## Identity involving $\sigma(i·j)$

Arithmetic, algebra, number theory, sequence and series, analysis, ...
castrate
Posts: 27
Joined: Mon Aug 19, 2019 2:34 pm

### Identity involving $\sigma(i·j)$

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 fakesson
Posts: 15
Joined: Thu Jul 12, 2018 1:06 am
Location: New York, NY

### Re: Identity involving $\sigma(i·j)$

Such an identity exists. However, I'm afraid that publishing it here would be a against the rule to provide hints to problems.
Last edited by fakesson on Wed Sep 02, 2020 7:01 pm, edited 1 time in total. castrate
Posts: 27
Joined: Mon Aug 19, 2019 2:34 pm

### Re: Identity involving $\sigma(i·j)$

I found the formula on wolframalpha and solved problem 439 on yesterday evening. Thank you all the same. pjt33
Posts: 70
Joined: Mon Oct 06, 2008 6:14 pm

### Re: Identity involving $\sigma(i·j)$

I think it's compatible with the aims and ethos of this site to say that it's easy to show that such an $F$ exists by strong induction: just take $\sigma(i\cdot j)=\sum_{x|i}\sum_{y|j}F(x,y)$ as defining $F(i, j)$. As to whether there's a more useful expression for it, one can use the inductive definition (and probably dynamic programming) to explore the value of $F$ for small $i, j$.