Take $q_0<q_1<...<q_k<q_{k+1}<...$ positive integers, $z$ complex



I would need to extract the first coefficient $q_0$

It is irrelevant if this procedure is not achievable in practice. I would like to know if there is a principal solution. We do not know anything special about $q$'s. They could be just any positive integers.

I feel that the coefficient would have to appear somewhere, but where?


Now I can explain the reason for this. I was trying to make something out of Euler function and Riemann to make the used sieving more explicit.

$$E_1(s)=\sum\limits_{n=1}^{\infty}\frac{1}{n^s}=\zeta (s)$$

Let us create a function that is capable of extracting the smallest coefficient from the reciprocal sum of any ordered integers


(this is where the answer fits)

$$ \lim\limits_{s \to \infty} \frac{1}{f(s)^\frac{1}{s}}=q_1$$

We have


which is eliminating all coefficients divisible by $p_n$.

$n^{th}$ prime number $p_n$ is then simply

$$ p_n=\lim\limits_{|s| \to +\infty} \frac{1}{(E_n(s)-1)^\frac{1}{s}}$$

I hope others will find this at least a little bit interesting as well.

$$ \ln(p_n)=-\lim\limits_{|s| \to +\infty} \frac{\ln\left (\zeta(s)\prod\limits_{k=1}^{n-1}(1-p_k^{-s})-1 \right )}{s} $$

Essentially the explicit form of the selection of the next prime is not given as we feel that is kind of obvious. Still, I find it missing in the sieving logic of the Euler product formula.

  • 1
    $\begingroup$ Assuming you mean positive integers, this is just a Dirichlet series in disguise. Successively take the limit, as $\Re z\to\infty$, of $T(z)$, $2^zT(z)$, $3^zT(z)$, etc.; the first limit that is nonzero will be $q_0^z T(z)$ (and the limit will equal $1$). $\endgroup$ – Greg Martin May 18 '18 at 21:00
  • $\begingroup$ @GregMartin: Fine. Thanks. I am thinking now if it can be merged into one formula without testing it all one by one. $\endgroup$ – alex.peter May 18 '18 at 21:03
  • $\begingroup$ @GregMartin Something like create $g(x)=\lim\limits_{z \to \infty} x^zT(z)$ This would essentially extract all coefficients. So I guess the second part would be to detect the smallest coefficient, but if $x$ is binary position that means detecting first $1$ in binary notation. Hm, I will think about it. $\endgroup$ – alex.peter May 18 '18 at 21:13
  • $\begingroup$ A formula would be $q_0 = \lim_z\rightarrow\infty T(z)^{-1/z}$. $\endgroup$ – Jan-Christoph Schlage-Puchta May 19 '18 at 13:16
  • $\begingroup$ @Jan-ChristophSchlage-Puchta That much is derived already in the accepted answer. An improvement would be to get rid of the requirement $z>1$ for example, although it creates no difference in what I needed it for. $\endgroup$ – alex.peter May 19 '18 at 14:00

For $z>1$, we have \begin{equation} q_0^{-z}\le T(z)\le\sum\limits_{q=q_0}^{\infty}q^{-z}\le q_0^{-z}+\int_{q_0}^\infty q^{-z}\,dq =q_0^{-z}(1+q_0/(z-1)). \end{equation} Hence, \begin{equation} q_0=\lim_{z\to+\infty}T(z)^{-1/z}. \end{equation}


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