Of Ramanujan's famous congruences for the partition function, $p(5k+4)\equiv0\mod 5$, $p(7k+5)\equiv0\mod7$, and so on, does the converse also hold? For example, if $p(n)\equiv0\mod5$, does that mean $n=5k+4$ for some $k$? If so, does this also hold for the other Ramanujan-style congruences, such as those relating to powers of primes?
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$\begingroup$ $p\left(7\right) \equiv 0 \mod 5$ but $7 \not\equiv 4 \mod 5$. $\endgroup$– darij grinbergCommented Oct 4, 2017 at 1:00
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No. For example, $p(7) = 15$ and $p(10) = 42$.
answered Oct 4, 2017 at 1:00