refinement of some partition inequalities partitions, the partition counting function restricted...

Refinement of Some Partition Inequalities

James Mc Laughlin

West Chester Universityjmclaughl@wcupa.edu

http://math.wcupa.edu/mclaughlin/

West Coast Number TheoryPacific Grove

December 18, 2015

Outline

q-series NotationInteger Partitions, The Partition Counting FunctionRestricted Partition FunctionsFerrers Diagram, Durfee SquarePartition Generating FunctionsPartition InequalitiesPartition Generating Functions that Track the Number ofPartsSome ExperimentationResultsConcluding Remarks

q-series Notation

q-products: (a;q)0 := 1 and for n 1,

(a;q)n := (1 a)(1 aq) (1 aqn1)(q;q)n := (1 q)(1 q2) (1 qn) ()

(a1, . . . ,aj ;q)n := (a1;q)n (aj ;q)n(a;q) := (1 a)(1 aq)(1 aq2)

(a1, . . . ,aj ;q) := (a1;q) (aj ;q)

The q-binomial theorem: if |z|, |q| < 1, then

n=0

(a;q)n(q;q)n

zn =(az;q)(z;q)

. (1)

Special Cases of the q-binomial theorem

Special Cases of the q-binomial theorem:

n=0

zn

(q;q)n=

1(z;q)

, |z| < 1, |q| < 1. (2)

n=0

(a)nqn(n1)/2

(q;q)n= (a;q), |q| < 1. (3)

Integer Partitions

Definition: A partition of a positive integer n is a way of writingn as a sum of positive integers, where order does not matter.

Example. The partitions of 5 are

54 + 13 + 23 + 1 + 12 + 2 + 12 + 1 + 1 + 11 + 1 + 1 + 1 + 1

The summands of a partition are called parts of the partition.

The number of partitions of n is given by the partition functionp(n).For example, p(5) = 7.

Restricted Partition Functions, I

Some well known examples of restricted partition functions arepO(n), the number of partitions of n into odd parts, and pD(n),the number of partitions of n into distinct parts.

pO(5) = 3 (5, 3 + 1 + 1, 1 + 1 + 1 + 1 + 1),pD(5) = 3 (5, 4 + 1, 3 + 2).

(PO(n) = PD(n), n N)

Restricted Partition Functions, IILet p2,3,5(n) denote the number of partitions of n into parts 2,3( mod 5), andP(n) denote the number of partitions of n where each partfrom 1 to the largest part occurs at least twice.

p2,3,5(10) = 4 2 + 2 + 2 + 2 + 23 + 3 + 2 + 27 + 38 + 2

P(10) = 4 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 12 + 2 + 1 + 1 + 1 + 1 + 1 + 12 + 2 + 2 + 1 + 1 + 1 + 12 + 2 + 2 + 2 + 1 + 1

(P2,3,5(n) = P(n), n N)

Ferrers Diagram, Durfee Square

Partition Generating Functions, ILet S be any set of positive integers, finite or infinite. Then thegenerating function for pS(n), the number of partitions of thepositive integer n with parts from S is

n=0

pS(n)qn =1

aiS 1 qai

= (1 + qa1 + q2a1 + q3a1 + . . . )

(1 + qa2 + q2a2 + q3a2 + . . . ) (1 + qa3 + q2a3 + q3a3 + . . . ) . . .

The generating function for pS(n), the number of partitions ofthe positive integer n with distinct parts from S is

n=0

pS(n)qn =

aiS

1 + qai

= (1 + qa1)(1 + qa2)(1 + qa3) . . .

Partition Generating Functions, II

Recall that p(n) is the number of (unrestricted) partitions of n.

n=0

p(n)qn =1

k=1 1 qk=

1(q;q)

=

k=0

qk

(q;q)k

=

k=0

qk2

(q;q)2k

= 1 +

k=1

qk

(qk ;q)= 1 +

1(q;q)

k=1

(q;q)k1qk

Partition Generating Functions, III

Recall that pD(n) is the number of partitions of n into distinctpositive integers.

n=0

pD(n)qn =

k=1

(1 + qk ) = (q;q)

= (1 + q)(1 + q2)(1 + q3) . . .

Partition Generating Functions, IV

Recall that p2,3,5(n) denote the number of partitions of n intoparts 2,3( mod 5).

n=0

p2,3,5(n)qn

=1

(1 q2)(1 q3)(1 q7)(1 q8)(1 q12)(1 q13) . . .

=1

(q2;q5)(q3;q5)=

1(q2,q3;q5)

Partition Inequalities, I

Fact: For each positive integer n,

p1,4,5(n) p2,3,5(n) 0.

Alternatively, if the sequence {cn} is defined by

n=0

cnqn =1

(q,q4;q5) 1

(q2,q3;q5),

then cn 0,n 0.

Partition Inequalities, II

Proof.By the Rogers-Ramanujan identities,

1(q,q4;q5)

1(q2,q3;q5)

=

k=0

qk2

(q;q)k

k=0

qk2+k

(q;q)k

=

k=1

qk2(1 qk )(q;q)k

=

k=1

qk2

(q;q)k1

Partition Inequalities, Variations and Extensions, I

Theorem (Berkovich and Garvan, 2005)

Suppose L > 0, and 1 < r < m 1. If the sequence {en} isdefined by

n=0

enqn =1

(q,qm1;qm)L 1

(qr ,qmr ;qm)L,

thenen 0, n 0 r - m r and m r - r .

Partition Inequalities, Variations and Extensions, II

Theorem (Andrews, 2011)

If L > 0, and the sequence {fn} is defined by

n=0

fnqn =1

(q,q5,q6;q8)L 1

(q2,q3,q7;q8)L,

thenfn 0,n 0.

Partition Inequalities, Variations and Extensions, III

Theorem (Berkovich and Grizzell, 2012)For any L > 0, and any odd y > 1, the q-series expansion of

1(q,qy+2,q2y ;q2y+2)L

1(q2,qy ,q2y+1;q2y+2)L

=

n=0

a(L, y ,n)qn

has only non-negative coefficients. Furthermore, the coefficienta(L, y ,n) is 0 if and only if either

n {2,4,6, . . . , y + 1} {y} or (L, y ,n) = (1,3,9).

Partition Inequalities, Variations and Extensions, IV

Theorem (Berkovich and Grizzell, 2012)For any L > 0, and any odd y > 1, and any x with1 < x y + 2, the q-series expansion of

1(q,qx ,q2y ;q2y+2)L

1(q2,qy ,q2y+1;q2y+2)L

=

n=0

a(L, x , y ,n)qn

has only non-negative coefficients. Furthermore, the coefficienta(L, y ,n) is 0 if and only if either . . . .

Partition Inequalities, Variations and Extensions, V

Theorem (Berkovich and Grizzell, 2013)

For any octuple of positive integers (L,m, x , y , z, r ,R, ), theq-series expansion of

1(qx ,qy ,qz ,qrx+Ry+z ;qm)L

1(qrx ,qRy ,qz ,qx+y+z ;qm)L

=

n=0

a(L, x , y , z, r ,R, ,n)qn

has only non-negative coefficients.

Partition Inequalities, Variations and Extensions, VI

Theorem (Berkovich and Grizzell, 2013)

For any positive integers m,n, y, and z, with gcd(n, y) = 1, andintegers K and L, with K L 0,

1(qz ;qm)K (qnyz ;qnm)L

1(qyz ;qm)K (qnz ;qnm)L

=

k=0

a(K ,L, x , y , z,n,m, k)qk

has only non-negative coefficients.

Partition Generating Functions that Track the Numberof Parts

Let S be any set of positive integers, finite or infinite. Then thegenerating function for pS(m,n), the number of partitions of thepositive integer n with exactly m parts from S is

n=0

pS(m,n)smqn =1

aiS 1 sqai

= (1 + sqa1 + s2q2a1 + s3q3a1 + . . . )

(1 + sqa2 + s2q2a2 + s3q3a2 + . . . ) (1 + sqa3 + s2q2a3 + s3q3a3 + . . . ) . . .

The generating function for pS(n), the number of partitions ofthe positive integer n with distinct parts from S is

n=0

pS(n)qn =

aiS

1 + sqai

= (1 + sqa1)(1 + sqa2)(1 + sqa3) . . .

Partition Inequalities that Track the Number of Parts

Q. If the polynomials {fn(s)} are defined by

n=0

fn(s)qn =1

(sq, sq4;q5) 1

(sq2, sq3;q5),

are there situations where the coefficients in fn(s) are allnon-negative?

Experimental Output

n fn(s)1 s

2 s + s2

3 s + s3

4 s s2 + s4

5 s5

6 s s2 + s6

7 s + s2 s3 + s4 + s7

8 s + s2 s4 + s5 + s8

9 s s2 + s6 + s9

10 s7 + s10

Experimental Output

11 s s2 + s3 s4 + s6 + s8 + s11

12 s + 2s2 s3 + s4 s5 + s7 + s9 + s12

13 s + s2 + s5 s6 + s7 + s8 + s10 + s13

14 s 2s2 + s3 + s6 s7 + s8 + s9 + s11 + s14

15 s7 + s9 + s10 + s12 + s15

16 s 2s2 + 2s3 s4 + s6 + s8 + s10 + s11 + s13 + s16

17 s + 2s2 3s3 + 3s4 s5 + s7 + 2s9 + s11 + s12

+ s14 + s17

18 s + 2s2 s3 + 2s5 2s6 + s7 + s8 + 2s10 + s12

+ s13 + s15 + s18

19 s 2s2 + 2s3 s4 + 2s6 s7 + s8 + s9 + s10 + 2s11

+ s13 + s14 + s16 + s19

20 s5 + s7 + s9 + s10 + s11 + 2s12 + s14 + s15 + s17 + s20

Experimentation, II

5 s5

10 s10 + s7

15 s15 + s12 + s10 + s9 + s7

20 s20 + s17 + s15 + s14 + 2s12 + s11 + s10 + s9 + s7 + s5

25 s25 + s22 + s20 + s19 + 2s17 + s16 + 2s15 + 2s14 + s13

+ 3s12 + s11 + 2s10 + 2s9 + 2s7 + s5

30 s30 + s27 + s25 + s24 + 2s22 + s21 + 2s20 + 2s19 + s18

+ 4s17 + 2s16 + 3s15 + 4s14 + s13 + 5s12 + 2s11 + 2s10

+ 3s9 + 3s7 + s5

35 s35 + s32 + s30 + s29 + 2s27 + s26 + 2s25 + 2s24 + s23

+ 4s22 + 2s21 + 4s20 + 5s19 + 2s18 + 7s17 + 4s16 + 5s15

+ 7s14 + 2s13 + 7s12 + 4s11 + 3s10 + 5s9 + 4s7 + s5

First Theorem

Theorem (Mc L. 2015)

Let M 5 be a positive integer, and let a and b be integerssuch that 1 a < b < M/2 and gcd(a,M) = gcd(b,M) = 1.Define the integers c(m,n) by

1(sqa, sqMa;qM)

1(sqb, sqMb;qM)

:=

m,n0c(m,n)smqn. (4)

(i) Then c(m,Mn) 0 for all integers m 0,n 0.(ii) If, in addition, M is even, then c(m,Mn + M/2) 0 for allintegers m 0,n 0.

Partitions Interpretation

Corollary

Let M, a and b be as in Theorem 7. Letpa,M,m(n) = # partitions of n into exactly m parts, each a( mod M),and letpb,M,m(n) = # partitions of n into exactly m parts, each b( mod M).Then

(i) pa,M,m(nM) pb,M,m(nM) for all integers n 1, and allintegers m, 1 m Mn.(ii) If M is even, then pa,M,m(nM + M/2) pb,M,m(nM + M/2)for all integers n 0, and integers m with 1 m Mn + M/2.

Proof of First Theorem

Proof.We recall a special case of the q-binomial theorem: