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Background

There are a few questions on this venue on functions that generate themselves:

  • In this question, a function is sought that is its own ordinary generating function. An answer is obtained through a linear combination of functions of the form $$\phi_i(z) = (-1)^i\frac{\sin \pi z}{\pi (z - i)}.$$
  • Over here, user Nikolaj-K inquires about a function that is its own exponential generating function. An answer is found by considering solutions to the equation $e^c = c$, which can be obtained through the Lambert W function.

What I'm interested in, is whether a function can be found that is its own Dirichlet generating function (DGF). In other words, I'm looking for a function $f(\cdot)$ such that $$ f(s) = \sum_{n=1}^{\infty} \frac{f(n)}{n^s}. \label{1}\tag{1} $$

I've tried applying the Dirichlet Inversion formula to obtain equalities for the coefficients, but didn't see a good way to do so.

Question

Are there any analytic solutions to equation \eqref{1}, which means the function is its own DGF?

Max Muller
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  • If $f(1)$ converges and is Nonzero, the arithmetic function $f$ has an inverse for the Dirichlet product, let's call it $f^{-1}$. Hence, $f(s) \cdot \sum_n \frac{f^{-1}(n)}{n^s} = \sum_n \frac{(f *f^{-1})(n)}{n^s} = 1$. This implies that, if $f$ is convergent and nonzero at $1$, it must equal $1/\sum_n \frac{f^{-1}(n)}{n^s}$. Now you could try if this satisfies equation (1). I will try it later :) – groupoid Nov 12 '23 at 12:53
  • In fact the condition that $f$ converges at $1$ can be dropped – groupoid Nov 12 '23 at 13:23
  • What have you tried? – jjagmath Nov 12 '23 at 13:38
  • @groupoid If $f(1)$ doesn't converge then the problem doesn't make sense since the first term of $\sum_{n=1}^\infty \frac{f(n)}{n^s}$ won't be defined. – jjagmath Nov 12 '23 at 13:41
  • @jjagmath I've tried using the Dirichlet Inversion formula to obtain expressions for the coefficients, but didn't see a good way to do this – Max Muller Nov 12 '23 at 13:42
  • @jjagmath you're right! – groupoid Nov 12 '23 at 13:52
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    Another observation is the following. If $f(s) = \sum_n \frac{f(n)}{n^s}$, then $f(1) = f(1) + \sum_{n \geq 2} \frac{f(n)}{n}$ and hence $\sum_{n \geq 2} \frac{f(n)}{n} = 0$. This implies that, if $f(n) \in \mathbb{R}_{\geq 0}$ for $n \geq 2$, the function $f$ is constant $f(s) = f(1)$. – groupoid Nov 12 '23 at 14:36
  • @groupoid Interesting observation. Let's try to find non-constant solutions as well! So we allow $f(n) \in \mathbb{R}$ or even $\mathbb{C}$ – Max Muller Nov 12 '23 at 14:42
  • @groupoid That gives $f(s)=0$ for all $s$ – jjagmath Nov 12 '23 at 14:50

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