Assume the following process:

  1. Let's start with the set of primes $\{p_k\}$
  2. Then we use the Euler product being equivalent to Riemann's Zeta function $$ \prod_{p \text{ prime}} \frac{1}{1-p^{-s}} = \sum_{n=1}^\infty\frac{1}{n^s} = \zeta(s). $$
  3. Now $\rho$, the non-trivial roots of $\zeta(s)$, contribute to the Prime Counting Function $\pi(x)$ in the following way $$\pi(x) = \operatorname{R}(x) - \sum_{\rho}\operatorname{R}(x^{\rho}) - \frac1{\ln x} + \frac1\pi \arctan \frac\pi{\ln x} , $$ with $ \operatorname{R}(x) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \operatorname{li}(x^{1/n})$. (Very nice demonstration can be found here.)
  4. The $k$th prime $p_k$ can now be calutated by using $\pi(p_k)=k$.
  5. So we get back to where we started: (1.) the set of primes $\{p_k\}$ and we now could start again.

My question is: What if a certain prime is missing at the beginning? Will the missing prime be generated automatically, if you iterate the process above?

It would also be interesting to see how the roots are distributed. Are they still lying on the critical line $1/2+iz$?

Is there an easy way to calculate the roots directly from the Euler Product?

Sorry for not going into details, but I think it's all common online knowledge from here and there.

  • $\begingroup$ Interesting question! It seems that you get $(1-p^{-s})\zeta(s)$ having the same nontrivial zeros as $\zeta$ as pointed out below. So you should get back the old $\pi(x)$ regardless. Similarly for leaving out finitely many $p_k$. Makes me wonder whether one might even leave out infinitely many primes? Does anyone know? $\endgroup$
    – Sam
    Jan 11, 2012 at 1:38
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    $\begingroup$ @Sam, the cases I am wondering about now are the Euler product of $4n+1$ primes and that of $4n+3$ primes. Do both of these functions have the same zeros as $\zeta(s)$? $\endgroup$ Jan 11, 2012 at 7:07
  • $\begingroup$ @Dan: I have no idea, I'm afraid. $\endgroup$
    – Sam
    Jan 11, 2012 at 20:40

1 Answer 1


Leaving some primes out of the Euler product won't affect the location of the zeroes, since you will end up with the Zeta function multiplied by a non-zero analytic function (which won't produce any more zeros), and the formula for the prime counting function depends only on the location of the zeroes. So as far as I understand your algorithm, yes, it will "regenerate" any (finite number of) primes that were initially missing.

EDIT: Corrected my erroneous description of the missing term as "constant".

EDIT: To show that the analytic continuation of the product is the same as the product of the analytic continuation, use the fact that the analytic continuation is unique: "Let $f_1$ and $f_2$ be analytic functions on domains $\Omega_1$ and $\Omega_2$, respectively, and suppose that the intersection $\Omega_1 \cap \Omega_2$ is not empty and that $f_1 = f_2$ on $\Omega_1 \cap \Omega_2$. Then $f_2$ is called an analytic continuation of $f_1$ to $\Omega_2$, and vice versa (Flanigan 1983, p. 234). Moreover, if it exists, the analytic continuation of $f_1$ to $\Omega_1$ is unique." We will also need the fact that $\zeta(s)$ and $1-p^{-s}$ are analytic (and the more basic fact that the product of two analytic functions is analytic).

COMMENT: I think your idea here is pretty interesting! I suspect that it may work even in some cases where an infinite number of primes are discarded.

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    $\begingroup$ Notice that $\rho$ are zeros of an analytic function $\zeta(s)$ obtained via analytic continuation of the function defined by Euler product for $\Re(s) > 1$. Leaving out few primes, gives a different function to analytically continue. I am unconvinced that zeros will be unaffected. Please back up your claims. $\endgroup$
    – Sasha
    Jan 10, 2012 at 22:58
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    $\begingroup$ If you leave out a prime, you get $\zeta(s)(1-p^{-s})$, not quite constant that factor. $\endgroup$ Jan 10, 2012 at 23:08
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    $\begingroup$ The product of two holomorphic functions is a holomorphic function, and if two holomorphic functions on a path connected subset $U$ of $\mathbb C$ agree on a subset including a limit point, then their also agree $U$ (IIRC). Anyway, the analytic continuation of the product is equal to the product of the analytic continuations. $\endgroup$ Jan 11, 2012 at 0:26
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    $\begingroup$ @DanBrumleve: The dynamic system idea just came to my mind, when I thought of removing primes (or adding non-primes) as a kind of excitation. And since we should get all primes back by the process, it seems to be stable... $\endgroup$
    – draks ...
    Jan 11, 2012 at 8:07
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    $\begingroup$ One can indeed leave out infinitely many primes without changing the zeros on the critical strip (although probably not all of the primes $\equiv 1$ or $\equiv 3 \pmod 4$): Just choose a sequence of primes $p_k$ such that $\prod_{k=1}^\infty (1-p_k^{-s})$ converges normally on $\{Re(s) > 1/4\}$. Then this product has no zeros for $s = 1/2 + it$ and therefore $\prod_{i = 1}^\infty(1-p_i^{-s})\zeta(s) = \prod_{p \ne p_k\forall k} (1-p^{-s})^{-1}$ has the same zeros as $\zeta(s)$ on the critical strip. $\endgroup$
    – Sam
    Jan 31, 2012 at 17:57

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