\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf New Congruences for Partitions where the \\ \vskip .1in Odd Parts are Distinct } \vskip 1cm \large Liuquan Wang\\ Department of Mathematics\\ National University of Singapore\\ Singapore, 119076\\ Singapore\\ \href{wangliuquan@nus.edu.sg}{\tt wangliuquan@u.nus.edu}\\ \href{mathlqwang@163.com}{\tt mathlqwang@163.com}\\ \end{center} \vskip .2 in \begin{abstract} Let $\mathrm{pod}(n)$ denote the number of partitions of $n$ wherein odd parts are distinct (and even parts are unrestricted). We find some new interesting congruences for $\mathrm{pod}(n)$ modulo 3, 5 and 9. \end{abstract} \section {Introduction and Main Results} Let $\psi(q)$ be one of Ramanujan's theta functions, namely \[\psi (q)=\sum\limits_{n=0}^{\infty }{{{q}^{n(n+1)/2}}}=\frac{({{q}^{2}};{{q}^{2}})_{\infty }^{2}}{{{(q;q)}_{\infty }}}.\] We let $\mathrm{pod}(n)$ (see \seqnum{A006950}) denote the number of partitions of $n$ wherein the odd parts are distinct (and even parts are unrestricted). For example, $\mathrm{pod}(4)=3$ 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 $\mathrm{pod}(n)$ is given by \begin{equation}\label{gen} \sum\limits_{n=0}^{\infty }{\mathrm{pod}(n){{q}^{n}}}=\frac{(-q;{{q}^{2}})_{\infty }}{({{q}^{2}};{{q}^{2}})_{\infty }}=\frac{1}{\psi {{(-q)}}}. \end{equation} The arithmetic properties of $\mathrm{pod}(n)$ were first studied by Hirschhorn and Sellers \cite{hisc} in 2010. They obtained some interesting congruences involving the following infinite family of Ramanujan-type congruences: for any integers $\alpha \ge 0$ and $n \ge 0$, \[\mathrm{pod}\Big({{3}^{2\alpha +3}}n+\frac{23\times {{3}^{2\alpha +2}}+1}{8}\Big)\equiv 0 \pmod{3}.\] Later on Radu and Sellers \cite{radu} obtained other deep congruences for $\mathrm{pod}(n)$ modulo $5$ and $7$, such as \[\mathrm{pod}(135n+8)\equiv \mathrm{pod}(135n+107)\equiv \mathrm{pod}(135n+116)\equiv 0 \pmod{5}, \quad \text{\rm{and}}\] \[\mathrm{pod}(567n+260)\equiv \mathrm{pod}(567n+449)\equiv 0 \pmod{7}.\] 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 $\mathrm{pod}(3n+2)$ found in \cite{hisc}, Lovejoy and Osburn \cite{lovejoy} discovered the following arithmetic relation: \begin{equation}\label{osburn} \mathrm{pod}(3n+2)\equiv {{(-1)}^{n}}{{r}_{5}}(8n+5) \pmod{3}. \end{equation} Following their steps, we will present some new congruences modulo 5 and 9 for $\mathrm{pod}(n)$. Firstly, we find that (\ref{osburn}) can be improved to a congruence modulo 9. \begin{theorem}\label{mod9} For any integer $n\ge 0$, we have \[\mathrm{pod}(3n+2)\equiv 2{{(-1)}^{n+1}}{{r}_{5}}(8n+5) \pmod{9}.\] \end{theorem} The following result will be a consequence of Theorem \ref{mod9} upon invoking some properties of $r_{5}(n)$. \begin{theorem}\label{modcor}\label{mod9cor} Let $p\ge 3$ be a prime, and $N$ be a positive integer such that $pN \equiv 5$ \text{\rm{(mod 8)}}. Let $\alpha$ be any nonnegative integer.\\ (1) If $p\equiv 1$ \text{\rm{(mod 3)}}, then \[\mathrm{pod}\Big(\frac{3{{p}^{6\alpha +5}}N+1}{8}\Big)\equiv 0 \pmod{3},\] and \[\mathrm{pod}\Big(\frac{3{{p}^{18\alpha +17}}N+1}{8}\Big)\equiv 0 \pmod{9}.\] (2) If $p\equiv 2$ \text{\rm{(mod 3)}}, then \[\mathrm{pod}\Big(\frac{3{{p}^{4\alpha +3}}N+1}{8}\Big)\equiv 0 \pmod{9}.\] \end{theorem} Secondly, with the same method used in proving Theorem \ref{mod9}, we can establish a similar congruence for $\mathrm{pod}(n)$ modulo 5. \begin{theorem}\label{mod5} For any integer $n\ge 0$, we have \[\mathrm{pod}(5n+2)\equiv 2{{(-1)}^{n}}{{r}_{3}}(8n+3) \pmod{5}.\] \end{theorem} Some miscellaneous congruences can be deduced from this theorem. \begin{theorem}\label{mod5cor} For any integers $n \ge 0$ and $\alpha \ge 1$, we have \[\mathrm{pod}\Big({{5}^{2\alpha +2}}n+\frac{11\cdot {{5}^{2\alpha +1}}+1}{8}\Big)\equiv 0 \pmod{5},\] and \[\mathrm{pod}\Big({{5}^{2\alpha +2}}n+\frac{19\cdot {{5}^{2\alpha +1}}+1}{8}\Big)\equiv 0 \pmod{5}.\] \end{theorem} \begin{theorem}\label{cubemod5} Let $p\equiv 4$ \text{\rm{(mod 5)}} be a prime, and $N$ be a positive integer which is coprime to $p$ such that $pN \equiv 3$ \text{\rm{(mod 8)}}. We have \[\mathrm{pod}\Big(\frac{5{{p}^{3}}N+1}{8}\Big)\equiv 0 \pmod{5}.\] \end{theorem} For example, let $p=19$ and $N=8n+1$ where $n\ge 0$ and $n \not\equiv 7$ (mod 19). We have \[\mathrm{pod}(34295n+4287)\equiv 0 \pmod{5}.\] \begin{theorem}\label{2case} Let $p\ge 3$ be a prime, and $N$ be a positive integer which is not divisible by $p$ such that $pN \equiv 3$ \text{\rm{(mod 8)}}. Let $\alpha $ be any nonnegative integer. \\ (1) If $p\equiv 1$ \text{\rm{(mod 5)}}, we have \[\mathrm{pod}\Big(\frac{5 {{p}^{10\alpha +9}}N+1}{8}\Big)\equiv 0 \pmod{5}.\] (2) If $p\equiv 2,3,4$ \text{\rm{(mod 5)}}, we have \[\mathrm{pod}\Big(\frac{5 {{p}^{8\alpha +7}}N+1}{8}\Big)\equiv 0 \pmod{5}.\] \end{theorem} \section{Preliminaries} \begin{lemma}\label{palpha} (Cf.\ \cite[Lemma 1.2]{radu}.) Let $p$ be a prime and $\alpha $ be a positive integer. Then \[(q;q)_{\infty }^{{{p}^{\alpha }}}\equiv ({{q}^{p}};{{q}^{p}})_{\infty }^{{{p}^{\alpha -1}}} \pmod{{{p}^{\alpha }}}.\] \end{lemma} \begin{lemma}\label{t4t8} For any prime $p \ge 3$, we have \[{{t}_{4}}\Big(pn+\frac{p-1}{2}\Big)\equiv {{t}_{4}}(n) \pmod{p}, \quad {{t}_{8}}(pn+p-1)\equiv {{t}_{8}}(n) \pmod{p^3}.\] \end{lemma} \begin{proof} By \cite[Theorem 3.6.3]{Bruce}, we know ${{t}_{4}}(n)=\sigma (2n+1).$ For any positive integer $N$, we have \[\sigma (N)=\sum\limits_{d|N, \, p|d}{d}+\sum\limits_{d|N, \,p \nmid d}{d}\equiv \sum\limits_{d|N, \, p \nmid d}{d} \pmod{p}.\] Let $N=2n+1$ and $N=p(2n+1)$, respectively. It is easy to deduce that $\sigma (p(2n+1))\equiv \sigma (2n+1)$ (mod $p$). This clearly implies the first congruence. From \cite[Eq.(3.8.3), page 81]{Bruce}, we know \[{{t}_{8}}(n)=\sum\limits_{\begin{smallmatrix} d|(n+1) \\ d \, \mathrm{odd} \end{smallmatrix}}{{{\Big(\frac{n+1}{d}\Big)}^{3}}}. \] By a similar argument we can prove the second congruence. \end{proof} \begin{lemma}\label{rsrelate} (Cf.\ \cite{relation}.) For $1\le k\le 7$, we have \[{{r}_{k}}(8n+k)={{2}^{k}}\Big(1+\frac{1}{2}\binom{k}{4}\Big){{t}_{k}}(n).\] \end{lemma} \begin{lemma}\label{r5relation} (Cf.\ \cite{Cooper}.) Let $p\ge 3$ be a prime and $n$ be a positive integer such that ${{p}^{2}} \nmid n$. For any integer $\alpha \ge 0$, we have \[{{r}_{5}}({{p}^{2\alpha }}n)=\Bigg(\frac{{{p}^{3\alpha +3}}-1}{{{p}^{3}}-1}-p\Big(\frac{n}{p}\Big)\frac{{{p}^{3\alpha }}-1}{{{p}^{3}}-1}\Bigg){{r}_{5}}(n),\] where $(\frac{\cdot }{p})$ denotes the Legendre symbol. \end{lemma} \begin{lemma}\label{r3relation} (Cf.\ \cite{3square}.) Let $p \ge 3$ be a prime. For any integers $n \ge 1$ and $\alpha \ge 0$, we have \[{{r}_{3}}({{p}^{2 \alpha }}n)=\Bigg(\frac{{{p}^{\alpha +1}}-1}{p-1}-\Big(\frac{-n}{p}\Big)\frac{{{p}^{\alpha }}-1}{p-1}\Bigg){{r}_{3}}(n)-p\frac{{{p}^{\alpha }}-1}{p-1}{{r}_{3}}(n/{{p}^{2}}),\] where we take ${{r}_{3}}(n/{{p}^{2}})=0$ unless ${{p}^{2}} | n$ . \end{lemma} \section{Proofs of the Theorems} \begin{proof}[Proof of Theorem \ref{mod9}] Let $p=3$ in Lemma \ref{t4t8}. We deduce that ${{t}_{8}}(3n+2)\equiv {{t}_{8}}(n)$ (mod 9). By (\ref{gen}) we have \[\psi {{(q)}^{9}}\sum\limits_{n=0}^{\infty }{\mathrm{pod}(n){{(-q)}^{n}}}=\psi {{(q)}^{8}}=\sum\limits_{n=0}^{\infty }{{{t}_{8}}(n){{q}^{n}}}.\] By Lemma \ref{palpha} we obtain $\psi {{(q)}^{9}}\equiv {\psi ({{q}^{3}})}^3$ (mod 9). Hence \[{\psi (q^3)}^3\sum\limits_{n=0}^{\infty }{\mathrm{pod}(n){{(-q)}^{n}}} \equiv \sum\limits_{n=0}^{\infty }{{{t}_{8}}(n){{q}^{n}}} \pmod{9}.\] If we extract those terms of the form ${{q}^{3n+2}}$ on both sides, we obtain \[{\psi (q^3)}^3 \sum\limits_{n=0}^{\infty }{\mathrm{pod}(3n+2){{(-q)}^{3n+2}}}\equiv \sum\limits_{n=0}^{\infty }{{{t}_{8}}(3n+2){{q}^{3n+2}}} \pmod{9}.\] Dividing both sides by ${{q}^{2}}$, then replacing ${{q}^{3}}$ by $q$, we get \[\psi {{(q)}^{3}}\sum\limits_{n=0}^{\infty }{\mathrm{pod}(3n+2){{(-q)}^{n}}}\equiv \sum\limits_{n=0}^{\infty }{{{t}_{8}}(3n+2){{q}^{n}}}\equiv \sum\limits_{n=0}^{\infty }{{{t}_{8}}(n){{q}^{n}}}=\psi {{(q)}^{8}} \pmod{9}.\] Hence \[\sum\limits_{n=0}^{\infty }{\mathrm{pod}(3n+2){{(-q)}^{n}}}\equiv \psi {{(q)}^{5}}\equiv \sum\limits_{n=0}^{\infty }{{{t}_{5}}(n){{q}^{n}}} \pmod{9}.\] Comparing the coefficients of $q^{n}$ on both sides, we deduce that $\mathrm{pod}(3n+2)\equiv {{(-1)}^{n}}{{t}_{5}}(n)$ (mod 9). Let $k=5$ in Lemma \ref{rsrelate}. We obtain ${{t}_{5}}(n)={{r}_{5}}(8n+5)/112$, and from this the theorem follows. \end{proof} \begin{proof}[Proof of Theorem \ref{mod9cor}] (1) Let $n=pN$ in Lemma \ref{r5relation}, and then we replace $\alpha $ by $3\alpha +2$. Since \[\frac{{{p}^{9\alpha +9}}-1}{{{p}^{3}}-1}=1+{{p}^{3}}+\cdots +{{p}^{3(3\alpha +2)}}\equiv 0 \pmod{3},\] we deduce that ${{r}_{5}}({{p}^{6\alpha +5}}N)\equiv 0$ (mod 3). Let $n=\frac{{{p}^{6\alpha +5}}N-5}{8}$ in Theorem \ref{mod9}. We deduce that $\mathrm{pod}(\frac{3{{p}^{6\alpha +5}}N+1}{8})\equiv 0$ (mod 3). Similarly, let $n=pN$ in Lemma \ref{r5relation} and we replace $\alpha $ by $9\alpha +8$. Since $p\equiv 1$ (mod 3) implies ${{p}^{3}}\equiv 1$ (mod 9), we have \[\frac{{{p}^{27\alpha +27}}-1}{{{p}^{3}}-1}=1+{{p}^{3}}+\cdots +{{p}}^{3(9\alpha +8)}\equiv 0 \pmod{9}.\] Hence ${{r}_{5}}({{p}^{18\alpha +17}}N)\equiv 0$ (mod 9). Let $n=\frac{{{p}^{18\alpha +17}}N-5}{8}$ in Theorem \ref{mod9}. We deduce that $\mathrm{pod}(\frac{3{{p}^{18\alpha +17}}N+1}{8})\equiv 0$ (mod 9). \medskip (2) Let $n=pN$ in Lemma \ref{r5relation}, and then we replace $\alpha $ by $2\alpha +1$. Note that $p\equiv 2$ (mod 3) implies ${{p}^{3}}\equiv -1$ (mod 9). Since ${{p}^{6\alpha +6}}\equiv 1$ (mod 9), we have ${{r}_{5}}({{p}^{4\alpha +3}}N)\equiv 0$ (mod 9). Let $n=\frac{{{p}^{4\alpha +3}}N-5}{8}$ in Theorem \ref{mod9}. We complete our proof. \end{proof} \begin{proof}[Proof of Theorem \ref{mod5}] Let $p=5$ in Lemma \ref{t4t8}. We deduce that ${{t}_{4}}(5n+2)\equiv {{t}_{4}}(n)$ (mod 5). By (\ref{gen}) we have \[\psi {{(q)}^{5}}\sum\limits_{n=0}^{\infty }{\mathrm{pod}(n){{(-q)}^{n}}}=\psi {{(q)}^{4}}=\sum\limits_{n=0}^{\infty }{{{t}_{4}}(n){{q}^{n}}}.\] By Lemma \ref{palpha} we obtain $\psi {{(q)}^{5}}\equiv \psi ({{q}^5})$ (mod 5). Hence \[\psi {{(q^{5})}}\sum\limits_{n=0}^{\infty }{\mathrm{pod}(n){{(-q)}^{n}}} \equiv \sum\limits_{n=0}^{\infty }{{{t}_{4}}(n){{q}^{n}}} \pmod{5}.\] If we extract those terms of the form ${{q}^{5n+2}}$ on both sides, we obtain \[\psi (q^5) \sum\limits_{n=0}^{\infty }{\mathrm{pod}(5n+2){{(-q)}^{5n+2}}}\equiv \sum\limits_{n=0}^{\infty }{{{t}_{4}}(5n+2){{q}^{5n+2}}} \pmod{5}.\] Dividing both sides by ${{q}^{2}}$, and then replacing ${{q}^{5}}$ by $q$, we get \[\psi (q)\sum\limits_{n=-\infty }^{\infty }{\mathrm{pod}(5n+2){{(-q)}^{n}}}\equiv \sum\limits_{n=0}^{\infty }{{{t}_{4}}(5n+2){{q}^{n}}}\equiv \sum\limits_{n=0}^{\infty }{{{t}_{4}}(n){{q}^{n}}}=\psi {{(q)}^{4}} \pmod{5}.\] Hence we have \[\sum\limits_{n=0}^{\infty }{\mathrm{pod}(5n+2){{(-q)}^{n}} }\equiv \psi {{(q)}^{3}}= \sum\limits_{n=0}^{\infty }{{{t}_{3}}(n){{q}^{n}}} \pmod{5}.\] Comparing the coefficients of $q^{n}$ on both sides, we deduce that $\mathrm{pod}(5n+2)\equiv {{(-1)}^{n}}{{t}_{3}}(n)$ (mod 5). Let $k=3$ in Lemma \ref{rsrelate}. We obtain ${{t}_{3}}(n)={{r}_{3}}(8n+3)/8$, from which the theorem follows. \end{proof} \begin{proof}[Proof of Theorem \ref{mod5cor}] Let $p=5$ and $n=5m+r$ ($r\in \{1,4\}$) in Lemma \ref{r3relation}. Since $\big(\frac{-r}{5}\big)=1$, we deduce that ${{r}_{3}}({{5}^{2\alpha }}(5m+r))\equiv 0$ (mod 5) for any integer $\alpha \ge 1$. Let $n=\frac{{{5}^{2\alpha }}(40m+a)-3}{8}$ ($a\in \{11,19\}$). By Theorem \ref{mod5}, we have \[{{r}_{3}}(8n+3)={{r}_{3}}({{5}^{2\alpha }}(40m+a))\equiv 0 \pmod{5}.\] Hence \[\mathrm{pod}\Big({{5}^{2\alpha +2}}m+\frac{a\cdot {{5}^{2\alpha +1}}+1}{8}\Big)=\mathrm{pod}(5n+2)\equiv 2{{(-1)}^{n}}{{r}_{3}}(8n+3)\equiv 0 \pmod{5}.\] \end{proof} \begin{proof}[Proof of Theorem \ref{cubemod5}] Let $\alpha =1$ and $n=pN$ in Lemma \ref{r3relation}. We have \[{{r}_{3}}({{p}^{3}}N)=(1+p){{r}_{3}}(pN)\equiv 0 \pmod{5}.\] Let $n=\frac{{{p}^{3}}N-3}{8}$ in Theorem \ref{mod5}. We have \[\mathrm{pod}\Big(\frac{5{{p}^{3}}N+1}{8}\Big)=\mathrm{pod}(5n+2)\equiv 2{{(-1)}^{n}}{{r}_{3}}(8n+3)=2{{(-1)}^{n}}{{r}_{3}}({{p}^{3}}N)\equiv 0 \pmod{5}.\] \end{proof} \begin{proof}[Proof of Theorem \ref{2case}] (1) Let $n=pN$ in Lemma \ref{r3relation}, and then we replace $\alpha $ by $5\alpha +4$. We have \[\frac{{{p}^{5\alpha +5}}-1}{p-1}=1+p+\cdots +{{p}^{5\alpha +4}}\equiv 0 \pmod{5}.\] Hence ${{r}_{3}}({{p}^{10\alpha +9}}N)\equiv 0$ (mod 5). Let $n=\frac{{{p}^{10\alpha +9}}N-3}{8}$ in Theorem \ref{mod5}. We have \[\mathrm{pod}\Big(\frac{5{{p}^{10\alpha +9}}N+1}{8}\Big)=\mathrm{pod}(5n+2) \equiv 2{{(-1)}^{n}}{{r}_{3}}({{p}^{10\alpha +9}}N)\equiv 0 \pmod{5}.\] \medskip (2) Let $n=pN$ in Lemma \ref{r3relation}, and then replace $\alpha $ by $4\alpha +3$. Since ${{p}^{4\alpha +4}}\equiv 1$ (mod 5), we deduce that ${{r}_{3}}({{p}^{8\alpha +7}}N)\equiv 0$ (mod 5). Let $n=\frac{{{p}^{8\alpha +7}}N-3}{8}$ in Theorem \ref{mod5}. We have \[\mathrm{pod}\Big(\frac{5{{p}^{8\alpha +7}}N+1}{8}\Big)=\mathrm{pod}(5n+2)\equiv 2{{(-1)}^{n}}{{r}_{3}}({{p}^{8\alpha +7}}N)\equiv 0 \pmod{5}.\] \end{proof} \begin{thebibliography}{9} \bibitem{relation} P. Barrucand, S. Cooper, and M. D. Hirschhorn, Relations between squares and triangles, \emph{Discrete Math.} \textbf{248} (2002), 245--247. \bibitem{Bruce}B. C. Berndt, \emph{Number Theory in the Spirit of Ramanujan}. Amer. Math. Soc., 2006. \bibitem{Cooper} S. Cooper, Sums of five, seven and nine squares, \emph{Ramanujan J}. \textbf{6} (2002), 469--490. \bibitem{hisc} M. D. Hirschhorn and J. A. Sellers, Arithmetic properties of partitions with odd parts distinct, \emph{Ramanujan J.} \textbf{22} (2010), 273--284. \bibitem{3square} M. D. Hirschhorn and J. A. Sellers, On representations of a number as a sum of three squares, \emph{Discrete Math.} \textbf{199} (1999), 85--101. \bibitem{lovejoy} J. Lovejoy and R. Osburn, Quadratic forms and four partition functions modulo 3, \emph{Integers} \textbf{11} (2011), Paper A4. \bibitem{radu} S. Radu and J. A. Sellers, Congruence properties modulo 5 and 7 for the pod function, \emph{Int. J. Number Theory} \textbf{7} (2011), 2249--2259. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 05A17; Secondary 11P83. \noindent \emph{Keywords: } congruence, partition, distinct odd parts, theta function, sum of squares. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A000041} and \seqnum{A006950}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received January 8 2015; revised version received February 22 2015. Published in {\it Journal of Integer Sequences}, May 12 2015. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}