Entry 5

Given the partition function \(p(n)\), Ramanujan found,
$$\begin{align}
p(5k+4) & \equiv 0 \pmod 5 \\
p(7k+5) & \equiv 0 \pmod 7 \\
p(11k+6) & \equiv 0 \pmod {11}
\end{align}$$ It was later proved in 2000 by Ken Ono, building on the work of A.O. Atkin, that such congruences exist modulo every integer \(m\) coprime to \(6\). Some examples from the link above, $$p(13 \cdot 11^3k + 237)\equiv 0 \pmod {13}$$ $$p(19 \cdot 101^4k + 815655)\equiv 0 \pmod {19}$$ $$p(31\cdot 107^4k + 30064597)\equiv 0\pmod{31}$$ Given a Ramanujan-type congruence \(p(Ak+B) \equiv 0 \pmod m\) with \(m>11\). Must the factorization of \(A\) involve an integer with a power \(> 1\)? Also, I think the one with \(m=47\) should involve smaller integers than the last three examples.