Can an odd perfect number be a nontrivial multiple of a triangular number?

Question

(Note: This question has been cross-posted to MO.)

Can an odd perfect number be a nontrivial multiple of a triangular number?

Answer 1

The answer turns out to be YES.

Claim

If $N = q^k n^2$ is an odd perfect number, then $N$ can be written in the form $$N = \dfrac{q(q+1)}{2}\cdot{d}$$ for some integer $d>1$.

Proof

By Slowak's 1999 result, every odd perfect number must have the form $$\dfrac{{q^k}\sigma(q^k)}{2}\cdot{d'}$$ for some integer $d' > 1$.

If $k=1$, then our claim readily follows from Slowak's result.

Otherwise, if $k>1$, since $q \mid q^k$ and $(q + 1) \mid \sigma(q^k)$ for all $k \equiv 1 \pmod 4$, then we have that every odd perfect number must have the form $$\dfrac{q(q+1)}{2}\cdot{d''}$$ for some integer $d'' = ({q^{k-1}}{\sigma(q^k)}{d'})/(q+1) > 1$.