The equation $\sigma(n)=\sigma(n+1)$ In OEIS, the solutions of $$\sigma(n)=\sigma(n+1)$$ where $\sigma(n)$ denotes the sum of the divisors of $n$ including $1$ and $n$ , are shown upto $n=10^{13}$
The entry can be found already by entering the first three solutions $14,206,957$
It is mentioned that it is unknown whether there are infinite many solutions.
My questions :


*

*Has $$\sigma(n)=\sigma(n+1)=\sigma(n+2)$$ a solution ? I checked the OEIS-entries and upto $10^{13}$ there is none. Can we perhaps show that this double-equation cannot have a solution ?

*Is a family of solutions known that is probably infinite (but not proven, since the problem whether infinite many solutions exist is open) ?

*Can we find the next solution more efficiently than by just checking all cases ?
 A: Just sharing some ideas that could be useful for resolving this problem, and which are too long to fit in the Comments section:


*

*The expression $n(n+1)(n+2)$ is divisible by $6$, which is an even perfect number.

*We can use the divisibility constraint $\gcd(n,n+1)=\gcd(n+1,n+2)=1$, and $\gcd(n,n+2)=1$ (if $n$ is odd).

*We can then apply the $\sigma$-function to the products
$$n(n+1)$$
$$(n+1)(n+2)$$
and $n(n+2)$ (if $n$ is odd in this last case).

*Then note that we have
$$\sigma(n(n+1))=\sigma(n)\sigma(n+1)=\sigma((n+1)(n+2))=\sigma(n+1)\sigma(n+2)=k^2$$
where $k=\sigma(n)=\sigma(n+1)=\sigma(n+2)$ is the common value.

*If $n$ is odd, note that we have
$$\sigma(n(n+2))=\sigma(n)\sigma(n+2)=k^2.$$

*So now, consider the entire product $n(n+1)(n+2)$.  This is a nontrivial multiple of $6$ if $n > 1$, so that
$$\frac{\sigma(n(n+1)(n+2))}{n(n+1)(n+2)}>2.$$
Assume to the contrary that $n>1$ is odd.  Then the last inequality implies
$$\frac{\sigma(n)}{n}\frac{\sigma(n+1)}{n+1}\frac{\sigma(n+2)}{n+2}>2,$$
so that we obtain
$$k^3=\sigma(n)\sigma(n+1)\sigma(n+2)>2n(n+1)(n+2).$$
We therefore get the inequality
$$\sigma(n)=\sigma(n+1)=\sigma(n+2)>\sqrt[3]{2n(n+1)(n+2)}.$$
I will stop here.
