

4

0 1 2 3 4 5 6 7 8 9 .0 0 0 0 0 0 0 0 0 0 .1 1 1 1 1 1 1 1 1 1 .1 1 2 3 5 8 13 21 34 55 .1 1 2 4 7 13 24 44 81 149 .1 1 2 4 8 15 29 56 108 208 .

. . . . . . . . . . . .

FFFFF

Definition. Formal power series ( ) ( )0

ms s

m

F x F m x∞

=

=∑

is generating function of Fibonacci sequence of order s. From definitions we have

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( )( )

0

0 1 1

11 1 1

1 1 0 0 0

1 1 11 1 1

0 0 0 0

0 1

1 1 1 1 1

1 1 1

1 ...

m m ms s s s s

m m m

sm m m

s s sm m m i

s s sm i i m i i i

s s si m i m i i

ss

F x F m x F x F m x F m x

F m x F m x F m i x

F m i x x F m i x x F x x

F x x x

∞ ∞ ∞

= = =

∞ ∞ ∞ −+ + +

+ = = = =

− ∞ − ∞ −− + − + +

= = = = =

= = + = + =

= + + = + + = + − =

⎛ ⎞= + − = + − = + =⎜ ⎟⎝ ⎠

= + + +

∑ ∑ ∑

∑ ∑ ∑∑

∑∑ ∑ ∑ ∑

Quasicompositions of natural number

233

Solve this equation by ( )sF x we have

( ) 20

11 ( ... )

ms s

mF m x

x x x

∞

=

=− + + +∑

Considering [7] then we have

( ) ( ) ( )

( )

22

0 0

1

0 0 0 0 01 1

0 0 1

1 ... 1 ...1 ...

1 ...

,

mm s ss s

m m

mm s m i m i

m m i m is s

m

m i s

F m x x x x x xx x x

m mx x x x x x

i i

m i m i mx

i m m

∞ ∞

= =

∞ ∞ ∞ ∞ ∞− +

= = = = =− −

∞ ∞

= = −

= = + + < = + + + =− + + +

⎛ ⎞ ⎛ ⎞= + + + = = =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠− + →⎛ ⎞

= ⎜ ⎟ → −⎝ ⎠

∑ ∑

∑ ∑ ∑ ∑∑

∑∑ i

equalizing the coefficients of x with the same power, we get

[14] ( )0 01 1

si is s

m i m iF m

i i

∞

= ≥− −

− −⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠∑ ∑

Identity of Vandermonde

From polynomial formula ( )

( ) ( ) ( )0 1

0 1 0 1

0 0 01 1 10 0 1 0 11

0 1 0 1 0 1

0 1

0

0 0 01 1 1

0 0 0 0 0 00 0 01 1 1

1 ... 1 ... 1 ...a a s

a a a ak s s s

k s

a s a s a sa s a s a st t t t tt

t t t t t tss s s

a ax x x x x x x

k

a a aa a ax x x x

t t tt t t

++

=

+ +

= = = = = =

+⎛ ⎞= + + + = + + + + + + =⎜ ⎟

⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∑

∑ ∑ ∑ ∑ ∑∑( )0 1

0 1

0 1

0 0 1

a a sk

k t t k ss

a ax

t t

+

= + =

=

⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠∑ ∑

from above we gate the relation so called Vandermonde identity of order s

0 1 0

0 0 110 1

00 0 01

k

t t k ts ss s s

a a aaa at t k ttk + = =

+ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ −⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠∑ ∑

denote 0 1 0 1, ka a m k ms t i I t k i= = = = ∈ ⇒ = − then considering [4.s] the last equation turn into

[15] 2

0 0 0

2 ms ms ms

i i is s s s s s

m m m m m mms i ms i i i i= = =

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠∑ ∑ ∑

in the same way we can prove that

0 1 1

0 110 1 1

... 0 11

......

m

mm

t t t k ms ss s

a aaa a at ttk

−

−−

+ + + = −

+ + + ⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠∑

that is called generalized Vandermonde identity.

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234

Generalized Pascal triangle An infinite table with rows m and columns k that in meting point of row m with column

k is placed number s

mk

⎛ ⎞⎜ ⎟⎝ ⎠

is called Pascal triangle of order s

0 10 0 0

00 1

1 1 11

0 1

0 1

s s s

s s s

s s s

k

k

k

m m mm

k

⋅ ⋅

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠⋅ ⋅ ⋅ ⋅ ⋅ ⋅

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠⋅ ⋅ ⋅ ⋅ ⋅ ⋅

Pascal triangle of order s

Basic properties of this table are. • First row 0m = begin with 1 from [1s] all other elements are 0 from [2s] • Each element who isn’t in first row following [3s] is the sum of his neighbor

up to him and min ,k s neighbors who’s are placed on left in the same row. Pascal triangle can be constructed row after row recursively. Below we make mention at 4 important properties of Pascal triangle of order s

1. The sum of elements in row m is equal at ( )1 ms + (from [8])

2. Alternative sum in row m is ( )1 12

ms⎛ ⎞+ −⎜ ⎟⎜ ⎟⎝ ⎠

(from [8a])

3. sum of squares of elements in row m is equal at 2

s

mms

⎛ ⎞⎜ ⎟⎝ ⎠

(from [14])

4. sum of elements in m –th small diagonal is equal at ( )1sF m+ (from [15]) For 0s = we have Pascal triangle of order 0 this triangle not has interesting properties, all elements of first column are 1 and all others are equal at 0. In the tables below we present the Pascal triangles of order, 1s = that is usual Pascal triangle, and Pascal triangles of orders 2s = and 3s = .

Quasicompositions of natural number

235

0 3 5 6 7 8 91 2 40 11 1 12 1 2 13 3 31 1

64 1 4 4 15 5 10 10 51 16 6 15 20 15 61 17 7 35 35 71 21 21 18 8 28 56 70 56 28 81 19 9 36 84 126 126 84 36 91 1

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

Pascal triangle of order 1

0 3 5 6 7 8 91 2 40 11 1 1 1

32 1 2 2 13 3 6 7 6 31 1

10 16 19 16 104 1 4 4 15 5 15 30 45 51 45 30 15 516 6 50 90 126 126 90 501 21 1417 7 28 77 161 266 357 393 357 26618 8 36 266 560 784 1016 1107 10161 1129 9 45 156 438 990 1610 1960 2907 31391

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ Pascal triangle of order 2

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236

0 3 5 6 7 8 9 .1 2 40 .1

.1 1 1 1 13 3 .2 1 2 4 2 1

3 3 6 10 10 6 3 .1 12 12 110 20 31 40 40 31 20 .4 1 4 44

5 5 15 35 65 101 135 155 155 135 .16 6 56 120 216 336 456 546 5801 217 7 28 84 203 413 728 1128 1554 1918 .18 8 36 120 322 728 1428 2472 3823 5328 .19 9 45 165 486 1206 2598 4950 81

⋅

451 13051 .. . . . . . . . . . . .

Pascal triangle of order 3 Theorem From [7], [13], and [5] we can conclude that number of quasicompositions of k in m parts of order s have the following properties

[16] 0 1

m

is s

m m ik i k i= −

⎛ ⎞ ⎛ ⎞⎛ ⎞=⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠⎝ ⎠∑

[17] 0 1

m

is s

m m m ik i k si= −

−⎛ ⎞ ⎛ ⎞⎛ ⎞=⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠⎝ ⎠∑

[18] ( ) ( )0

1 11

1

mi

is

m m m k s ik i m=

⎛ + − + − ⎞⎛ ⎞ ⎛ ⎞= − ⎜ ⎟⎜ ⎟ ⎜ ⎟ −⎝ ⎠ ⎝ ⎠⎝ ⎠∑

Proof: From

( ) ( )( )

( ) ( )

2

0

2 1

0 0

0 0 0 01 1

1 ... 1 ...

... 1 ...

mmk s s

k s

m mi is i s

i i

m mi j k

i j k is s

mx x x x x x

k

m mx x x x x x

i i

m i m ix x

i j i k i

∞

=

−

= =

∞ ∞+

= = = =− −

⎛ ⎞= + + + = + + + + =⎜ ⎟

⎝ ⎠

⎛ ⎞ ⎛ ⎞= + + + = + + + =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞

= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

∑

∑ ∑

∑ ∑ ∑∑

we get [16], similarly from

Quasicompositions of natural number

237

( ) ( )( )

( )

1

0

1

0 0 0 1

0 0 0 01 1

1 ... 1 ...

1 ...

mmk s s s

k s

m mm is si si j

i i j s

m msi j k

i j k is s

mx x x x x x

k

m m m ix x x x x

i i j

m m i m m ix x

i j i k si

∞−

=

∞−−

= = = −

∞ ∞+

= = = =− −

⎛ ⎞= + + + = + + + + =⎜ ⎟

⎝ ⎠−⎛ ⎞ ⎛ ⎞ ⎛ ⎞

= + + + = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

∑

∑ ∑ ∑

∑ ∑ ∑∑

we get [17] and finally from

( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( )( ) ( )

11

0

11

0 0 0 0

0 0

1 1 11

1 11

1 1

1 11 1

1

ms m mk s

k s

m mi i j s is j

i j i j

mi ik

i k

m xx x xk x

m m j m m jx x x

i j i j

j s i k j k s i

m k s im mx

k s ii i

+∞−+

=

∞ ∞+ ++

= = = =

∞

= =

⎛ ⎞⎛ ⎞ −= = − − =⎜ ⎟⎜ ⎟ −⎝ ⎠ ⎝ ⎠+ − + −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞

− = − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠+ + = ⇒ = − +

+ − + −⎛ ⎞⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟⎜ ⎟ ⎜− +⎝ ⎠ ⎝⎝ ⎠

∑

∑ ∑ ∑ ∑

∑ ∑ ( )0 0

1 11

mk

k i

m k s ix

m

∞

= =

⎛ + − + − ⎞⎜ ⎟⎟ −⎠⎝ ⎠

∑∑we get [18] Example 1. Count the number of quasicompositions of 4 in 4 parts of order 2 a) From [9] for 2s = we have

( )1 2 1 20 1 2

1 2 1 20 1 2 1 2 1 22

! !! ! ! ! ! !t t k t t k

t t t m

m m mk t t t m t t t t+ = + =

+ + =

⎛ ⎞= =⎜ ⎟ − −⎝ ⎠∑ ∑

( )

( ) ( ) ( )

1 2

1 2

1 21 2 4 1 2 1 22

1 2

1 4 2 0 4; 4, 04 4! 1 2 2 1 4; 2, 14 4 ! ! !

1 0 2 2 4; 0, 24! 4! 4! 19

4 4 0 !4!0! 4 2 1 !2!1! 4 0 2 !0!2!

t t

t tt t

t t t tt t+ =

⋅ + ⋅ = = =⎛ ⎞

= = ⋅ + ⋅ = = = =⎜ ⎟ − −⎝ ⎠ ⋅ + ⋅ = = =

= + + =− − − − − −

∑

b) from [16],[17] and [18] for 2s = we have respectively

02

m

i

m m ik i k i=

⎛ ⎞ ⎛ ⎞⎛ ⎞=⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠⎝ ⎠∑

02 2

m

i

m m m ik i k i=

−⎛ ⎞ ⎛ ⎞⎛ ⎞=⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠⎝ ⎠∑

( )02

3 11

1

mi

i

m m m k ik i m=

+ − −⎛ ⎞ ⎛ ⎞⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠⎝ ⎠∑

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238

that give 4

02

4 4 4 2 4 3 4 46 1 4 3 1 1 6 12 1 19

4 4 2 2 3 1 4 0i

ii i=

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= = + + = ⋅ + ⋅ + ⋅ = + + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠∑

4

02

4 4 4 4 4 4 3 4 21 4 3 6 1 1 12 6 19

4 4 2 0 4 1 2 2 0i

ii i=

−⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= = + + = + ⋅ + ⋅ = + + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠∑

( )4

02

4 4 7 3 4 7 4 41 35 16 19

4 3 0 3 1 3i

i

ii=

−⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= − = − = − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠∑

as is shown above counting with formula [18] is more effective. In continuation we give the list of all 19 quasicompositions of number 4 in 4 parts of order 2

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

0,0,2,2 , 0, 2,0, 2 , 0, 2, 2,0 , 2,0,0,2 , 2,0,2,0 , 2,2,0,0

0,1,1, 2 , 0,1, 2,1 , 0, 2,1,1 , 1,0,1,2 , 1,0,2,1 , 1,1,0,2 , 1,1,2,0

1,2,0,1 , 1, 2,1,0 , 2,0,1,1 , 2,1,0,1 , 2,1,1,0 , 1,1,1,1

Example 2. From Pascal triangle of order 3 we can see that in meting point of row 3 with column 5 is placed number 12, that means number of quasicompositions of 5 in 3 parts of order 3 is 12. That number also we can get from [18]

( )3

03

3 3 7 4 3 7 3 31 21 9 12

5 2 0 2 1 2i

i

ii=

−⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= − = − = − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠∑

All 12 quasicompositions of 5 in 3 parts of order 3 are ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

0,2,3 , 0,3,2 , 2,0,3 , 2,3,0 , 3,0,2 , 3,2,0

1,1,3 , 1,3,1 , 3,1,1

1,2,2 , 2,1, 2 , 2, 2,1

LITERATURE

[1] Сачков В.Н., Введение в комбинаторньiе методьi дискретной матема-тики, Мockвa 1982 (384 p.) [2] I.P.Goulden D.M.Jackson, Combinatorial enumeration, ( пeрeчиcлиteльhaя komбиhatopиka) in russian Mockвa 1990 (504 p.) [3] С.В. Яaблoнсkий, Введение в дискретную математику, Mockвa 1986 (384 p.) [4] Б.А.Бoндаpeнko, Обoбщehыe tpeугoльники и пиpamиди пackaля, Тaшkeнt 1990 (192 p.) [5] B.И.Бapaнob Б.C.Cтeчkин, Эkctpemaльhыe komбиhatophыe зaдaчи и иx пpилoжehия, Мockвa 1989 (160 p.) [6] D. Cvetković, Diskretne matematičke strukture,,beograd 1978 (152 p.) [7] H. Wilff, Generatingfunctionology, Philadelphia 1994 (226 p.)

1Secondary school ,,Niko Nestor,, 6330 Struga, Macedonia e-mail: inadida@yahoo.com