TITLE: Has it been proved that odd perfect numbers cannot be triangular?
QUESTION [14 upvotes]: (Note:  This question has been cross-posted from MSE.)
Euclid and Euler proved that every even perfect number is of the form $m = \frac{{M_p}\left(M_p + 1\right)}{2}$ where $M_p = 2^p - 1$ is a prime number, called a Mersenne prime.  Thus, an even perfect number is triangular.
On the other hand, Euler showed that an odd perfect number, if one exists, takes the form $N = q^k n^2$, where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n) = 1$.  (Descartes, Frenicle and subsequently Sorli conjectured that $k = 1$ always holds.)
Here is my question:

Has it been proved that odd perfect numbers cannot be triangular?

Added March 26 2016
If $\sigma(q) = 2n^2$, then it would follow that $n < q$, which implies that $k = 1$.  The odd perfect number $N = q^k n^2$ then takes the form $N = \frac{q(q + 1)}{2}$.  Unfortunately, it is known that $\sigma(q^k) \leq \frac{2n^2}{3}$.
Any pointers to the existing literature containing such a proof would be most appreciated.

REPLY [3 votes]: Not an answer, but I just want to point out some thoughts that recently occurred to me, which are related to this problem.
By this answer, we know that every odd perfect number $N = q^k n^2$ can be written in the form
$$N = \dfrac{q(q+1)}{2} \cdot d$$
where $d > 1$.  (That is, an odd perfect number is a nontrivial multiple of the triangular number
$$T(q) = \dfrac{q(q+1)}{2},$$
where $q$ is the Euler prime of $N$.)
If $k=1$, then it is easy to show that
$$d = D(n^2)$$
where $D(n^2) = 2n^2 - \sigma(n^2)$ is the deficiency of the non-Euler part $n^2$.
(I have asked a separate question for the form of $d'$ here, if $N = q^k n^2$ is an odd perfect number with $k > 1$, and $N = \dfrac{q(q+1)}{2} \cdot d'$ with $d' > 1$.)