# Can the partition function $p(n)$ take perfect power values?

Recall that the perfect powers are those integers $$m^k$$ with $$k,m\in\{2,3,\ldots\}$$. I don't consider $$0$$ or $$1$$ as a perfect power.

Y. Bugeaud, M. Mignotte and S. Siksek [Annals of Math., 2006] proved that the only perfect powers in the Fibonacci sequence are $$F_6=2^3$$ and $$F_{12}=12^2$$.

Let $$p(n)$$ be the partition function in number theory. By the Hardy-Ramanujan formula, $$\lim_{n\to\infty}\frac{\log p(n)}{\sqrt n}=\pi\sqrt{\frac23}.$$ So, $$p(n)$$ eventually grows faster than any polynomial but slower than exponential functions.

QUESTION: Can $$p(n)$$ be a perfect power?

In 2013 I conjectured that this question has a negative answer and verified this for $$n\le 15000$$.

Remark. I also conjecture that no Bell number (or Franel number $$f_n=\sum_{k=0}^n\binom nk^3$$, or Apery number $$A_n=\sum_{k=0}^n\binom nk^2\binom{n+k}k^2$$) is a perfect power. For the prime-counting function $$\pi(x)$$, I guess that $$\pi(2^3)=4$$ is the only perfect power in the increasing sequence $$\pi(2^n)\ (n=1,2,3,\ldots)$$.