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Let $t\in\Bbb{N}$ and consider the sequences $p_t(n)$ defined by $$\sum_{n\geq0}p_t(n)x^n=\prod_{i\geq1}\frac1{(1-x^i)^t}=(x;x)_{\infty}^{-t}.$$ The numbers $p_t(n)$ can be regarded as enumerating partitions of $n$ into parts that come with $t$ colors. Furthermore, $p_t(n)=\sum_{\lambda\vdash n}\prod_{j\geq1}\binom{k_j+t-1}{t-1}$ where $\lambda=1^{k_1}2^{k_2}\cdots$ and each $k_j\geq0$. Note also that $p_1(n)=p(n)$ is the usual number of (unrestricted) integer partitions of $n$. Ramanujan's famous congruences state $$\begin{cases} p(5n+4)\equiv0\mod 5, \\ p(7n+5)\equiv0\mod 7, \\ p(11n+6)\equiv0\mod 11. \end{cases}$$

In the same spirit, the following appear to be true. Are they? $$\begin{cases} p_t(5n+4)\equiv0\mod 5, \qquad t\equiv0,1,2,4\mod 5 \\ \,p_t(7n+5)\equiv0\mod 7, \qquad \,\,t\equiv0,1,4 \,\,\,\, \mod 7\\ p_t(11n+6)\equiv0\mod 11, \qquad t\equiv0,1,10\mod 11. \end{cases}$$

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    $\begingroup$ Nice. You could add the "generating-functions" and the "ramanujan" tag. $\endgroup$
    – Wolfgang
    May 4, 2018 at 19:00
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    $\begingroup$ The cases $t=0,1 \mod p$ ($p=5,7,11$) follow from finite field arithmetic. Indeed, denoting the generating function of $p_1$ by f, we have: $f^{pn_1 + n_0} (x) = f(x^ {p})^{n_1} f(x)^{n_0} \mod p$ for any prime p and $n_0<p$. By comparing coefficients, $p_{p n_1 + n_0}(p m_1 + m_0) = \sum_{i=0}^{m_1} p_{n_1}(i)* p_{n_0}(p(m_1 -i)+ m_0) \mod p$ for $m_0 < p$. In the case of $p=5$ for instance, we may take $n_0=1, m_0=4$ and use Ramamujan's congruences to obtain your result for t=1. For t=0 it is even easier, as it corresponds to $n_0 = 0$, and $p_{0}(j)$ is indicator of 0. $\endgroup$
    – Ofir Gorodetsky
    May 4, 2018 at 20:47
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    $\begingroup$ See also: (Freeman) Dyson's crank. It's a great story. $\endgroup$
    – shreevatsa
    May 5, 2018 at 4:49

1 Answer 1

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More general versions of this have been established: see in particular Theorem 2 of Kiming and Olsson, and for other work see (for example) Locus and Wagner.

To answer the question fully, as Ofir Gorodetsky observed the problem is trivial when $t$ is $0$ or $1$ modulo the prime modulus. The other cases correspond to $t \equiv \ell-1$ or $t\equiv \ell-3$ modulo $\ell$ (with $\ell$ being $5$, $7$, or $11$). These cases are the "non-exceptional congruences" covered by Theorem 2 of the paper by Kiming and Olsson (and the argument there is essentially elementary following easily from Euler's pentagonal number theorem in the case of $\ell-1$, and an identity of Jacobi in the case of $\ell-3$). The real interest is in the situation of exceptional congruences, and Theorem 4 of Kiming and Olsson gives some examples of these. For instance, if $t\equiv 3 \mod 11$ then $$ p_{t}(11n+7) \equiv 0 \pmod{11}. $$

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  • $\begingroup$ I appreciate the details $\endgroup$
    – T. Amdeberhan
    May 5, 2018 at 13:47

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