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Hello everyone

In connection with calculating the Fourier coefficients of some quasi-modular forms which I have been looking at lately, I have come across the following type of sum

$$ S_{a,b}(N) := \sum_{t=1}^{N-1} \ \ \sum_{(n,m) \in I_{N-t,t}} \frac{1}{m^a n^b} $$
where $$ I_{k,l} = \{ \ (m,n) \ \big| \ \ m|k \ , \ n| l \ , \ m>n \}$$ Note the final condition in $I_{k,l}$ which is of course what prevents this sum from simply being a sum of products of divisor functions. My questions are

1.) Can anybody think of a way of calculating this sum rapidly for large $N$. Perhaps by somehow expressing it as a sum of some modified divisor functions or something similar.

2.) Has series of this type occurred elsewhere in math?

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  • $\begingroup$ What happens if you use J_{k,l} which is your I_{k,l} except with m <= n? Can you say anything about either sum? Gerhard "Ask Me About System Design" Paseman, 2010.12.09 $\endgroup$ Dec 9, 2010 at 17:15
  • $\begingroup$ I think the sum you suggest would just be my sum plus $\sigma_{a+b}(N)$, though I haven't thought it through thoroughly. Do you have any idea's in that case? $\endgroup$
    – torben
    Dec 10, 2010 at 21:00
  • $\begingroup$ Ahh, no sorry, read your suggestion as $m\geq n $ for some reason. The sum you actually suggest would be something like $\sum_{t=1}^{N-1} sigma_a(t)\sigma_b(N-t) - S_{a,b}(N)$ right? Either way I am still interested if you have some thoughts/references in this case. $\endgroup$
    – torben
    Dec 10, 2010 at 21:03
  • $\begingroup$ I am not allowed to edit comments? $\endgroup$
    – torben
    Dec 10, 2010 at 21:05
  • 1
    $\begingroup$ The only way to "edit" a comment is to delete it and re-type it. $\endgroup$ Dec 10, 2010 at 21:29

2 Answers 2

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This function seems related to Percy MacMahon's generalized sum-of-divisors functions. A generating function can be given as follows:

Let $A_k(q) = \sum_n a_{n,k}q^n$ where $a_{n,k}$ is given by the sum $\sum s_1 \cdots s_k$ taken over all ways of writing $k = m_1s_1 + \cdots m_ks_k$ with $0 < m_1 < \cdots < m_k$.

Specifically, $a_{2,k}$ looks similar to what you've written above: You're summing over all ways of partitioning $N$ into two numbers, and summing over the divisors of those numbers... or at least some power of their reciprocals.

I don't see exactly how to write what you've done in terms of these functions (note the condition that you have that $m > n$ is the reverse of the inequality in MacMahon's function i.e. you're summing over the product of the $m_k$ in his function instead of the $s_k$), but they're not too far off.

There is some literature on these functions; there's MacMahon's original paper "Divisors of numbers and their continuations in the theory of partitions", as well as a few papers by George Andrews more recently (one co-authored by me). These generating functions are quasi-modular forms, so it seems to be at least on a similar tack...

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  • $\begingroup$ Very nice, even if it is not exactly the type of sums I am looking at. I trust some of the idea's used in connection with the MacMahon functions will be usefull. $\endgroup$
    – torben
    Dec 17, 2010 at 23:50
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The sum in question can be rewritten as

$\sum_{m\leq N-1}\sum_{n\leq N-1,m\leq n-1}\frac{1}{m^an^b}\sum_{t\leq N-1,t=0(\bmod m),t=N(\bmod n)}1,$

that is

$\sim (N-1)\sum_{m\leq N-1}\sum_{n\leq N-1,m\leq n-1,(m,n)|N}\frac{(m,n)}{m^{a+1}n^{b+1}}.$

Of course, there will come out a so-called "error" term, I haven't check it is really an error.

Sorry, I am not quite familar with the latex on mathoverflow.

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  • $\begingroup$ In fact, the sum in question is bounded if $a,b>0$, in which case it is "certain constant" + $N^{-\delta},\delta>0$, maybe? $\endgroup$
    – arithboy
    Dec 14, 2010 at 0:56
  • $\begingroup$ $O(N^{-\delta})$ $\endgroup$
    – arithboy
    Dec 14, 2010 at 1:01

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