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Kyungpook Mathematical Journal 2021; 61(1): 49-59

Published online March 31, 2021

Copyright © Kyungpook Mathematical Journal.

Some Congruences for Andrews' Partition Function EO¯(n)

Utpal Pore and Syeda Noor Fathima*

Department of Mathematics, Ramanujan School of Mathematical Sciences, Pondicherry University, Puducherry - 605 014, India
e-mail : utpal.mathju@gmail.com and dr.fathima.sn@gmail.com

Received: March 14, 2019; Revised: October 7, 2020; Accepted: October 8, 2020

Recently, Andrews introduced partition functionsEO(n) and EO¯(n) where the function EO(n) denotes the number of partitions of n in which every even part is less than each odd part and the function EO¯(n) denotes the number of partitions enumerated by EO(n) in which only the largest even part appears an odd number of times. In this paper we obtain some congruences modulo 2, 4, 10 and 20 for the partition function EO¯(n). We give a simple proof of the first Ramanujan-type congruences EO¯10n+80mod5 given by Andrews.

Keywords: partitions, partitions with even parts below odd parts, congruences.

A partition of a positive integer n is a nonincreasing sequence of positive integers λ1λ2λk such that λ1+λ2+λk=n. Let p(n) be the number of partitions of n. For example p(5)=7. The seven partitions of 5 are 5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1. The generating function for p(n) is given by

n=0p(n)qn=1(q;q) ,

where throughout this paper, for any complex numbers a and q<1 we define

(a;q)n=(1a)(1aq)(1aqn1),(a;q)= k=0 (1aqk).

Almost a century back Ramanujan established the following identity [7],

n=0p(5n+4)qn=5(q5;q5) 5(q;q) 6,

which in fact implies Ramanujan's congruences for p(n) modul 5,

p(5n+4)0mod5.

Recently, Andrews [2] introduced the partition function EO(n) which counts the number of partitions of n in which every even part is less than each odd part. For example, EO(6)=7. The seven partitions of 6 it enumerates are 6, 5+1, 4+2, 3+3, 3+1+1+1, 2+2+2, 1+1+1+1+1+1. In [2], Andrews shows that the generating function for EO(n) is

n=0EO(n)qn:=1(1q)(q2;q2) .

Andrews [2], also defined the partition function EO¯(n) which counts the number of partitions enumerated by EO(n) in which only the largest even part appears an odd number of times. For example, EO¯(6)=4. The four partitions of 6 it enumerates are 6, 3+3, 2+2+2, 1+1+1+1+1+1. In [2], Andrews shows that the generating function for EO¯(n) is

n=0EO¯(n)qn=(q4;q4) 3(q2;q2) 2.

In Section 3 of this paper, we prove some congruences modulo 2 and 4 for the partition function EO¯(n). In Section 4, we give a simple proof of Andrews' congruences

EO¯10n+80mod5,

and we prove some interesting congruences modulo 10 and 20. In the Section 5, we consider

n=0EOe(n)qn:=(q4;q4) 2(q2;q2) 2,

where the function EOe(n) counts the elements in the set of partitions which are enumerated by EO¯(n) together with the partitions enumerated by EO(n) where all parts are odd and number of parts is even, i.e, EOe(n) denotes the number of partitions enumerated by EO(n) in which only the largest even part appears an odd number of times except when parts are odd and number of parts is even. For example, EOe(6)=6. The six partitions of 6 it enumerates are 6, 3+3, 2+2+2, 1+1+1+1+1+1 (which are counted by EO¯(n)) and 5+1 and 3+1+1+1 (only counted by EO(n) in which all parts are odd and the number of parts is even). We prove some arithmetic properties modulo 2 satisfied by EOe(n). All of the proofs will follow from elementary generating function considerations and q-series manipulations. The paper concludes with a conjecture on EO¯(n).

We require the following definitions and lemmas to prove the main results in the next three sections. For ab<1, Ramanujan's general theta function f(a,b) is defined as

f(a,b)= n=an(n+1)/2bn(n1)/2.

Using Jacobi's triple product identity [1, Theorem 2.8], (2.1) takes the shape

f(a,b)=(a;ab)(b;ab)(ab;ab).

The special cases of f(a,b) are

ϕ(q):=f(q,q)= n=qn2=(q;q2)2(q2;q2)=(q2;q2) 5(q;q) 2(q4;q4) 2,
ψ(q):=f(q,q3)= n=0qn(n+1)/2=(q2;q2) (q;q2) =(q2;q2) 2(q;q) ,
ϕ(q):= n=(1)nqn2=(q;q) 2(q2;q2) .

Lemma 2.1.

(Hirschhorn [6, p. 14, Eqn. 1.9.4]) We have the following 2-dissection of ϕ(q),

ϕ(q)=ϕ(q4)+2qψ(q8).

Lemma 2.2.

(Hirschhorn [5] or Hirschhorn [6, p. 36, Eqn. 3.6.4]) we have

(q;q)3=n=0(1)n(2n+1)q(n2+n)/2f(q10,q15)3qf(q5,q20)mod5.

Lemma 2.3.

(Hirschhorn [6, p. 105, Eqn. 10.7.6]) We have the following beautiful identity due to Ramanujan,

(q;q)2(q4;q4)2(q2;q2)= n=(3n+1)q3n2+2n.

From the Binomial Theorem, for any positive integer, k,

(qk;qk)5(q5k;q5k)mod5.

In this section we prove some congruences modulo 2 and 4 satisfied by EO¯(n).

We require the following generating functions to prove congruences for EO¯(n).

Theorem 3.1.

We have,

n=0EO¯(4n)qn =(q4;q4) 5(q;q) 2(q8;q8) 2,
n=0EO¯(4n+2)qn =2(q2;q2) 2(q8;q8) 2(q;q) 2(q4;q4) ,
n=0EO¯(8n)qn =(q2;q2) 5(q4;q4) 3(q;q) 5(q8;q8) 2,
n=0EO¯(8n+2)qn =2(q4;q4) 7(q;q) 3(q2;q2) (q8;q8) 2,
n=0EO¯(8n+4)qn =2(q2;q2) 7(q8;q8) 2(q;q) 5(q4;q4) 3,
n=0EO¯(8n+6)qn =4(q2;q2) (q4;q4) (q8;q8) 2(q;q) 3.

Proof. From (1.4), we have

n=0EO¯(n)qn=(q4;q4) 3(q2;q2) 2,

since there are no terms on the right in which the power of q is odd, we have

EO¯(2n+1)=0,

thus by using (2.6), we obtain

n=0EO¯(2n)qn=(q2;q2)3(q;q)2=(q2;q2)3(q4;q4)2(q2;q2)5ϕ(q)=(q4;q4)2(q2;q2)2ϕ(q4)+2qψ(q8).

It follows that

n=0EO¯(4n)qn=(q2 ;q2 ) 2(q;q) 2ϕ(q2)=(q4 ;q4 ) 5(q;q) 2(q8 ;q8 ) 2

and

n=0EO¯(4n+2)qn=2(q2 ;q2 ) 2(q;q) 2ψ(q4)=2(q2 ;q2 ) 2(q8 ;q8 ) 2(q;q) 2(q4 ;q4 ) ,

which is our (3.1) and (3.2). We have

n=0EO¯(4n)qn=(q2;q2)2(q;q)2ϕ(q2)=(q2;q2)2ϕ(q2)(q4;q4)2(q2;q2)5ϕ(q)=(q4;q4)2(q2;q2)3ϕ(q2)ϕ(q)=(q4;q4)2(q2;q2)3ϕ(q2)ϕ(q4)+2qψ(q8).

It follows that

n=0 EO¯(8n)qn= (q2;q2) 2 (q;q) 3ϕ(q)ϕ(q2)= (q2;q2) 2 (q;q) 3 (q2;q2) 5 (q;q) 2 (q4;q4) 2 (q4;q4) 5 (q2;q2) 2 (q8;q8) 2= (q2;q2) 5 (q4;q4) 3 (q;q) 5 (q8;q8) 2

and

n=0EO¯(8n+4)qn=2(q2 ;q2 ) 2(q;q) 3ϕ(q)ψ(q4)=2(q2 ;q2 ) 7(q8 ;q8 ) 2(q;q) 5(q4 ;q4 ) 3,

which is our (3.3) and (3.5). We have

n=0EO¯(4n+2)qn=2(q2;q2)2(q;q)2ψ(q4)=2(q2;q2)2ψ(q4)(q4;q4)2(q2;q2)5ϕ(q)=2(q4;q4)2(q2;q2)3ψ(q4)ϕ(q)=2(q4;q4)2(q2;q2)3ψ(q4)ϕ(q4)+2qψ(q8).

It follows that

n=0EO¯(8n+2)qn=2(q2 ;q2 ) 2(q;q) 3ψ(q2)ϕ(q2)=2(q4 ;q4 ) 7(q;q) 3(q2 ;q2 ) (q8 ;q8 ) 2

and

n=0EO¯(8n+6)qn=4(q2 ;q2 ) 2(q;q) 3ψ(q2)ψ(q4)=4(q2 ;q2 ) (q4 ;q4 ) (q8 ;q8 ) 2(q;q) 3,

which is our (3.4) and (3.6).

We have the following congruences.

Corollary 3.2.

For all n0,

EO¯(2n+1) =0,
EO¯(4n+2) 0mod2,
EO¯(8n+4) 0mod2,
EO¯(8n+6) 0mod4.

Remark 3.3.

The congruences (3.11)-(3.13) were obtained earlier by Andrews et al. [4]. Andrews et al. [3] introduced a partition function pν(n) which counts the number of partitions of n in which the parts are distinct and all odd parts are less than twice the smallest part.

n=0pν(n)qn=ν(q),

where ν(q) is a mock theta function. Andrews [2, Corollary 5.2] noted that

pν(2n)=EO¯(2n).

He proved the congruences using the properties of mock theta function, whereas we use the q-series identities.

In this section we prove some congruences modulo 5, 10 and 20 for EO¯(n). In the next theorem, we give a simple proof of the Andrews' result [2, Eqn. 1.6], which can be tracked back to [3, Thrm. 6.7]. He used the properties of mock theta functions to prove the congruence, whereas we manipulate the q-series identities to get the result.

Theorem 4.1.

For all n0,

EO¯10n+80mod5.

Proof. Applying (2.9) in (1.4), we obtain

n=0 EO¯(2n)qn= (q2;q2) 3 (q;q) 2= (q2;q2) 3 (q;q) 3 (q;q) 5 (q2;q2) 3 (q;q) 3 (q5;q5) mod5.

From (2.7), we have

(q;q)3J0+J1mod5,

where Ji contains terms in which the power of q is congruent to i modulo 5, then

(q2;q2)3J0*+J2*mod5,

where J*i contains terms in which the power of q is congruent to i modulo 5. Substituting (4.3) and (4.4) in (4.2), we have

n=0EO¯(2n)qn1(q5;q5) J0+J1J0*+J2*mod5.

There are no terms on the right in which the power of q is 4 modulo 5, so

n=0EO¯(2(5n+4))q5n+40mod5,

from which we deduce (4.1).

In the next theorem, we derive two congruences modulo 10 from the generating Łinebreak functions (3.2) and (3.5).

Theorem 4.2.

For all n0,

EO¯(20n+18) 0mod10,
EO¯(40n+28) 0mod10.

Proof. Using (2.9) in (3.2), we have

n=0 EO¯(4n+2)qn=21 (q;q) 2 (q2;q2) 2 (q8;q8) 2 (q4;q4) =2 (q;q) 3 (q;q) 5 (q2;q2) 2 (q8;q8) 2 (q4;q4) 2 (q;q) 3 (q5;q5) (q2;q2) 2 (q8;q8) 2 (q4;q4) mod10.

Replacing q by q2 in (2.8), we have

(q2;q2)2(q8;q8)2(q4;q4)= n=(3n+1)q6n2+4nR0*+R1*+R2*mod5,

where R*i contains terms in which the power of q is congruent to i modulo 5.

Substituting (4.3) and (4.9) in (4.8), we obtain

n=0EO¯(4n+2)qn21(q5;q5) J0+J1R0*+R1*+R2*mod10.

There are no terms on the right in which the power of q is 4 modulo 5, so

n=0EO¯(4(5n+4)+2)q5n+40mod10,

from which we deduce (4.6). Using (2.9) in (3.5), we have

n=0EO¯(8n+4)qn=2(q2;q2)5(q;q)5(q4;q4)2(q2;q2)2(q8;q8)2(q4;q4)=2(q4;q4)3(q2;q2)5(q;q)5(q4;q4)5(q2;q2)2(q8;q8)2(q4;q4)2(q4;q4)3(q10;q10)(q5;q5)(q20;q20)(q2;q2)2(q8;q8)2(q4;q4)mod10.

From (2.7), we have

(q4;q4)3J0**+J4**mod5,

where Ji** contains terms in which the power of q is congruent to i modulo 5. Substituting (4.9) and (4.12) in (4.11), we obtain

n=0 EO¯(8n+4)qn2 (q 10;q 10) (q5;q5) (q 20;q 20) J0 **+J4 **R0*+R1*+R2*mod10.

There are no terms on the right in which the power of q is 3 modulo 5, so

n=0EO¯(8(5n+3)+4)q5n+30mod10,

from which we deduce (4.7).

In the next theorem, we derive a congruences modulo 20 from the generating Łinebreak function (3.6).

Theorem 4.3.

For all n0,

EO¯(40n+38)0mod20.

Proof. Using (2.9) in (3.6), we have

n=0 EO¯(8n+6)qn=41 (q;q) 3 (q4;q4) 2 (q2;q2) (q2;q2) 2 (q8;q8) 2 (q4;q4) =41 (q;q) 5 (q;q) 2 (q2;q2) 4 (q2;q2) (q2;q2) 2 (q8;q8) 2 (q4;q4) 41 (q5;q5) (q;q) 2 (q2;q2) 4 (q2;q2) (q2;q2) 2 (q8;q8) 2 (q4;q4) mod20.

From (2.8), we have

(q;q)2(q4;q4)2(q2;q2)= n=(3n+1)q3n2+2nR0+R2+R3mod5,

where Ri contains terms in which the power of q is congruent to i modulo 5. Substituting (4.9) and (4.16) in (4.15), we obtain

n=0 EO¯(8n+6)qn41 (q5;q5) R0+R2+R3R0*+R1*+R2*mod20.

There are no terms on the right in which the power of q is 4 modulo 5, so

n=0EO¯(8(5n+4)+6)q5n+40mod20,

from which we deduce (4.14).

In this section we prove some congruences modulo 2 for EOe(n).

Theorem 5.1.

n=0EOe(4n)qn =(q4;q4) 5(q;q) 3(q8;q8) 2,
n=0EOe(4n+2)qn =2(q2;q2) 2(q8;q8) 2(q;q) 3(q4;q4) .

Proof. From (1.5), we have

n=0EOe(n)qn=(q4;q4) 2(q2;q2) 2,

since there are no terms on the right in which the power of q is odd, we have

EOe(2n+1)=0,

by using (2.6), we obtain

n=0EOe(2n)qn=(q2;q2)2(q;q)2=(q2;q2)2(q4;q4)2(q2;q2)5ϕ(q)=(q4;q4)2(q2;q2)3ϕ(q4)+2qψ(q8).

It follows that

n=0EOe(4n)qn=(q2 ;q2 ) 2(q;q) 3ϕ(q2)=(q4 ;q4 ) 5(q;q) 3(q8 ;q8 ) 2

and

n=0EOe(4n+2)qn=2(q2 ;q2 ) 2(q;q) 3ψ(q4)=2(q2 ;q2 ) 2(q8 ;q8 ) 2(q;q) 3(q4 ;q4 ) ,

which is our (5.1) and (5.2).

We have the following congruences.

Corollary 5.2.

For all n0,

EOe(2n+1)=0,
EOe(4n+2)0mod2.

Andrews [2, Problem 4], proposed to further investigate the properties of EO¯n. We conclude the paper with the following conjecture. Using maple, we found the following congruences hold up to n = 2000.

Conjecture 6.1.

Andrews [2, Problem 4], proposed to further investigate the properties of EO¯n. We conclude the paper with the following conjecture. Using maple, we found the following congruences hold up to n = 2000.

For all n0,

EO¯50n+180mod20,
EO¯50n+280mod20,
EO¯50n+380mod20,
EO¯50n+480mod20.
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