Triangulate the divisors and divide the triangulars
Can it be made even faster?


Sure.

Fact number one. An n'th triangular number is \$\frac{n(n+1)}{2}\$

Fact number two. A number of divisors (aka \$\sigma(n)\$) is multiplicative, that is \$\sigma(m*n) = \sigma(m)*\sigma(n)\$ for coprime m and n.

Fact number three. \$n\$ and \$n + 1\$ are coprime.

Which means you don't have to factorize (fairly large) triangular numbers. Instead, you only need to factorize \$n\$ and \$n+1\$; in fact, since one of them is even, and contributes as a number twice that small, at most one number needs to be factorized at each iteration. You need to memoize factorizations though.

And of course a precomputed prime numbers table may help.