[seqfan] Re: Euler product and prime zeta function

[Tomasz Ordowski]

P.S. Note that
P(n)  >~ 1 - 1/zeta(n),
where P(n) = Sum_{prime p} 1/p^n.

Conjecture:
0 < 1 - 1/zeta(n) - 1/2^n - 1/3^n < 1/5^n,
for every n > 1.

T. Ordowski

śr., 14 gru 2022 o 17:43 Tomasz Ordowski <tomaszordowski at gmail.com>
napisał(a):

> Dear readers!
>
> By the Euler product,
> Product_{prime p} (1-1/(2p^n-1))/(1+1/(2p^n-1)) =
> = Product_{prime p} (1-1/p^n) = 1/zeta(n), for n > 1.
>
> Note that w(n) = (1-1/zeta(n)/(1+1/zeta(n) = (zeta(n)-1)/(zeta(n)+1)
> is the relativistic sum of the velocities v = 1/(2p^n-1) over all primes
> p,
> in units where the speed of light c = 1. Cf. A348829 / A348830.
>
> So, according to the above physical interpretation,
> 1/(2*2^n-1) < w(n) = (zeta(n)-1)/(zeta(n)+1) < Sum_{prime p} 1/(2p^n-1).
> The prime zeta P(n) = Sum_{prime p} 1/p^n ~ 2(zeta(n)-1)/(zeta(n)+1),
> (but P(n) < 2w(n) for n > 1). Hence P(n) ~ log(zeta(n)) ~ zeta(n)-1.
>
> Let a(n) be the smallest prime q such that
> Sum_{prime p <= q} 1/(2p^n-1) > (zeta(n)-1)/(zeta(n)+1).
>
> Is a(n) = Q for every n >= N (until the end of the calculation)?
> If so, what are the hypothetical values of Q and N ?
>
> Best regards,
>
> Thomas Ordowski
> __________________
> Cf. A348829 / A348830.
> See my draft with a new conjecture in the formula section:
> The On-Line Encyclopedia of Integer Sequences® (OEIS®)
> <https://oeis.org/history/view?seq=A348829&v=94>
>
>