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− aqn−1)

(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)nzn =

(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(n−1)/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∏

ai∈S 1− qai

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

× (1 + qa2 + q2a2 + q3a2 + . . . )

× (1 + qa3 + q2a3 + q3a3 + . . . ) . . .

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

∞∑n=0

p∗S(n)qn =

∏ai∈S

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)k−1qk

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)k−1

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,qm−1;qm)L− 1

(qr ,qm−r ;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, ρ), theq-series expansion of

1(qx ,qy ,qz ,qrx+Ry+ρz ;qm)L

− 1(qrx ,qRy ,qρz ,qx+y+z ;qm)L

=∞∑

n=0

a(L, x , y , z, r ,R, ρ,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∏

ai∈S 1− sqai

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

× (1 + sqa2 + s2q2a2 + s3q3a2 + . . . )

× (1 + sqa3 + s2q2a3 + s3q3a3 + . . . ) . . .

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

∞∑n=0

p∗S(n)qn =

∏ai∈S

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, sqM−a;qM)∞

− 1(sqb, sqM−b;qM)∞

:=∑

m,n≥0

c(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:

1(z;q)∞

=∞∑

n=0

zn

(q;q)n. (5)

Hence

1(sqa, sqM−a;qM)∞

− 1(sqb, sqM−b;qM)∞

=∑

j,k≥0

sj+kqa(j−k)+kM

(qM ;qM)j(qM ;qM)k−

∑j,k≥0

sj+kqb(j−k)+kM

(qM ;qM)j(qM ;qM)k(6)

Set j + k =: m, so that j = m − k and the right side of (6)becomes

Proof, Continued

Proof Continued.∑m≥0

smm∑

k=0

qa(m−2k)+kM − qb(m−2k)+kM

(qM ;qM)m−k (qM ;qM)k(7)

Next, we restrict the values of k so that when the inner sum isexpanded as a power series, it contains only those powers of qwhose exponents are multiples of M(so that the series multiplying sm is

∑∞n=0 c(m,Mn)qMn).

Since gcd(a,M) = gcd(b,M) = 1, this means restricting k sothat M|(m − 2k).

If m is even, then k = m/2 is such a value, andqa(m−2k)+km − qb(m−2k)+km = 0 in this case.

Hence we need only consider those k in the intervals0 ≤ k < m/2 and m/2 < k ≤ m satisfyingm − 2k ≡ 0( mod M).

Proof, Continued

Proof Continued.

sm

∑0≤k<m/2

+∑

m/2<k≤m

qa(m−2k)+kM − qb(m−2k)+kM

(qM ;qM)m−k (qM ;qM)k

Note that(1) every k ′ in the upper interval may be expressed ask ′ = m − k , for some k in the lower interval;(2) every k in the lower interval can be similarly matched with ak ′ in the upper interval;

(3) m − 2k ≡ 0( mod M)⇐⇒ m − 2(m − k) ≡ 0( mod M);

(4) the denominators of the summands remain invariant underthe transformation k ↔ m − k .

Proof, Continued

Proof Continued.

∑m,n≥0

c(m,Mn)smqMn =∑m≥0

sm∑

0≤k<m/2M|m−2k

qa(m−2k)+kM − qb(m−2k)+kM

+ q−a(m−2k)+(m−k)M − q−b(m−2k)+(m−k)M

(qM ;qM)m−k (qM ;qM)k

=∑m≥0

sm∑

0≤k<m/2M|m−2k

qa(m−2k)+kM(1− q(m−2k)(b−a))(1− q(m−2k)(M−b−a))

(qM ;qM)m−k (qM ;qM)k

Proof, Continued

Proof Continued.Finally,- M|(m − 2k)(b − a) and M|(m − 2k)(M − b − a);

- the conditions on a and b give that they are different multiplesof M, each less than (m − k)M;

- the factors (1− q(m−2k)(b−a)) and (1− q(m−2k)(M−b−a)) arecancelled by two different factors in the q-product (qM ;qM)m−k ;

- the remaining factors in the denominators may be expandedas geometric series with only non-negative coefficients, and theclaim at (i) above follows;

- The claim at (ii) follows similarly, upon noting that

m − 2k ≡ M/2( mod M)

⇐⇒ m − 2(m − k) ≡ −M/2 ≡ M/2( mod M).

Second 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 d(m,n) by

(−sqa,−sqM−a;qM)∞ − (−sqb,−sqM−b;qM)∞

:=∑

m,n≥0

d(m,n)smqn. (8)

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

Partitions Interpretation

Corollary

Let M, a and b be as in Theorem 9. Let

p∗a,M,m(n) denote the number of partitions of n into exactly mdistinct parts ≡ ±a( mod M), and let

p∗b,M,m(n) denote the number of partitions of n into exactly mdistinct parts ≡ ±b( mod M).

Then

(i) p∗a,M,m(nM) ≥ p∗b,M,m(nM) for all integers n ≥ 1, and allintegers m, 1 ≤ m ≤ Mn.

(ii) If M is even, then p∗a,M,m(nM + M/2) ≥ p∗b,M,m(nM + M/2)for all integers n ≥ 0, and integers m with 1 ≤ m ≤ Mn + M/2.

Sketch of Proof

Sketch of Proof.Recall

(−a;q)∞ =∞∑

n=0

anqn(n−1)/2

(q;q)n. (9)

The application of this to the infinite products leads to

∑m,n≥0

d(m,n)smqn =∑m≥0

smm∑

k=0

(qa(m−2k)+kM − qb(m−2k)+kM)qM[(m−k)(m−k−1)/2+k(k−1)/2]

(qM ;qM)m−k (qM ;qM)k

The proof now follows similarly.

No Finite Analogues

It does not appear that replacing “∞” with a positive integer L inTheorems 7 and 9 “works”. In other words, if L is a positiveinteger, and

1(sqa, sqM−a;qM)L

− 1(sqb, sqM−b;qM)L

:=∑

m,n≥0

c(m,n)smqn, (10)

it does not appear to be the case that c(m,Mn) ≥ 0 for all mand n, and likewise for the other theorem.Perhaps there is some restricted version of such a theorem thatis valid?

Injective Proofs

Let M ≥ 5 and 1 ≤ a < b < M/2 be integers. For integers1 ≤ m ≤ n let pa,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).Can you find an injection from the partitions counted bypb,M,m(Mn) to those counted by pa,M,m(Mn)?Mind Floss: (1) Consider the partition of kMb consisting kMparts of size b.Find a partition of kMb into kM parts, where each part is ≡ ±a(mod M).

(2) Consider the partition of kM(M − b) consisting kM parts ofsize M − b.Find a partition of kM(M − b) into kM parts, where each part is≡ ±a( mod M).

The End.