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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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# Some congruences for 3-component multipartitions

Tao Yan Zhao
/ Lily J. Jin
/ C. Gu
Published Online: 2016-10-16 | DOI: https://doi.org/10.1515/math-2016-0067

## Abstract

Let p3(n) denote the number of 3-component multipartitions of n. Recently, using a 3-dissection formula for the generating function of p3(n), Baruah and Ojah proved that for n ≥ 0, p3(9n + 5) ≡ 0 (mod 33) and p3 (9n + 8) ≡ 0 (mod 34). In this paper, we prove several congruences modulo powers of 3 for p3(n) by using some theta function identities. For example, we prove that for n ≥ 0, p3 (243n + 233) ≡ p3 (729n + 638) ≡ 0 (mod 310).

Keywords: Congruences; Multipartitions; Theta functions

MSC 2010: 11P83; 05A17

## 1 Introduction

The objective of this paper is to prove several congruences modulo powers of 3 for 3-component multipartitions by employing theta function identities.

Recall that the k-tuple λ = (λ1, λ2,…, λk) of integers λi (i = 1,2,…, k) is a partition of the natural number n if $n={\sum }_{i=1}^{k}{\lambda }_{i}$ and λ1λ2 ≥ ··· ≥ λk > 0. The partition function p(n) is defined to be the number of partitions of n. A multipartition of n with r-components, as called by Andrews [1], also referred to as an r-colored partition, see, for example [2, 3], is an r-tuple λ = (λ(1), λ(2), ··· λ(r)) of partitions whose weights sum to n. Let pr(n) denote the number of r-component multipartitions of n. Multipartitions arise in combinatorics, representation theory and physics, see, for example Bouwknegt [4] and Fayers [5]. As usual, set pr(0) = 1. The generating function of pr(n) is $∑n=0∞prnqn=1q;q∞r,$(1)

where $(a;q)∞=∏n=1∞1−aqn.$(2)

A number of congruences satisfied by pr(n) were discovered, see, for example Andrews [1], Eichhorn and Ono [3], Atkin [6], Baruah and Ojah [7], Boylan [8], Cheema and Haskell [9], Gordon [10], Kiming and Olsson [11], Newman [12], Sinick [13], Treneer [14], Xia [15] and Yao [16]. Recently, Baruah and Ojah [7] established a 3-dissection formula for the generating function for p3(n) and proved that for n ≥ 0, $p39n+5≡0mod33$(3)

and $p39n+8≡0mod34.$(4)

In this paper, we prove several congruences modulo powers of 3 for p3(n). The main results of this paper can be stated as follows.

For n > 0, $p327n+17≡0mod35,$(5) $p327n+26≡0mod37,$(6) $p381n+44≡0mod36,$(7) $p381n+71≡0mod37,$(8) $p3243n+152≡0mod38,$(9) $p3243n+233≡0mod310,$(10) $p3729n+395≡0mod39,$(11) $p3729n+638≡0mod310.$(12)

## 2 Some lemmas

In this section, we collect three lemmas which are needed to prove the main results of this paper.

The following 3-dissection formulas are true: $aq=aq3+6qq9;q9∞3q3;q3∞$(13)

and $bq=aq3−3qq9;q9∞3q3;q3∞$(14)

where a(q) and b(q) are defined by $aq=∑m,n=−∞∞qm2+mn+n2$(15)

and $bq=∑m,n=−∞∞ωm−nqm2+mn+n2,ω=exp2πi/3.$(16)

Lemma 2.1 was proved by Borwein, Borwein and Garvan [18].

We have $bq=q;q∞3q3;q3∞.$(17)

Lemma 2.2 was also proved by Borwein, Borwein and Garvan [18].

The following 3-dissection formula holds: $1q;q∞3=q9;q9∞3q3;q3∞10a2q3+3qaq3q9;q9∞3q3;q3∞+9q2q9;q9∞6q3;q3∞2.$(18)

Baruah and Ojah [7] also deduced the following 3-dissection formula for $\frac{1}{{\left(q;q\right)}_{\mathrm{\infty }}^{3}},$ which is different from (18): $1(q;q)∞3=(q9;q9)∞9(q3;q3)∞121w2(q3)+3qw(q3)+9q2+8q3w(q3)+12q4w2(q3)+16q6w4(q3),$(23)

where $w(q)=(q;q)∞(q6;q6)∞3(q2;q2)∞(q3;q3)∞3.$(24)

## 3 Proof of Theorem 1.1

Setting r = 3 in (1) and using (18), we obtain $∑n=0∞p3(n)qn=(q9;q9)∞3(q3;q3)∞10a2(q3)+3qa(q3)(q9;q9)∞3(q3;q3)∞+9q2(q9;q9)∞6(q3;q3)∞2,$(25)

which yields $∑n=0∞p3(3n+2)qn=9q3;q3∞9q;q∞12.$(26)

Substituting (18) into (26), we deduce that $∑n=0∞p33n+2qn=9q9;q9∞12q3;q3∞31a2q3+3qaq3q9;q9∞3q3;q3∞+9q2(q9;q9)∞6(q3;q3)∞24≡9q9;q9∞12q3;q3∞31a8q3+108qq9;q9∞15q3;q3∞32a7q3+810q2q9;q9∞18q3;q3∞33a6q3+3888q3q9;q9∞21q3;q3∞34a5q3+13851q4q9;q9∞24q3;q3∞35a4q3+34992q5q9;q9∞27q3;q3∞36a3q3+6561q6q9;q9∞30q3;q3∞37a2q3+19683q7q9;q9∞33q3;q3∞38aq3mod310,$(27)

which yields $∑n=0∞p39n+8qn≡810q3;q3∞18q;q∞33a6q+34992qq3;q3∞27q;q∞36a3qmod310.$(28)

Substituting (13) and (18) into (28), we see that $∑n=0∞p39n+8qn≡810q9;q9∞33q3;q3∞92a2q3+3qq9;q9∞3q3;q3∞aq3+9q2q9;q9∞6q3;q3∞211×aq3+6qq9;q9∞3q3;q3∞6+34992qq9;q9∞36q3;q3∞93a2q3+3qq9;q9∞3q3;q3∞aq3+9q2q9;q9∞6q3;q3∞212×aq3+6qq9;q9∞3q3;q3∞3≡810q9;q9∞33q3;q3∞92a28q3+31833qq9;q9∞36q3;q3∞93a27q3+50301q2q9;q9∞39q3;q3∞94a26q3+52488q3q9;q9∞42q3;q3∞95a25q3+39366q4q9;q9∞45q3;q3∞96a24q3mod310,$(29)

which implies that $∑n=0∞p327n+17qn≡131×35q3;q3∞36q;q∞93a27q+2×39qq3;q3∞45q;q∞96a24qmod310$(30)

and $∑n=0∞p327n+26qn≡23×37q3;q3∞39q;q∞94a26qmod310.$(31)

Congruences (5) and (6) follow from (30) and (31).

Substituting (13) and (18) into (30), we get $∑n=0∞p327n+17qn≡131×35q9;q9∞93q3;q3∞274a2q3+3qq9;q9∞3q3;q3∞aq3+9q2q9;q9∞6q3;q3∞231×aq3+6qq9;q9∞3q3;q3∞27+2×39qq9;q9∞96q3;q3∞275a2q3+3qq9;q9∞3q3;q3∞aq3+9q2q9;q9∞6q3;q3∞232×aq3+6qq9;q9∞3q3;q3∞24≡31833q9;q9∞93q3;q3∞274a89q3+8019qq9;q9∞96q3;q3∞275a88q3+30618q2q9;q9∞99q3;q3∞276a87q3+52488q3q9;q9∞102q3;q3∞277a86q3+39366q4q9;q9∞105q3;q3∞278a85q3mod310.$(32)

It follows from (32) that $∑n=0∞p381n+44qn≡11×36q3;q3∞96q;q∞275a88q+2×39qq3;q3∞105q;q∞278a85qmod310$(33)

and $∑n=0∞p381n+71qn≡14×37q3;q3∞99q;q∞276a87qmod310.$(34)

Congruences (7) and (8) follow from (33) and (34).

By the binomial theorem, $q3;q3∞90q;q∞270≡1mod27.$(35)

Combining (34) and (35), we have $∑n=0∞p381n+71qn≡14×37q3;q3∞9q;q∞6a87qmod310.$(36)

Substituting (13) and (18) into (36), we find that $∑n=0∞p381n+71qn≡14×37q9;q9∞6q3;q3∞11a2q3+3qq9;q9∞3q3;q3∞aq3+9q2q9;q9∞6q3;q3∞22×aq3+6qq9;q9∞3q3;q3∞87≡14×37q9;q9)∞6q3;q3∞11a91q3+7×38qq9;q9∞9q3;q3∞12a90q3mod310.$(37)

Congruences (9) and (10) follow from (37).

Congruence (37) also implies that $∑n=0∞p3243n+152qn≡7×38q3;q3∞9q;q∞12a90qmod310.$(38)

By (13) and the binomial theorem, $a3q≡a3q3mod9.$(39)

Thanks to (38) and (39), $∑n=0∞p3243n+152qn≡7×38q3;q3∞9q;q∞12a90q3mod310.$(40)

Substituting (18) into (40), we see that $∑n=0∞p3243n+152qn≡7×38a90q3q9;q9∞12q3;q∞31a2q3+3qaq3q9;q9∞3q3;q3∞+9q2q9;q9∞6q3;q3∞24≡7×38q9;q9∞12q3;q3∞31a98q3+39qq9;q9∞15q3;q3∞32a97q3mod310.$(41)

Congruences (11) and (12) follow from (41). This completes the proof of Theorem 1.1.

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## About the article

Accepted: 2016-08-22

Published Online: 2016-10-16

Published in Print: 2016-01-01

Citation Information: Open Mathematics, Volume 14, Issue 1, Pages 783–788, ISSN (Online) 2391-5455,

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