In this post we try to relate a sequence from The On-Line Encyclopedia of Integer Sequences and a sequence that solves an equation that involves the number-of-divisors function $\sigma_0(n)=\sum_{1\leq d\mid n}1$.
I don't know if my conjecture is easy to get (other conjectures can be done when one thinks in similar equations for other figurate numbers, for example the heptagonal pyramidal numbers in relation to A111398).
Conjecture. The integers $y\geq 1$ and $x\geq 1$ satisfy $$\sigma_0(y)=\frac{x}{3}(2x^2+1)\tag{1}$$ if and only if $y$ belongs to the sequence A162947 from the OEIS, that are the integers $y$ such that $$\prod_{1\leq d\mid y}d=y^3.$$
The expression in the RHS of $(1)$ is the formula for octahedral numbers, Wikipedia has the article with title Octahedral number.
Question. Can you to prove or provide me hints to prove the conjecture? Many thanks.
Edited. Now I see that the $\Leftarrow$ is easy, since from exercise 10 of Chapter 2 of Tom M. Apostol, Introduction to Analytic Number Theory, Springer (1976), it is possible to deduce that each term $y$ of A162947 has six divisors whenever $y>1$, and from here $x=2$ gives the solution of the equation $(1)$.
I don't know if it is easy to get (at least a part of the conjecture should be easy to get), since this sequence of integers has a very special spatial configuration. If you know it from the literature, feel free to answer as a reference request and I try to search and read the solution from the literature.
References:
[1] John Horton Conway and Richard K. Guy, The Book of Numbers, Springer-Verlag (1996).
If $x=3$, then $\frac{x}{3}(2x^2+1)=\frac{3}{3}(2(3)^2+1)=2(9)+1=18+1=19$, so $y=2^{18}$ (which has exactly $19$ divisors) is a counterexample to the conjecture.
I do not know the smallest counterexample, however.