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.