I was reading this post about the Bell Numbers where users Lucian and Vladimir Reshetnikov give us Dobiński's formula for the Bell numbers

$$ B(x) = \frac{1}{e} \sum_{k=1}^{\infty} \frac{k^x}{k!}. $$

Now I was trying to reason about this function on the complex plane. It's easy to reason that for no value of $x$ such that $|x| < \infty$ will this expression have a singularity so this function definitely doesn't have any poles.

Now the other question I wanted to turn my attention to are, can we characterize the zeroes of this function? Are there only finitely many or countably many? Is there some natural characterization of the zeroes (like they all live on some well known curve(s))?

There also appears to be some Riemann surface structure attached to this wherein $k^x$ is multivalued for complex $x$ as it is $e^{\ln(k)x}$ where the choice of branch of $\ln(k)$ needs to be made.

  • $\begingroup$ Did you mean to link specifically to your comment, or perhaps to @VladimirReshetnikov's answer or @‍Lucian's comment, or to the MSE question A generalization of Bell numbers to arbitrary complex arguments? $\endgroup$
    – LSpice
    Oct 4, 2022 at 19:25
  • $\begingroup$ Just @Vladimirs answer, I have linked to the wrong thing $\endgroup$ Oct 4, 2022 at 19:31
  • $\begingroup$ I have edited accordingly. $\endgroup$
    – LSpice
    Oct 4, 2022 at 19:32
  • $\begingroup$ Looking at the maximal term in the series above (where $\log k$ is the principal one) for $x>0$ large, one sees that it happens around $k=x/\log x$ and that for any $\epsilon>0$ we have $k_x^x/k_x! >>x^{(1-\epsilon)x}=e^{(1-\epsilon)x\log x}$ for large $x>x_{\epsilon}, k_x=[x/\log x]$ say, hence $B(x)>>_{\epsilon}e^{(1-\epsilon)x\log x}$; now it is straightforward to show that $|B(x)| <<_{\epsilon} e^{|x|^{1+\epsilon}}$ so $B$ is entire order $1$ and maximal type, hence it must have infinitely many zeroes; $\endgroup$
    – Conrad
    Oct 4, 2022 at 22:02

1 Answer 1


The function $B(z)$ is an example of an almost periodic function. The zeroes of an almost periodic function that is holomorphic on some strip are also almost periodic, so such a function either has no zeroes at all or infinitely many zeros.

Let $f(z)=\sum_{k=0}^{\infty}a_ke^{b_kz}$ where each $b_k$ is real and each $a_k$ is complex. Suppose that $\alpha,\beta\in[-\infty,\infty]$ and that the sum converges uniformly and absolutely on compact subsets of the strip $\{z:\alpha<\text{Re}(z)<\beta\}$.

Now, there is a sequence of real numbers $r_n$ with $r_n\rightarrow\infty$ where $$\lim_{n\rightarrow\infty}(e^{ir_nb_k})_{k=0}^{\infty}=(1)_{k=0}^{\infty}.$$

This ensures that $f(z+ir_n)\rightarrow f(z)$ and $f(z-ir_n)\rightarrow f(z)$ uniformly on compact subsets of $\{z:\alpha<\text{Re}(z)<\beta\}$. Therefore, by Hurwitz' theorem, the locations of the zeroes of $f(z+ir_n)$ will converge to the locations of the zeros of $f(z)$.


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