New Congruences for Partitions where the Odd Parts are Distinct

Liuquan Wang Department of Mathematics National University of Singapore Singapore, 119076 Singapore

                     wangliuquan@u.nus.edu
                      mathlqwang@163.com

Abstract Let pod(n) denote the number of partitions of n wherein odd parts are distinct (and even parts are unrestricted). We find some new interesting congruences for pod(n) modulo 3, 5 and 9.

1 Introduction and Main Results

Let ψ(q) be one of Ramanujan’s theta functions, namely

ψ(q) = ∑_{n=0}^{∞} q^{n(n+1)/2} = \frac{(q^{2}; q^{2})_{∞}^{2}}{(q; q)_{∞}} .

We let pod(n) (see \underline{A006950}) denote the number of partitions of n wherein the odd parts are distinct (and even parts are unrestricted). For example, pod(4) = 3 since there are 3 different partitions of 3 such that the odd parts are distinct, namely 4 = 3 + 1 = 2 + 2. The generating function of pod(n) is given by

∑_{n=0}^{∞} pod(n)q^{n} = \frac{(−q; q^{2})_{∞}}{(q^{2}; q^{2})_{∞}} = \frac{1}{ψ(−q)}.  (1)

The arithmetic properties of pod(n) were first studied by Hirschhorn and Sellers [4] in 2010. They obtained some interesting congruences involving the following infinite family of Ramanujan-type congruences: for any integers α ≥ 0 and n ≥ 0,

pod(3^{2α+3}n + \frac{23 × 3^{2α+2} + 1}{8}) ≡ 0 (mod 3).

Later on Radu and Sellers [7] obtained other deep congruences for pod(n) modulo 5 and 7, such as

pod(135n + 8) ≡ pod(135n + 107) ≡ pod(135n + 116) ≡ 0 (mod 5), and

pod(567n + 260) ≡ pod(567n + 449) ≡ 0 (mod 7).

For nonnegative integers n and k, let r_{k}(n) (resp., t_{k}(n)) denote the number of representations of n as sum of k squares (resp., triangular numbers). In 2011, based on the generating function of pod(3n + 2) found in [4], Lovejoy and Osburn [6] discovered the following arithmetic relation:

pod(3n + 2) ≡ (−1)^{n}r_{5}(8n + 5) (mod 3).  (2)

Following their steps, we will present some new congruences modulo 5 and 9 for pod(n). Firstly, we find that (2) can be improved to a congruence modulo 9.

Theorem 1. For any integer n ≥ 0, we have

pod(3n + 2) ≡ 2(−1)^{n+1}r_{5}(8n + 5) (mod 9).

The following result will be a consequence of Theorem 1 upon invoking some properties of r_{5}(n).

Theorem 2. Let p ≥ 3 be a prime, and N be a positive integer such that pN ≡ 5 (mod 8). Let α be any nonnegative integer.

(1) If p ≡ 1 (mod 3), then

pod(\frac{3p^{6α+5}N + 1}{8}) ≡ 0 (mod 3),

and

pod(\frac{3p^{18α+17}N + 1}{8}) ≡ 0 (mod 9).

(2) If p ≡ 2 (mod 3), then

pod(\frac{3p^{4α+3}N + 1}{8}) ≡ 0 (mod 9).

Secondly, with the same method used in proving Theorem 1, we can establish a similar congruence for pod(n) modulo 5.

Theorem 3. For any integer n ≥ 0, we have

pod(5n + 2) ≡ 2(−1)^{n}r_{3}(8n + 3) (mod 5).

Some miscellaneous congruences can be deduced from this theorem.

Theorem 4. For any integers n ≥ 0 and α ≥ 1, we have

pod(5^{2α+2}n + \frac{11 · 5^{2α+1} + 1}{8}) ≡ 0 (mod 5),

and

pod(5^{2α+2}n + \frac{19 · 5^{2α+1} + 1}{8}) ≡ 0 (mod 5).

Theorem 5. Let p ≡ 4 (mod 5) be a prime, and N be a positive integer which is coprime to p such that pN ≡ 3 (mod 8). We have

pod(\frac{5p^{3}N + 1}{8}) ≡ 0 (mod 5).

For example, let p = 19 and N = 8n + 1 where n ≥ 0 and n ≢ 7 (mod 19). We have

pod(34295n + 4287) ≡ 0 (mod 5).

Theorem 6. Let p ≥ 3 be a prime, and N be a positive integer which is not divisible by p such that pN ≡ 3 (mod 8). Let α be any nonnegative integer.

(1) If p ≡ 1 (mod 5), we have

pod(\frac{5p^{10α+9}N + 1}{8}) ≡ 0 (mod 5).

(2) If p ≡ 2, 3, 4 (mod 5), we have

pod(\frac{5p^{8α+7}N + 1}{8}) ≡ 0 (mod 5).

2 Preliminaries

Lemma 7. (Cf. [7, Lemma 1.2].) Let p be a prime and α be a positive integer. Then

(q; q)_{∞}^{p^{α}} ≡ (q^{p}; q^{p})_{∞}^{p^{α−1}} (mod p^{α}).

Lemma 8. For any prime p ≥ 3, we have

t_{4}(pn + \frac{p − 1}{2}) ≡ t_{4}(n) (mod p), t_{8}(pn + p − 1) ≡ t_{8}(n) (mod p^{3}).

Proof. By [2, Theorem 3.6.3], we know t_{4}(n) = σ(2n + 1). For any positive integer N, we have

σ(N) = ∑_{d|N, p|d} d + ∑_{d|N, p∤d} d ≡ ∑_{d|N, p∤d} d (mod p).

Let N = 2n + 1 and N = p(2n + 1), respectively. It is easy to deduce that σ(p(2n + 1)) ≡ σ(2n + 1) (mod p). This clearly implies the first congruence.

From [2, Eq.(3.8.3), page 81], we know

t_{8}(n) = ∑_{\begin{matrix}d|(n+1) \\ d odd \end{matrix}} (\frac{n + 1}{d})^{3}.

By a similar argument we can prove the second congruence.

Lemma 9. (Cf. [1].) For 1 ≤ k ≤ 7, we have

r_{k}(8n + k) = 2^{k}(1 + \frac{1}{2}\binom{k}{4})t_{k}(n).

Lemma 10. (Cf. [3].) Let p ≥ 3 be a prime and n be a positive integer such that p^{2} ∤ n. For any integer α ≥ 0, we have

r_{5}(p^{2α}n) = (\frac{p^{3α+3} − 1}{p^{3} − 1} − p(\frac{n}{p})\frac{p^{3α} − 1}{p^{3} − 1})r_{5}(n),

where (\frac{·}{p}) denotes the Legendre symbol.

Lemma 11. (Cf. [5].) Let p ≥ 3 be a prime. For any integers n ≥ 1 and α ≥ 0, we have

r_{3}(p^{2α}n) = (\frac{p^{α+1} − 1}{p − 1} − (\frac{−n}{p})\frac{p^{α} − 1}{p − 1})r_{3}(n) − p\frac{p^{α} − 1}{p − 1}r_{3}(n/p^{2}),

where we take r_{3}(n/p^{2}) = 0 unless p^{2}|n .

3 Proofs of the Theorems

Proof of Theorem 1. Let p = 3 in Lemma 8. We deduce that t_{8}(3n + 2) ≡ t_{8}(n) (mod 9). By (1) we have

ψ(q)^{9} ∑_{n=0}^{∞} pod(n)(−q)^{n} = ψ(q)^{8} = ∑_{n=0}^{∞} t_{8}(n)q^{n}.

By Lemma 7 we obtain ψ(q)^{9} ≡ ψ(q^{3})^{3} (mod 9). Hence

ψ(q^{3})^{3} ∑_{n=0}^{∞} pod(n)(−q)^{n} ≡ ∑_{n=0}^{∞} t_{8}(n)q^{n} (mod 9).

If we extract those terms of the form q^{3n+2} on both sides, we obtain

ψ(q^{3})^{3} ∑_{n=0}^{∞} pod(3n + 2)(−q)^{3n+2} ≡ ∑_{n=0}^{∞} t_{8}(3n + 2)q^{3n+2} (mod 9).

Dividing both sides by q^{2}, then replacing q^{3} by q, we get

ψ(q)^{3} ∑_{n=0}^{∞} pod(3n + 2)(−q)^{n} ≡ ∑_{n=0}^{∞} t_{8}(3n + 2)q^{n} ≡ ∑_{n=0}^{∞} t_{8}(n)q^{n} = ψ(q)^{8} (mod 9).

Hence

∑_{n=0}^{∞} pod(3n + 2)(−q)^{n} ≡ ψ(q)^{5} ≡ ∑_{n=0}^{∞} t_{5}(n)q^{n} (mod 9).

Comparing the coefficients of q^{n} on both sides, we deduce that pod(3n + 2) ≡ (−1)^{n}t_{5}(n) (mod 9).

Let k = 5 in Lemma 9. We obtain t_{5}(n) = r_{5}(8n + 5)/112, and from this the theorem follows.

Proof of Theorem 2. (1) Let n = pN in Lemma 10, and then we replace α by 3α + 2. Since

\frac{p^{9α+9} − 1}{p^{3} − 1} = 1 + p^{3} + ⋯ + p^{3(3α+2)} ≡ 0 (mod 3),

we deduce that r_{5}(p^{6α+5}N) ≡ 0 (mod 3).

Let n = \frac{p^{6α+5}N−5}{8} in Theorem 1. We deduce that pod(\frac{3p^{6α+5}N+1}{8}) ≡ 0 (mod 3).

Similarly, let n = pN in Lemma 10 and we replace α by 9α + 8. Since p ≡ 1 (mod 3) implies p^{3} ≡ 1 (mod 9), we have

\frac{p^{27α+27} − 1}{p^{3} − 1} = 1 + p^{3} + ⋯ + p^{3(9α+8)} ≡ 0 (mod 9).

Hence r_{5}(p^{18α+17}N) ≡ 0 (mod 9).

Let n = \frac{p^{18α+17}N−5}{8} in Theorem 1. We deduce that pod(\frac{3p^{18α+17}N+1}{8}) ≡ 0 (mod 9).

(2) Let n = pN in Lemma 10, and then we replace α by 2α + 1. Note that p ≡ 2 (mod 3) implies p^{3} ≡ −1 (mod 9). Since p^{6α+6} ≡ 1 (mod 9), we have r_{5}(p^{4α+3}N) ≡ 0 (mod 9).

Let n = \frac{p^{4α+3}N−5}{8} in Theorem 1. We complete our proof.

Proof of Theorem 3. Let p = 5 in Lemma 8. We deduce that t_{4}(5n + 2) ≡ t_{4}(n) (mod 5). By (1) we have

ψ(q)^{5} ∑_{n=0}^{∞} pod(n)(−q)^{n} = ψ(q)^{4} = ∑_{n=0}^{∞} t_{4}(n)q^{n}.

By Lemma 7 we obtain ψ(q)^{5} ≡ ψ(q^{5}) (mod 5). Hence

ψ(q^{5}) ∑_{n=0}^{∞} pod(n)(−q)^{n} ≡ ∑_{n=0}^{∞} t_{4}(n)q^{n} (mod 5).

If we extract those terms of the form q^{5n+2} on both sides, we obtain

ψ(q^{5}) ∑_{n=0}^{∞} pod(5n + 2)(−q)^{5n+2} ≡ ∑_{n=0}^{∞} t_{4}(5n + 2)q^{5n+2} (mod 5).

Dividing both sides by q^{2}, and then replacing q^{5} by q, we get

ψ(q) ∑_{n=−∞}^{∞} pod(5n + 2)(−q)^{n} ≡ ∑_{n=0}^{∞} t_{4}(5n + 2)q^{n} ≡ ∑_{n=0}^{∞} t_{4}(n)q^{n} = ψ(q)^{4} (mod 5).

Hence we have

∑_{n=0}^{∞} pod(5n + 2)(−q)^{n} ≡ ψ(q)^{3} = ∑_{n=0}^{∞} t_{3}(n)q^{n} (mod 5).

Comparing the coefficients of q^{n} on both sides, we deduce that pod(5n + 2) ≡ (−1)^{n}t_{3}(n) (mod 5).

Let k = 3 in Lemma 9. We obtain t_{3}(n) = r_{3}(8n + 3)/8, from which the theorem follows.

Proof of Theorem 4. Let p = 5 and n = 5m + r (r ∈ \{1, 4\}) in Lemma 11. Since (\frac{−r}{5}) = 1, we deduce that r_{3}(5^{2α}(5m + r)) ≡ 0 (mod 5) for any integer α ≥ 1.

Let n = \frac{5^{2α}(40m+a)−3}{8} (a ∈ \{11, 19\}). By Theorem 3, we have

r_{3}(8n + 3) = r_{3}(5^{2α}(40m + a)) ≡ 0 (mod 5).

Hence

pod(5^{2α+2}m + \frac{a · 5^{2α+1} + 1}{8}) = pod(5n + 2) ≡ 2(−1)^{n}r_{3}(8n + 3) ≡ 0 (mod 5).

Proof of Theorem 5. Let α = 1 and n = pN in Lemma 11. We have

r_{3}(p^{3}N) = (1 + p)r_{3}(pN) ≡ 0 (mod 5).

Let n = \frac{p^{3}N−3}{8} in Theorem 3. We have

pod(\frac{5p^{3}N + 1}{8}) = pod(5n + 2) ≡ 2(−1)^{n}r_{3}(8n + 3) = 2(−1)^{n}r_{3}(p^{3}N) ≡ 0 (mod 5).

Proof of Theorem 6. (1) Let n = pN in Lemma 11, and then we replace α by 5α + 4. We have

\frac{p^{5α+5} − 1}{p − 1} = 1 + p + ⋯ + p^{5α+4} ≡ 0 (mod 5).

Hence r_{3}(p^{10α+9}N) ≡ 0 (mod 5). Let n = \frac{p^{10α+9}N−3}{8} in Theorem 3. We have

pod(\frac{5p^{10α+9}N + 1}{8}) = pod(5n + 2) ≡ 2(−1)^{n}r_{3}(p^{10α+9}N) ≡ 0 (mod 5).

(2) Let n = pN in Lemma 11, and then replace α by 4α + 3. Since p^{4α+4} ≡ 1 (mod 5), we deduce that r_{3}(p^{8α+7}N) ≡ 0 (mod 5). Let n = \frac{p^{8α+7}N−3}{8} in Theorem 3. We have

pod(\frac{5p^{8α+7}N + 1}{8}) = pod(5n + 2) ≡ 2(−1)^{n}r_{3}(p^{8α+7}N) ≡ 0 (mod 5).

References

[1] P. Barrucand, S. Cooper, and M. D. Hirschhorn, Relations between squares and triangles, Discrete Math. 248 (2002), 245–247.

[2] B. C. Berndt, Number Theory in the Spirit of Ramanujan. Amer. Math. Soc., 2006.

[3] S. Cooper, Sums of five, seven and nine squares, Ramanujan J. 6 (2002), 469–490.

[4] M. D. Hirschhorn and J. A. Sellers, Arithmetic properties of partitions with odd parts distinct, Ramanujan J. 22 (2010), 273–284.

[5] M. D. Hirschhorn and J. A. Sellers, On representations of a number as a sum of three squares, Discrete Math. 199 (1999), 85–101.

[6] J. Lovejoy and R. Osburn, Quadratic forms and four partition functions modulo 3, Integers 11 (2011), Paper A4.

[7] S. Radu and J. A. Sellers, Congruence properties modulo 5 and 7 for the pod function, Int. J. Number Theory 7 (2011), 2249–2259.

2010 Mathematics Subject Classification: Primary 05A17; Secondary 11P83. Keywords: congruence, partition, distinct odd parts, theta function, sum of squares.

(Concerned with sequences \underline{A000041} and \underline{A006950}.)

Received January 8 2015; revised version received February 22 2015. Published in Journal of Integer Sequences, May 12 2015.

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