Inspired by this question
The linked question conjectures that $\frac{\sigma(n)}{n+1}$ (where $\sigma(n)$ denotes the divisor-sum function) is not an integer for any squarefree composite number.
If we drop the condition that $n$ is squarefree , there are still not many solutions. There is a (perhaps infinite) family of solutions , namely $$n=2^m\cdot (2^{m+1}-3)=2^{2m+1}-3\cdot 2^m$$ where $2^{m+1}-3$ is prime. In this case we have $$\sigma(n)=(2^{m+1}-1)(2^{m+1}-2)=2^{2m+2}-3\cdot 2^{m+1}+2=2(n+1)$$
The only solution I found so far that does not fit into this family is $$650=2\cdot 5^2\cdot 13$$. Current search limit is $n=10^9$
Is this the only such solution ?
Since the linked question is still open , this won't be easy , but can we at least solve the case that $n$ is not squarefree ? Has this perhaps already been verified ? If yes, a reference would be nice.