Generating function for the characteristic function of primes

Question

What do we know about the generating function of $\chi(n)$ (A010051) $$ f(x) = \sum_{n=0}^\infty \chi(n)x^n = \sum_{p\text{ prime}} x^p $$ for $\chi(n)$ the characteristic function of the primes:

$$ \chi(n) = \begin{cases} 1, & \text{if $n$ is prime}\\ 0, & \text{otherwise} \end{cases} $$ Are there some references I could take a look at?

Maybe you are interested in this: Summing over General Functions of Primes — draks ..., Dec 11 '13 at 13:41

score 6 · Answer 1 · answered Dec 11 '13 at 14:00

6

This is essentially the inverse Mellin transform of the Prime zeta function. Perhaps you'd like to look in that direction.

answered Dec 11 '13 at 14:00

Bruno Joyal

54,711

So you say that $\left{\mathcal{M}^{-1}P\right}(x) = f(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} x^{-s} P(s), ds, $ where $P(s)$ is Prime $\zeta$? – draks ... Dec 11 '13 at 20:25
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I'm really interested. Could you answer to my comment? – draks ... Mar 06 '14 at 20:23
@draks...: Sorry for overlooking your question the first time. If $f(x) = \sum_{n=1}^\infty a_n x^n$, then $\mathcal M(f(e^{-t}))(s) = \Gamma(s)L(f,s)$, where $L(f, s) = \sum_{n=1}^\infty a_n n^{-s}$. – Bruno Joyal Mar 06 '14 at 20:34

score 2 · Answer 2 · answered Dec 11 '13 at 13:46

2

We know that, $\chi(n)$ can be represented as $\pi(n)-\pi(n-1)$.

answered Dec 11 '13 at 13:46

draks ...

18,449

Generating function for the characteristic function of primes

2 Answers2