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

Solution 1:

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 $n$.