### Identity involving $\sigma(i·j)$

Posted:

**Sat Aug 29, 2020 2:23 am**As far as I know, there is an identity involving d(i·j), where d(n) denotes the number of divisors of n:

$d(i·j)=\sum_{x|i}\sum_{y|j}[gcd(x,y)=1]$

where [...]=1 if the condition holds, otherwise 0.

However, when I was looking for similar identities involving $\sigma(i·j)$ in the form of $\sigma(i·j)=\sum_{x|i}\sum_{y|j}F(x,y)$, I encountered some difficulties. I guessed several possible forms, but through calculations of small numbers, they all turned out to be wrong. Functions d(n) and $\sigma(n)$ are similar in many ways (eg. both multiplicative, both belong to the more general function $\sigma_k(n)$), so I assume there should be a similar identity involving $\sigma(i·j)$.

So does the similar identity exist? If so, could anyone provide it? Thanks

$d(i·j)=\sum_{x|i}\sum_{y|j}[gcd(x,y)=1]$

where [...]=1 if the condition holds, otherwise 0.

However, when I was looking for similar identities involving $\sigma(i·j)$ in the form of $\sigma(i·j)=\sum_{x|i}\sum_{y|j}F(x,y)$, I encountered some difficulties. I guessed several possible forms, but through calculations of small numbers, they all turned out to be wrong. Functions d(n) and $\sigma(n)$ are similar in many ways (eg. both multiplicative, both belong to the more general function $\sigma_k(n)$), so I assume there should be a similar identity involving $\sigma(i·j)$.

So does the similar identity exist? If so, could anyone provide it? Thanks