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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.

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    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?2012-01-11
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    @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 $\zeta(s)$?2012-01-11
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    @Dan: I have no idea, I'm afraid.2012-01-11

1 Answers 1

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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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    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.2012-01-10
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    If you leave out a prime, you get $\zeta(s)(1-p^{-s})$, not quite constant that factor.2012-01-10
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    Raskolnikov and Sasha, oops! I guess we need a theorem that says the analytic continuation of the product is equal to the product of analytic continuations... I am not even sure if that is true.2012-01-10
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    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.2012-01-11
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    @Thomas, thanks I will incorporate your comment into my answer when I have some more time (and after I have finished understanding it myself).2012-01-11
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    @DanBrumleve: Thx so far. So, when I think of primes as a dynamic system, they are kind of stable? Could one also use this approach as prime generator?2012-01-11
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    @Andreas, I do not think this would be an efficient prime generator, because removing more primes just makes it _harder_ to compute the zeros. I don't think about it as a dynamical system (not very familiar with them), but rather more like a topology where the set of all primes is "closed" and when we remove a few of them we have an "open" set whose "closure" (your algorithm) is all of them.2012-01-11
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    @DanBrumleve: What if add non-primes to the "closed" set. Would they been kicked out?2012-01-11
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    @Andreas, yes, by the same argument ($1-k^{-s}$ is analytic). In that sense it is _not_ like a topology... I'm curious about your dynamical systems analogy?2012-01-11
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    @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...2012-01-11
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    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.2012-01-31
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    @Sam What would be such a sequence? And why does it need to converge normally on $Re(s)>1/4$? How would a sequence look like that doesn't fulfill your criteria?2012-04-09
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    @draks: I can't give you such a sequence explicitely, of course. For $\prod_{k} (1-p_k^{-s})$ to converge normally for $Re(s) > 1/4$, it is sufficient for example to make sure that $2^{k+1}> p_k > 2^{k}$. The product doesn't *need* to converge normally on $Re(s) > 1/4$, that's just one possibility of many (and rather arbitrary). $k^2$ is a sequence, that doesn't fullfill this, since the product diverges for $s = 1/2$.2012-04-09