New infinite families of congruences modulo 8 for partitions with even parts distinct

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1 New infinite families of congruences modulo for partitions with even parts distinct Ernest X.W. Xia Department of Mathematics Jiangsu University Zhenjiang, Jiangsu , P. R. China Submitted: Jan 1, 2014; Accepted: Sep 2, 2014; Published: Oct 9, 2014 Mathematics Subject Classifications: 05A17, 11P3 Abstract Let pedn denote the number of partitions of an integer n wherein even parts are distinct. Recently, Andrews, Hirschhorn and Sellers, Chen, and Cui and Gu have derived a number of interesting congruences modulo 2, 3 and 4 for pedn. In this paper we prove several new infinite families of congruences modulo for pedn. For example, we prove that for α 0 and n 0, ped 3 4α+4 n α mod. Keywords: partition; congruence; regular partition 1 Introduction Let pedn denote the function which enumerates the number of partitions of n wherein even parts are distinct and odd parts are unrestricted. For a positive integer t we say that a partition is t-regular if no part is divisible by t. Andrews, Hirschhorn and Sellers [1] found the generating function for pedn: pednq n = n=1 1 + q 2n 1 q 2n 1 = f 4 f 1, 1 Supported by the National Natural Science Foundation of China No the electronic journal of combinatorics , #P4. 1

2 where here and throughout this paper, and for any positive integer k, f k is defined by f k := 1 q kn. 2 n=1 From 1 it is easy to see that pedn equals the number of 4-regular partitions of n. In recent years many congruences for the number of regular partitions have been discovered see for example, Cui and Gu [3, 4], Dandurand and Penniston [5], Furcy and Penniston [7], Gordon and Ono [], Keith [10], Lin and Wang [11], Lovejoy and Penniston [12], Penniston [13, 14], Webb [15], Xia and Yao [1, 17], and Yao[1]. Numerous congruence properties are known for the function pedn. For example, Andrews, Hirschhorn and Sellers [1] proved that for α 1 and n 0, ped3n mod 2, 3 ped9n mod 4, 4 ped9n mod 12, 5 ped 3 2α+2 n α mod 2, ped 3 2α+1 n α 1 0 mod, 7 ped 3 2α+2 n α mod. Recently, Chen [2] obtained many interesting congruences modulo 2 and 4 for pedn using the theory of Hecke eigenforms and Cui and Gu [3] found infinite families of wonderful congruences modulo 2 for the function pedn. The aim of this paper is to establish several new infinite families of congruences modulo for pedn by employing some results of Andrews, Hirschhorn and Sellers [1], and Cui and Gu [3]. The main results of this paper can be stated as the following theorems. Theorem 1. For α 0 and n 0, ped 3 2α n + 32α 1 pedn mod 4, 9 ped 3 4α n + 34α 1 5 α pedn mod, 10 ped 3 4α+4 n α mod, 11 ped 3 4α+4 n α mod. 12 the electronic journal of combinatorics , #P4. 2

3 In view of 9 and the facts ped1 = 1, ped2 = 2, ped3 = 3, ped4 = 4, we obtain the following corollary. Corollary 2. For α 0 and i = 0, 1, 2, 3 we have that ti 3 2α 1 ped i mod 4, 13 where t 0 = 33, t 1 = 9, t 2 = 17 and t 3 = 25. Replacing α by 2α in 10, we find that for α 0, ped 3 α n + 3α 1 pedn mod. 14 Employing 14 and the facts ped1 = 1, ped2 = 2, ped3 = 3, ped4 = 4, ped10 = 29, ped5 =, ped253 = and ped = 1, we obtain the following congruences modulo. Corollary 3. For α 0 and 0 j 7 we have that sj 3 α 1 ped j mod, 15 where s 0 = 5, s 1 = 9, s 2 = 17, s 3 = 25, s 4 = 33, s 5 = 1, s = 41 and s 7 = Utilizing the generating functions of ped9n + 4, ped9n + 7 discovered by Andrews, Hirschhorn and Sellers [1] and the p-dissection identities of two Ramanujan s theta functions due to Cui and Gu [3], we will prove the following theorem. Theorem 4. Let p be a prime such that p 5, 7 mod and 1 i p 1. Then for n 0 and α 1, ped 9p 2α n + 72i + 33pp2α mod 1 and ped 9p 2α n + 72i + 57pp2α mod Proof of Theorem 1 Andrews, Hirschhorn and Sellers [1] established the following results for ped3n + 1: ped9n + 1q n = f 2 2 f3 4 f 4 + q f 2 3 f3 3 f 4 f 3, 1 f1 5 f 2 f1 10 ped9n + 4q n = 4 f 2f 3 f 4 f f q f 2 2 f 4 f f the electronic journal of combinatorics , #P4. 3

4 and ped9n + 7q n = 12 f 2 4 f3 f f1 11 By the binomial theorem it is easy to see that for all positive integers m and k, By 21 we see that f 2m k f m 2k mod which yields f 2 1 f 2 f 2 f mod 2, 22 f2 2 f1 4 f 4 3 f 2 1 mod It follows from 1 and 23 that ped9n + 1q n f 4 mod 4. f 1 In view of 1 and we see that for n 0, ped9n + 1 pedn mod Congruence 9 follows from 25 and mathematical induction. Andrews, Hirschhorn and Sellers [1] also established the following 3-dissection formula of the generating function of pedn: pednq n = f 12f 4 1 f 3 3 f q f 2 f 3 9 f 3 f 4 3 f q 2 f f 1 f 3. 2 f3 3 Fortin, Jacob and Mathieu [], and Hirschhorn and Sellers [9] independently derived the following 3-dissection formula of the generating function of overpartitions: f 2 f1 2 = f 4 f9 + 2q f 3 f9 3 f3 f1 3 f3 7 Combining 1, 1, 2, 27 we deduced that ped9n + 1q n f 4 3 f 2 f 4 3 f 2 f 4 f 9 f 3 f 3 1 f2 2 f1 4 f 4 f 1 + 2q f 3 f 3 9 f q 2 f 2 f f3 + 4q 2 f 2 f1 3 2 f12 f1 4 + q f 2 f9 3 f 3 + 2q 2 f f 1 f 3 f3 f3 3 f3 2 f3 4 f1 2 f3 3 f f 12 9 f 12 f 15 3 f 2 1f q f f 15 9 f 3 f 1 3 f 1 + 4q f 5 f 9 9 f 12 f 1 f 14 3 f q 2 f 7 f 12 9 f 3 f 15 3 f q 2 f 4 f 9 f 12 f 4 1 f 13 3 f q 3 f f 9 9 f 3 f 14 3 f 2 1 mod. 2 the electronic journal of combinatorics , #P4. 4

5 Extracting those terms associated with powers q 3n+1 on both sides of 2, then dividing by q and replacing q 3 by q, we find that ped27n + 10q n f 2 f 15 3 f 12 f 1 1 f + 4 f 5 2 f 9 3 f 4 f f 14 1 f 2 12 mod. 29 By the binomial theorem and 22 we have f 2 f 1 1 f 3 1 f 1 mod, 30 which yields f 2 f 15 3 f 12 f 1 1 f f 12 f 3 mod. 31 It follows from 21 that f 5 2 f 9 3 f 4 f f 14 1 f 2 12 f 12 f 3 mod Substituting 31 and 32 into 29, we see that which implies that and for n 0, Thanks to 1 and 34, we see that for n 0, ped27n + 10q n 5 f 12 f 3 mod, 33 ped1n + 10q n 5 f 4 f 1 mod 34 ped1n mod, 35 ped1n mod. 3 ped1n pedn mod. 37 Congruence 10 follows from 37 and mathematical induction. Replacing n by 1n + 37 in 10 and employing 35, we obtain 11. Replacing n by 1n + 4 in 10 and using 3, we deduce 12. The proof is complete. the electronic journal of combinatorics , #P4. 5

6 3 Proof of Theorem 4 Thanks to 19 and 21, we have where ψq is defined by ped9n + 4q n 4f 2 ψq 3 mod, 3 ψq := f 2 2 f In their nice paper [3], Cui and Gu established p-dissection formulas for f 1 and ψq. They proved that for any odd prime p, ψq = p 3 k=0 q k2 +k 2 f q p2 +2k+1p 2, q p2 2k+1p 2 + q p2 1 ψq p2 40 and for any prime p 5, f 1 = p 1 k= 1 p 2, k ±p 1 1 k q 3k2 +k 2 f q 3p2 +k+1p 2, q 3p2 k+1p ±p 1 q p2 1 fp 2, 41 where ±p 1 : = p 1 p 1 if p 1 mod, and the Ramanujan theta function fa, b is defined by if p 1 mod 42 fa, b := a nn+1/2 b nn 1/2, 43 n= where ab < 1. Let an be defined by anq n := f 2 ψq It follows from 3 and 44 that for n 0, ped9n + 4 4an mod. 45 the electronic journal of combinatorics , #P4.

7 Substituting 40 and 41 into 44, we see that for any prime p 5, 7 mod, anq n = p 1 m= 1 p 2, m ±p 1 p 3 k=0 1 m q 3m2 +m f q 3p2 +m+1p, q 3p2 m+1p + 1 ±p 1 q p f2p 2 4 q 3k2 +k 2 f q 3p2 +2k+1p 2, q 3p2 2k+1p 2 + q 3p2 1 ψq 3p2. 47 Now, we consider the congruence 3m 2 + m + 3k2 + k 2 11p2 1 mod p, 4 where p 1 m p 1 and 0 k p 1. Congruence 4 can be rewritten as follows m k mod p. 49 Since p 5, 7 mod, we have that 2 is a ratic nonresidue modulo p and hence 49 is equivalent to m + 1 k mod p. 50 Thus, m = ±p 1 and k = p 1 11p 2 1. Extracting those terms associated with powers qpn+ 2 on both sides of 4 and employing the fact that Congruence 4 holds if and only if m = ±p 1 and k = p 1, we have 2 a pn + 11p2 1 q pn+ 11p2 1 = 1 ±p 1 q 11p2 1 f 2p 2ψq 3p2. 51 Dividing q 11p2 1 on both sides of 51 and then replacing q p by q, we get a pn + 11p2 1 q n = 1 ±p 1 f 2p ψq 3p, 52 which implies that a p 2 n + 11p2 1 q n = 1 ±p 1 f 2 ψq 3 53 the electronic journal of combinatorics , #P4. 7

8 and a ppn + i + 11p2 1 = 0 54 for n 0 and 1 i p 1. Combining 44 and 53, we have a p 2 n + 11p2 1 an mod By 55 and mathematical induction, we find that for n 0 and α 0, a p 2α n + 11p2α 1 an mod 2. 5 Replacing n by ppn + i + 11p2 1 1 i p 1 in 5 and using 54, we deduce that for n 0 and α 1, a p 2α n + i + 11pp2α mod Finally, replacing n by p 2α n + i+11pp2α i p 1 in 45 and using 57, we get 1. We conclude the paper by proving 17. In view of 20 and 21, we find that ped9n + 7q n 4f 1 ψq mod, 5 where ψq is defined by 39. Let bn be defined by bnq n := f 1 ψq. 59 By 5 and 59, we find that for n 0, ped9n + 7 4bn mod. 0 Substituting 40 and 41 into 59, we see that for any prime p 5, 7 mod, bnq n = p 1 m= 1 p 2, m ±p 1 1 m q 3m2 +m 2 f q 3p2 +m+1p 2, q 3p2 m+1p ±p 1 q p2 1 q 3k2 +k f p 3 k=0 q 3p2 +2k+1p, q 3p2 2k+1p + q 3p2 1 fp 2 4 ψq p2. 1 the electronic journal of combinatorics , #P4.

9 As above, for any prime p 5, 7 mod, p 1 m p 1 and 0 k p 1, the congruence relation 3m 2 + m 2 + 3k 2 + k 19p2 1 mod p 2 holds if and only if m = ±p 1 and k = p 1. This implies that 2 b pn + 19p2 1 q n = 1 ±p 1 f p ψq p. 3 Thanks to 3, we find that b p 2 n + 19p2 1 q n = 1 ±p 1 f 1 ψq 4 and b ppn + i + 19p2 1 = 0 5 for n 0 and 1 i p 1. It follows from 59 and 4 that for n 0, b p 2 n + 19p2 1 bn mod 2. By and mathematical induction, we deduce that for n 0 and α 0, b p 2α n + 19p2α 1 bn mod 2. 7 Replacing n by ppn + i + 19p2 1 1 i p 1 in 7 and employing 5, we find that b p 2α n + i + 19pp2α mod 2 for n 0, α 1 and 1 i p 1. Congruence 17 follows from 0 and. This completes the proof of Theorem 4. Acknowledgements We wish to thank the referee for helpful comments. the electronic journal of combinatorics , #P4. 9

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