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##  eISSN 0454-8124
pISSN 1225-6951

### Article

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

Published online March 31, 2021

### 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 functions$EO(n)$ 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 $EO¯10n+8≡0 mod5$ 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 $λ1≥λ2≥…≥λk$ 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

$∑ n=0∞p(n)qn=1(q;q) ∞,$

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

$(a;q)n=(1−a)(1−aq)⋯(1−aqn−1), (a;q)∞=∏ k=0 ∞(1−aqk).$

Almost a century back Ramanujan established the following identity ,

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

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

$p(5n+4)≡0 mod5.$

Recently, Andrews  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 , Andrews shows that the generating function for $EO(n)$ is

$∑ n=0∞EO(n)qn:=1(1−q)(q2;q2) ∞.$

Andrews , 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 , Andrews shows that the generating function for $EO¯(n)$ is

$∑ n=0∞EO¯(n)qn=(q4;q4) ∞3(q2;q2) ∞2.$

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

$EO¯10n+8≡0 mod5,$

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

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

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

$f(a,b)=∑ n=−∞∞an(n+1)/2bn(n−1)/2.$

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

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

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

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

### 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,

$(qk;qk)∞5≡(q5k;q5k)∞ mod5.$

### 3. Congruences Modulo 2 and 4 for $E O ¯ ( n )$

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,

Proof. From (1.4), we have

$∑ n=0∞EO¯(n)qn=(q4;q4) ∞3(q2;q2) ∞2,$

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

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

It follows that

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

and

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

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

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

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

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

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

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

It follows that

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

and

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

which is our (3.4) and (3.6).

We have the following congruences.

### Corollary 3.2.

For all $n≥0$,

### Remark 3.3.

The congruences (3.11)-(3.13) were obtained earlier by Andrews et al. . Andrews et al.  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.

$∑ n=0∞pν(n)qn=ν(−q),$

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.

### 4. Congruences Modulo 5, 10 and 20 for $E O ¯ ( n )$

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 $n≥0$,

$EO¯10n+8≡0 mod5.$

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

$(q;q)∞3≡J0+J1 mod5,$

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

$(q2;q2)∞3≡J0*+J2* mod5,$

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

$∑ n=0∞EO¯(2n)qn≡1(q5;q5) ∞J0+J1J0*+J2* mod5.$

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

$∑ n=0∞EO¯(2(5n+4))q5n+4≡0 mod5,$

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 $n≥0$,

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

$(q2;q2)∞2(q8;q8)∞2(q4;q4)∞=∑ n=−∞∞(3n+1)q6n2+4n≡R0*+R1*+R2* mod5,$

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

$∑ n=0∞EO¯(4n+2)qn≡21(q5;q5) ∞J0+J1R0*+R1*+R2* mod10.$

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

$∑ n=0∞EO¯(4(5n+4)+2)q5n+4≡0 mod10,$

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

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

From (2.7), we have

$(q4;q4)∞3≡J0**+J4** mod5,$

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

$∑n=0∞ EO¯(8n+4)qn≡2 (q 10;q 10) ∞ (q5;q5) ∞ (q 20;q 20) ∞J0 **+J4 **R0*+R1*+R2* mod10.$

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

$∑ n=0∞EO¯(8(5n+3)+4)q5n+3≡0 mod10,$

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 $n≥0$,

$EO¯(40n+38)≡0 mod20.$

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

$(q;q)∞2(q4;q4)∞2(q2;q2)∞=∑ n=−∞∞(3n+1)q3n2+2n≡R0+R2+R3 mod5,$

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

$∑n=0∞ EO¯(8n+6)qn≡41 (q5;q5) ∞R0+R2+R3R0*+R1*+R2* mod20.$

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

$∑ n=0∞EO¯(8(5n+4)+6)q5n+4≡0 mod20,$

from which we deduce (4.14).

### 5. Congruences for $E O e ( n )$

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

### Theorem 5.1.

Proof. From (1.5), we have

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

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

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

It follows that

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

and

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

which is our (5.1) and (5.2).

We have the following congruences.

### Corollary 5.2.

For all $n≥0$,

$EOe(2n+1)=0,$
$EOe(4n+2)≡0 mod2.$

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 $n≥0$,

$EO¯50n+18≡0 mod20,$
$EO¯50n+28≡0 mod20,$
$EO¯50n+38≡0 mod20,$
$EO¯50n+48≡0 mod20.$

The authors would like to thank Professor Michael Hirschhorn for his valuable comments and helpful suggestions.

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