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I have learnt $f(x,y)=\sum_{m,n \in \mathbb{N}}x^ny^m$ has a closed form $\frac{1}{1-x}.\frac{1}{1-y}$, but if I modified a little bit by defining a charateristic function for coprime indexes: $a(m,n)=1$ if $(m,n)=1$ and $a(m,n)=0$, otherwise, then the situation is completely different. I am completely newbie in this area and it seems there are not much research on multivariate generating functions. So my question is: Is there a closed form for the generating function as follows:

$g(x,y)=\sum_{\gcd(m,n)=1}x^ny^m$

If it exists, could you please give me the general methods to get there?

After searching for a while, I found a related open question in Does there exist a generating function for the rational numbers?

Markiff
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  • following your msg. I added my vote to reopen this post. Will think about – G Cab Jan 04 '21 at 18:07
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    $=\sum_{d\ge 1} \frac{\mu(d)}{(x^{-d}-1)(y^{-d}-1)}$ which doesn't have a closed-form. Letting $x=e^{-u},y=e^{-v}$ then its Mellin transform is $\frac{\zeta(s)\zeta(z)}{\zeta(s+z)}$ from which we can find the asymptotic near $x=1$ and $y=1$ – reuns Jan 04 '21 at 20:44

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I don't believe this generating function has a closed form in any usual sense; already setting $x = y$ produces

$$f(x) = \sum_{\gcd(m, n) = 1} x^{m+n} = \sum_{n \ge 1} \varphi(n) x^n$$

(since $\gcd(m, n) = 1$ is equivalent to $\gcd(m, m+n) = 1$) and I don't believe this generating function has a closed form in any usual sense; in particular it is not rational because $\varphi(n)$ has the wrong growth rate. It's known that

$$\liminf_{n \to \infty} \frac{\varphi(n) \log \log n}{n} = e^{- \gamma}$$

but this limit is either $0$ or $\infty$ for the coefficients of a rational generating function, because the coefficients of a rational generating function necessarily grow like $Cn^r \lambda^n$ where $r$ is a non-negative integer. Actually this asymptotic behavior even implies that $f(x)$ is neither meromorphic nor algebraic.

Qiaochu Yuan
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  • Thanks for your answer! How about the derivative of the function $f$, the limit will be $\infty$ I think. Is there a closed form for $f'$? – Markiff Jan 04 '21 at 20:47
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    For $f'(x) = \sum_{n \ge 1} n \varphi(n) x^n$ the same argument applies, just divide by $n$. It still has the wrong growth rate. – Qiaochu Yuan Jan 04 '21 at 21:05
  • that's strange, because $$ \sum\limits_{0, \le ,n,,m} {\left[ {\gcd \left( {n,m} \right) = 1} \right]x^{,n} y^{,m} } $$ is a sub-sum of $$ \sum\limits_{0, \le ,n,,m} {x^{,n} y^{,m} } = {1 \over {\left( {1 - x} \right)\left( {1 - y} \right)}} $$ which is convergent for $|x|,|y|<1$, and in any case we are speaking of formal series – G Cab Jan 04 '21 at 23:22
  • @G Cab: I'm not making any claims about convergence or divergence. The question is whether the generating function has a closed form and I'm saying it can't be rational, meromorphic, or algebraic. (This doesn't rule out a closed form like $e^x$ but the growth rate is wrong for that also. I think with some effort we can rule out the possibility that $f(x)$ satisfies a linear differential equation with polynomial coefficients.) – Qiaochu Yuan Jan 04 '21 at 23:24
  • I see. Sorry I did not know that from the asymptotics of the coefficients one can infer about being or not the formal series meromorphic, algebraic, etc.. Interesting! – G Cab Jan 04 '21 at 23:29
  • @QiaochuYuan: please allow me to the take the chance and ask you to have a look at this question: you might possibly give me a good hint :) – G Cab Jan 04 '21 at 23:35