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Kyungpook Mathematical Journal 2023; 63(2): 155-166

Published online June 30, 2023 https://doi.org/10.5666/KMJ.2023.63.2.155

Copyright © Kyungpook Mathematical Journal.

Congruences for Partition Functions EO¯(n) and EOe(n)

Riyajur Rahman and Nipen Saikia*

Department of Mathematics, Rajiv Gandhi University, Rono Hills, Doimukh 791112, Arunachal Pradesh, India
e-mail : riyajurrahman@gmail.com and nipennak@yahoo.com

Received: February 11, 2022; Revised: October 13, 2022; Accepted: January 17, 2023

In 2018, Andrews introduced the partition functions EO(n) and EO¯(n). The first of these denotes the number of partitions of n in which every even part is less than each odd part, and the second counts the number of partitions enumerated by the first in which only the largest even part appears an odd number of times. In 2021, Pore and Fathima introduced a new partition function EOe(n) which counts the number of partitions of n which are enumerated by EO¯(n) together with the partitions enumerated by EO(n) where all parts are odd and the number of parts is even. They also proved some particular congruences for EO¯(n) and EOe(n). In this paper, we establish infinitely many families of congruences modulo 2, 4, 5 and 8 for EO¯(n) and modulo 4 for EOe(n). For example, if p5 is a prime with Legendre symbol 3p=1, then for all integers n≥ 0 and α0, we have
EO¯8p2α+1(pn+j)+19p2α+2130mod8;  1j(p1).

Keywords: Partitions of integer, congruences, q-series identities

A partition of a positive integer n is a non-increasing sequence of positive integers whose sum equals n. The number of partitions of a non-negative integer n is usually denoted by p(n) (with p(0)=1). For example, p(4)=5 with the relevant partitions 4, 3+1, 2+2, 2+1+1 and 1+1+1+1. The generating function of p(n) is given by

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

where, for any complex number a,

(a;q)= n=0 (1aqn),|q|<1.

We will use the notation, for any positive integer k,

fk:=(qk;qk).

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(4) = 4 with the relevant partitions 4, 3+1, 2+2 and 1+1+1+1. The generating function for EO(n) [2] is given by

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

Andrews [2] also introduced another 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¯(4)=3, with the relevant partitions 4, 3+1 and 1+1+1+1. The generating function of EO¯(n) [2] is given by

n=0EO¯(n)qn=f43f22.

Andrews [2, p. 434, (1.6)] established that

EO¯(10n+8)0mod5.

Recently, Pore and Fathima [8] proved some particular congruences for EO¯(n) modulo 2,4,10 and 20. For example, they proved that

EO¯(4n+2)0mod4,EO¯(20n+18)0mod10,EO¯(40n+38)0mod20.

Pore and Fathima [8] also defined a new partition function EOe(n) which counts the number of partitions of n which are enumerated by EO¯(n) together with the partitions enumerated by EO(n) where all parts are odd and the 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(4) = 3 with the relevant partitions 4, 3+1 and 1+1+1+1. The generating function of EOe(n) [8] is given by

n=0EOe(n)qn=f42f22.

Pore and Fathima [8, Corollary 5.2] proved that

EOe(2n+1)=0 and EOe(4n+2)0mod2.

Motivated by the above work, in Section 3 of this paper, we prove infinite families of congruences modulo 2, 4, 5 and 8 for EO¯(n). In Section 4, we prove infinite families of congruences modulo 2 and 4 for EOe(n). To prove our results, we employ some known q-series identities which are listed in Section 2.

Lemma 2.1. ([3, p. 39, Entry 24(ii)]) We have

f13=n=0(1)n(2n+1)qn(n+1)/2.

Lemma 2.2. We have

1f12=f85f25f162+2qf42f162f25f8,
f12=f2f85f42f1622qf2f162f8.

The identity (2.2) is the 2-dissection of ϕ(q) [6, (1.9.4)]. The equation (2.3) can be obtained

from the equation (2.2) by replacing q by -q.

Lemma 2.3. ([1, Lemma 2.3]) For any prime p ≥ 3, we have

f13= k=0 k(p1)/2 (p1)(1)kqk(k+1)/2 n=0(1)n(2pn+2k+1)qpn(pn+2k+1)/2
+p(1)(p1)/2q(p21)/8fp23.

Furthermore, if k(p1)2,0kp1, then

(k2+k)2(p21)8modp.

Lemma 2.4. ([4, Theorem 2.2]) For any prime p ≥ 5, we have

f1= k=(p1)/2 k(±p1)/6 (p1)/2(1)kq(3k2+k)/2fq(3p2+(6k+1)p)/2,q(3p2(6k+1)p)/2
+(1)(±p1)/6q(p21)/24fp2,

where

±p16=(p1)6,if p1mod6,(p1)6,if p1mod6.

Furthermore, if

(p1)2k(p1)2andk(±p1)6,

then

(3k2+k)2(p21)24modp.

Lemma 2.5. ([7]) We have

f1=f25Rq5qq2Rq5,

where R(q) is the Roger-Ramanujan continued fraction defined by

R(q):=q1/51+q1+q21+q3 1+=(q2;q5)(q3;q5)(q;q5)(q4;q5),|q|<1.

Hirschhorn and Hunt [5, Lemma 2.2] proved that, if R is a series in powers of q5, then

η=q1R1qR1,

where

η=f1qf25.

Hirschhorn and Hunt [5] showed that

H5(η3)=5,

where H5 is an operator which acts on a series of positive and negative powers of a single variable and simply picks out the term in which the power is congruent to 0 modulo 5.

In addition to above q-series identities, we will be using following congruence properties which follow from binomial theorem and (1.2): For positive integers k and m,

fk2mf2kmmod2,
fk4mf2k2mmod4,
f15f5mod5,

Theorem 3.1. Let p5 be a prime with 3p=1 and 1j(p1). Then for all integers n0 and α0, we have

n=0EO¯8p 2αn+ 19p 2α13qn4f16f13mod8,
EO¯8p2α+1(pn+j)+19p2α+2130mod8,

where, here and throughout the paper () denotes the Legendre symbol.

Proof. From (1.5), we note that

n=0EO¯(n)qn=f43f22.

Extracting the terms involving q2n and using (2.2), we obtain

n=0EO¯(2n)qn=f85f22f162+2qf42f162f22f8.

Extracting the terms involving q2n+1, we obtain

n=0EO¯(4n+2)qn=2f22f82f12f4.

Using (2.2) in (3.5), we obtain

n=0EO¯(4n+2)qn=2f87f23f4f162+4qf4f8f162f23.

Extracting the terms involving q2n+1 from (3.6), we obtain

n=0EO¯(8n+6)qn=4f2f4f82f13.

Using (2.10) in (3.7), we obtain

n=0EO¯(8n+6)qn4f16f13mod8.

Congruence (3.8) is the α=0 case of (3.1). Suppose that congruence (3.1) is true for all α0. Utilizing (2.4) and (2.5) in (3.1), we obtain

n=0 EO¯ 8p 2αn+ 19p 2α13qn    4{ k=(p1)/2 k(±p1)/6 (p1)/2 (1)kq16(3k2+k)/2fq 16(3p2+(6k+1)p)/2,q 16(3p2(6k+1)p)/2    +(1)(±p1)/6q16(p21)/24f16p2}    ×{ m=0 m(p1)/2 (p1) (1)mqm(m+1)/2n=0 (1)n(2pn+2m+1)qpn(pn+2m+1)/2    +p(1)(p1)/2q(p21)/8fp23}mod8.

Consider the congruence

16(3k2+k)2+(m2+m)219(p21)24modp,

which is equal to

(24k+4)2+3(2m+1)20modp.

For 3p=1, the above congruence has only solution m=p12 and k=±p16. Therefore, extracting the terms involving qpn+19(p21)/24 from both sides of (3.9), dividing throughout by q19(p21)/24 and then replacing qp by q, we obtain

n=0EO¯8p 2α+1n+ 19p 2α+213qn4f16pfp3mod8.

Extracting the terms involving qpn from (3.10) and replacing qp by q, we obtain

n=0EO¯8p 2(α+1)n+ 19p 2(α+1)13qn4f16f13mod8,

which is the α + 1 case of (3.1). Thus, by the principle of mathematical induction, we arrive at (3.1). Extracting the coefficients of terms involving qpn+j for 1jp1, from both sides of (3.10), we complete the proof of (3.2).

Theorem 3.2. Let p5 be a prime with 3p=1 and 1j(p1). Then for all integers n0 and α0, we have

n=0EO¯8p 2αn+ 7p 2α13qn2f4f13mod4,
EO¯8p2α+1(pn+j)+7p2α+2130mod4.

Proof. From (3.5), we note that

n=0EO¯(4n+2)qn=2f22f82f12f4.

Employing (2.10) in (3.14), we have

n=0EO¯(4n+2)qn2f27mod4.

Extracting the terms involving q2n from both side of (3.15), we obtain

n=0EO¯(8n+2)qn2f17mod4.

Employing (2.10) in (3.16), we obtain

n=0EO¯(8n+2)qn2f4f13mod4.

Congruence (3.17) is the α=0 case of (3.12). Suppose that congruence (3.12) is true for all α0. Utilizing (2.4) and (2.5) in (3.12), we obtain

n=0 EO¯ 8p 2αn+ 7p 2α13qn    2{ k=(p1)/2 k(±p1)/6 (p1)/2 (1)kq4(3k2+k)/2fq 4(3p2+(6k+1)p)/2,q 4(3p2(6k+1)p)/2    +(1)(±p1)/6q4(p21)/24f4p2}    ×{ m=0 m(p1)/2 (p1) (1)mqm(m+1)/2n=0 (1)n(2pn+2m+1)qpn(pn+2m+1)/2    +p(1)(p1)/2q(p21)/8fp23}mod4.

Consider the congruence

4(3k2+k)2+(m2+m)27(p21)24modp,

which is equal to

(12k+2)2+3(2m+1)20modp.

For 3p=1, the above congruence has only solution m=p12 and k=±p16. Therefore, extracting the terms involving qpn+7(p21)/24 from both sides of (3.18), dividing throughout by q7(p21)/24 and then replacing qp by q, we obtain

n=0EO¯8p 2α+1n+ 7p 2α+213qn2f4pfp3mod4.

Extracting the terms involving qpn from (3.19) and replacing qp by q, we obtain

n=0EO¯8p 2(α+1)n+ 7p 2(α+1)13qn2f4f13mod4,

which is the α + 1 case of (3.12). Thus, by the principle of mathematical induction, we arrive at (3.12). Extracting the coefficients of terms involving qpn+j for 1jp1, from both sides of (3.19), we complete the proof of (3.13).

Theorem 3.3. Let p5 be a prime with 3p=1 and 1j(p1). Then for all integers n0 and α0, we have

n=0EO¯8p 2αn+ 13p 2α13qn2f1f43mod4,
EO¯8p2α+1(pn+j)+13p2α+2130mod4. 

Proof. Extracting the terms involving q2n from (3.4) and using (2.11), we obtain

n=0EO¯(4n)qnf4f12mod4.

Employing (2.2) in (3.23) and extracting the terms involving q2n+1, we obtain

n=0EO¯(8n+4)qn2f23f82f15f4mod4.

Using (2.10) in (3.24), we obtain

n=0EO¯(8n+4)qn2f1f43mod4.

Congruence (3.25) is the α=0 case of (3.21). Suppose that congruence (3.21) is true for all α0. Utilizing (2.4) and (2.5) in (3.21), we obtain

n=0 EO¯ 8p 2αn+ 13p 2α13qn      2{ k=(p1)/2 k(±p1)/6 (p1)/2 (1)kq(3k2+k)/2fq (3p2+(6k+1)p)/2,q (3p2(6k+1)p)/2      +(1)(±p1)/6q(p21)/24fp2}      ×{ m=0 m(p1)/6 (p1) (1)mq2m(m+1)n=0 (1)n(2pn+2m+1)q2pn(pn+2m+1)      +p(1)(p1)/2q(p21)/2fp23}mod4.

Consider the congruence

(3k2+k)2+2m(m+1)13(p21)24modp,

which is equal to

(6k+1)2+3(4m+2)20modp.

For 3p=1, the above congruence has only solution m=p12 and k=±p16. Therefore, extracting the terms involving qpn+13(p21)/24 from both sides of (3.26), dividing throughout by q13(p21)/24 and then replacing qp by q, we obtain

n=0EO¯8p 2α+1n+ 13p 2α+213qn2fpf4p3mod4.

Extracting the terms involving qpn from (3.27) and replacing qp by q, we obtain

n=0EO¯8p 2(α+1)n+ 13p 2(α+1)13qn2f1f43mod4,

which is the α + 1 case of (3.21). Thus, by the principle of mathematical induction, we arrive at (3.21). Extracting the coefficients of terms involving qpn+j for 1jp1, from both sides of (3.27), we complete the proof of (3.22).

Theorem 3.4. Let p5 be a prime and 1jp1. Then for all integers α,n0, we have

EO¯8n+k0mod2,for1k7.
n0EO¯8p2αn+p 2α13qnf1mod2,
EO¯8p2α+1(pn+j)+p2α+2130mod2.

Proof. From (1.5), we note that

n=0EO¯(n)qn=f43f22.

Employing (2.10) in (3.32), we have

n=0EO¯(n)qnf8mod2.

Extracting the terms involving q8n+k, we obtain the proof of (3.29). Again extracting the terms involving q8n and then replacing q8 by q, we obtain

n=0EO¯(8n)qnf1mod2.

Congruence (3.34) is the α=0 case of (3.30). Suppose that congruence (3.30) is true for all integer α0. Employing (2.5) in (3.30), we obtain

n0 EO¯8p 2αn+ p 2α13qn    { k=(p1)/2 k(±p1)/6 (p1)/2 (1)kq(3k2+k)/2fq (3p2+(6k+1)p)/2,q (3p2(6k+1)p)/2      +(1)(±p1)/6q(p21)/24fp2}mod2.

Extracting the term involving qpn+(p21)/24 from both sides of (3.35), dividing throughout by q(p21)/24 and then replacing qp by q, we obtain

n0EO¯8p2α+1n+p 2α+213qnfpmod2.

Extracting the terms involving qpn from (3.36) and replacing qp by q, we obtain

n0EO¯8p2(α+1)n+p 2(α+1)13qnf1mod2,

which is the α + 1 case of (3.30). Thus, by the principle of mathematical induction, we arrive at (3.30). Extracting the coefficients of terms involving qpn+j for 1jp1, from both sides of (3.36), we complete the proof of (3.31).

Theorem 3.5. For any non-negative integers n and k, where l=k(k+1) and k2mod5, we have

EO¯(10n+6+2l)0mod5.

Proof. Extracting q2n from (3.3) and then employing (2.12), we obtain

n=0EO¯(2n)qnf23f13f5mod5.

Employing (2.1) and (2.8) in (3.39), we obtain

n=0EO¯(2n)qnq3η3f253f5 k=0(1)k(2k+1)qk(k+1)mod5.

Extracting the terms involving the powers of q5n+l+3 from both sides of (3.40) and then using (2.9), we prove (3.38).

Theorem 4.1. Let p5 be a prime and 1j(p1). Then for all integers n0 and α0, we have

n=0EOe4p 2αn+ p 2α16qnf1mod2,
EOe4p2α+1(pn+j)+p2α+2160mod2.

Proof. From (1.6), we note that

n=0EOe(n)qn=f42f22.

Employing (2.10) in (4.3), we obtain

n=0EOe(n)qnf4mod2.

Extracting the terms involving q4n from (4.4) and replacing q4 by q, we obtain

n=0EOe(4n)qnf1mod2.

The remaining part of the proof is similar to proofs of the identities (3.30) and (3.31).

Theorem 4.2. Let p5 be a prime with 3p=1 and 1j(p1). Then for all integers n0 and α0, we have

n=0EOe4p 2αn+ 13p 2α16qn2f43f1mod4,
EOe4p2α+1(pn+j)+13p2α+2160mod4.

Proof. From (1.6), we note that

n=0EOe(n)qn=f42f22.

Extracting the terms involving q2n from (4.8) and replacing q2 by q, we obtain

n=0EOe(2n)qn=f22f12.

Multiplying the numerator and denominator by f12, we obtain

n=0EOe(2n)qn=f22f12f14.

Using (2.11) in (4.10), we obtain

n=0EOe(2n)qnf12mod4.

Using (2.3) in (4.11), we obtain

n=0EOe(2n)qnf2f85f42f1622qf2f162f8mod4.

Extracting the terms involving q2n+1 from (4.12) and replacing q2 by q, we obtain

n=0EOe(4n+2)qn2f1f82f4mod4.

Using (2.11) in (4.13), we obtain

n=0EOe(4n+2)qn2f1f43mod4.

The remaining part of the proof is similar to proofs of the identities (3.21) and (3.22).

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