How do I solve this: $\sigma(n)=\frac{3n(n-1)}{2}$?

Question

I would like to know the relationship between power of sum divisor function $\displaystyle\sigma(n)=\sum_{d|n} d$ and pentongonal numbers then i'm seeking to solve this equation :$$\displaystyle\sigma(n)=\frac{3n(n-1)}{2}$$

Note: For instance i have $n=2$ is a solution ,are there other ?

Thank you for any help !!!

Answer 1

Since the sum of divisors function $\sigma(n)$ must be less than or equal to the sum of all the numbers less than or equal to $n$, and $\sum_{i=1}^n i=\frac{n(n+1)}{2}$ and $3n(n-1) > n(n+1)$ if $n>2$ there are no solutions other than 2.

Perhaps you meant to ask when $\sigma(n)=\frac{3k(k-1)}{2}$ for some $k$. In other words, when is $\sigma(n)$ pentagonal?

Answer 2

Since $\sigma(n) < e^{\gamma}n\log\log n +\dfrac{0.6483 n}{\log\log n}$, there are at most a finite number, probably none, of solutions.

See https://en.m.wikipedia.org/wiki/Divisor_function