Refinement of Some Partition Inequalities Partitions, The Partition Counting Function Restricted Partition Functions Ferrers Diagram, Durfee Square Partition Generating Functions Partition

Refinement of Some Partition Inequalities

James Mc Laughlin

West Chester [email protected]

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.