# Finding $1/x^2 + 1/x^3 + 1/x^5 + \dots$

The following function came up in my work: $$f(x)=\sum_{p\text{ prime}}\frac{1}{x^p}=\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^5}+\frac{1}{x^7}+\frac{1}{x^{11}}+\cdots.$$

Naturally, this converges for $x>1$ since the geometric series does. Does this function have a name? Is there a better way to calculate it than the straightforward sum? In my application I can bound $x$ away from 1 if it helps.

• It is related with an exponential sum over primes (see Vinogradov) through the substitution $x=e^{\pm u}$ or $x=e^{\pm iu}$. Jul 1, 2016 at 15:10
• @JackD'Aurizio: Yes -- but crucially in my case, there is no cancellation and the magnitudes are not 1. Jul 1, 2016 at 16:53
• Naturally, I'm curious, where did this function come up in your work? I can't guess any application for it. Especially if you know no other form (i.e. an integral) Aug 17, 2016 at 12:58
• Has anyone calculated f(2) and compare it to common constants? Jan 13, 2017 at 13:01
• @ypercubeᵀᴹ It's oeis.org/A051006 for what it's worth. Jan 13, 2017 at 16:56

The values of your function $f$ at positive integers $n$ correspond to the base-$n$ representations of the prime constant.

Indeed, $f$ is closely related to the characteristic function of the prime numbers. For instance, $f(2)$ evaluates to the prime constant $\rho$, defined as:

$$\rho =\sum _{{p}}{\frac {1}{2^{p}}}=\sum _{{n=1}}^{\infty }{\frac {\chi _{{{\mathbb {P}}}}(n)}{2^{n}}},$$

where $\chi_\mathbb{P}$ is the characteristic function of the primes, i.e., the function such that for positive integer $n$:

$${\displaystyle \chi_\mathbb{P}(n):={\begin{cases}1&{\text{if }}x\in \mathbb{P},\\0&{\text{if }}x\notin \mathbb{P},\end{cases}}}$$

where $\mathbb{P}$ denotes the set of prime numbers.

The decimal expansion of $\rho$ begins with: \begin{align} \rho&=0.414682509851111660248109622\ldots \\ &=0.011010100010100010_2. \end{align}

and is included in the OEIS as sequence A051006.

The values of $f$ for other integers $n$ correspond simply to the base-$n$ representations of the prime constant. If we denote by $\rho_n$ the base-$n$ representation of $\rho$, we have:

\begin{align} f(3)=\sum _{{p}}{\frac {1}{3^{p}}}&=\rho_3 \\ &=0.011010100010100010_3 \\ &=0.152726266\ldots... \end{align}

Therefore $f(n)=\rho_n$ for positive integers $n$.