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I wish to compute $$\sum_{n=1}^{\infty} f(n)e^{-nz}$$ where $f(n)= |\{(a,b,c)| abc=n\}|$ and $z>0$. Its easy to compute that if $n = \prod p_{i}^{\alpha_i}$ where $p_i$ are distinct primes then $$f(n)= \prod \binom{\alpha_i+2}{2}$$ For instance $6=2\times 3$ and we can check that the set is $$\{(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1),(6,1,1),(1,6,1),(1,1,6)\}$$ and so $f(6)=9$. I am a novice to mathematica, can anyone help me figure out how I should compute this on mathematica. Your help will be appreciated.

Edit: As suggested by J. M., I put the code Sum[DivisorSigma[0, n] Exp[-nz]/(1 - Exp[-nz]), {n, 1, ∞}], but Mathematica gives same sum as output.

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    – bbgodfrey
    Dec 13, 2021 at 15:11
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    $\begingroup$ I do not understand your syntax: f(n)=|{(a,b,c)|abc=n}| However, note: if we call the z-transform of f: fz, then your sum is: sum= fx[Exp[z]]-f[0]. fz can be obtained from the function ZTransform $\endgroup$ Dec 13, 2021 at 15:19
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    $\begingroup$ It would be super helpful if you unpacked the math notation for me. Express it in words perhaps. For instance, I am not sure what operation the curly brace in $f(n)$ indicates. A list of the three values? Something else? $\endgroup$
    – MarcoB
    Dec 13, 2021 at 15:19
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    $\begingroup$ Are you aware that your sum can be expressed as a Lambert series? With[{q = Exp[-z]}, Sum[DivisorSigma[0, n] q^(n - 1)/(1 - q^n), {n, 1, ∞}] - 1] $\endgroup$ Dec 13, 2021 at 15:22
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    $\begingroup$ Right, Mathematica is not aware of a closed form for the Lambert series I gave. Thus, you need to evaluate numerically with e.g. NSum[]. $\endgroup$ Dec 13, 2021 at 16:07

1 Answer 1

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(This got too long for a comment.)

What the OP calls $f(n)$ counts what could be called a $3$-multiplicative partition of $n$. This is sometimes referred to in the literature as a Piltz function of order $3$ (see e.g. this paper).

In particular, the OEIS gives a convenient formula for the $3$-Piltz function:

$$f(n)=\tau_3(n)=\sum_{d \mid n} \sigma_0(d)$$

where $\sigma_0(n)$ counts the number of divisors of $n$, and is implemented in Mathematica as DivisorSigma[0, n]. This is then summed all over the divisors $d$ of $n$.

In fact, this sum over all divisors can be alternately expressed as a Dirichlet convolution (see my previous discussion here): DirichletConvolve[DivisorSigma[0, k], 1, k, n].

On the other hand, one might notice the following discrepancy:

With[{n = 1},
     {Apply[Times, Binomial[FactorInteger[n][[All, 2]] + 2, 2]], 
      DirichletConvolve[DivisorSigma[0, k], 1, k, n]}]
   {3, 1}

which can lead to different results depending on which form is used. To make things easy for myself, I will continue with the second representation.

As is customary, I let $q=\exp(-z)$ and consider the generating function

$$\sum_{n=1}^\infty \tau_3(n)q^n$$

An important identity that can be exploited here relates normal generating functions and so-called Lambert series:

$$\sum_{n=1}^\infty c_n q^n=\sum_{n=1}^\infty a_n \frac{q^n}{1-q^n}\quad\text{if}\quad c_n=\sum_{d \mid n} a_d$$

which means the generating function of $\tau_3(n)$ can be expressed as the Lambert series

$$\sum_{n=1}^\infty \sigma_0(n) \frac{q^n}{1-q^n}$$

(I mentioned the discrepancy between the two different versions of $\tau_3(n)$ earlier; one only needs to add or subtract a $2q$ term as needed.)

Some limited numerical experiments I did suggest that the Lambert form is more tractable than the OP's original generating function form, especially for small $q$ (corresponding to large $z$); perhaps someone with more time and inclination than me can do more detailed comparisons.

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