Analytic continuation of primality function

Question

(There is my initial question, but by advice of @Charles I'm splitting it)

For integers we have a primality function: $$ isprime(n)=\begin{cases}1,&\text{$n$ is prime}\\0,&\text{$n$ is not prime}\end{cases} $$ (in initial question I used $np(n)=1-isprime(n)$).

My question: can we construct an analytic continuation of primality function and get entire function?

I think that we can, and there is should be smth like $isprime(z)\sim \pi'(z)$ (because $\pi(n)=\sum_{i=1}^n isprime(i)$).

Answer 1

it is an exercise in Ahlfors that one may specify an infinite sequence $a_n$ ( such as the positive integers), with requirement $a_n \rightarrow \infty,$ and the values $A_n$ there, and build an entire function with $f(a_n) = A_n.$ Page 196 in the second edition.

No reason to expect this to have any information, any such function is far from unique, since one may add in some favorite value at, say, $3/2$ or at $i$ and change the thing.

Answer 2

Here is a sketch of a highly impractical method: one can use Wilson's theorem along with Iverson brackets to define $\mathrm{isprime}(n)$; having done so, you can construct an ordinary or exponential generating function for $\mathrm{isprime}(n)$, call it $P(z)$, and then use Cauchy's differentiation formula to evaluate $P^{(\alpha)}(0)$ (or $\Gamma(\alpha+1)P^{(\alpha)}(0)$) for arbitrary complex $\alpha$, in a manner similar to this other answer of mine. The problem is now in determining where $P(z)$ is analytic, and from that determining a suitable contour for Cauchy.