A381456 Decimal expansion of Product_{p prime} p^(1/(p^2-1)).

1, 7, 6, 8, 1, 9, 8, 0, 7, 8, 1, 5, 3, 2, 4, 4, 9, 8, 4, 1, 3, 0, 8, 5, 3, 0, 7, 7, 2, 3, 1, 4, 9, 6, 5, 5, 2, 3, 1, 2, 9, 4, 2, 2, 8, 5, 9, 1, 2, 5, 8, 9, 7, 6, 1, 2, 5, 3, 0, 1, 4, 1, 3, 7, 5, 8, 6, 1, 0, 7, 9, 1, 4, 6, 0, 0, 0, 0, 4, 3, 0, 0, 9, 3, 0, 3, 1, 5, 7, 1, 7, 1, 0, 7, 2, 8, 5, 1, 5, 6, 1, 9, 3, 8, 0, 6, 6, 6

Offset: 1

Author

Jwalin Bhatt, Feb 24 2025

Keywords

• cons
• nonn

Comments

The geometric mean of the zeta distribution with parameter value 2 (A381522) approaches this constant.

In general, for parameter value `s` it approaches e^(-zeta'(s)/zeta(s)). - Jwalin Bhatt, Feb 26 2025

Examples

			1.768198078153244984130853077...
		

Crossrefs

Cf. A000040, A001620, A074962, A013661, A085548, A306016, A381522, A381898.

Programs

Mathematica

N[Exp[-Zeta'[2]/Zeta[2]], 120]

PARI

exp(-zeta'(2)/zeta(2)) \\ Amiram Eldar, Feb 24 2025

Python

from mpmath import zeta, diff, exp, mp
mp.dps = 120
const = exp(-diff(zeta, 2)/zeta(2))
A381456 = [int(d) for d in mp.nstr(const, n=mp.dps)[:-1] if d != '.']  # Jwalin Bhatt, Apr 08 2025

Sage

N(exp(-diff(zeta(s:=var('s')), s).subs(s==2) / zeta(2)), 120)

Formula

Equals Product_{p>=2} p^(1/(p^2-1)) where p is prime.

Equals (A^12)/(2*Pi*(e^gamma)) where A = A074962 is the Glaisher-Kinkelin constant and gamma = A001620 is the Euler-Mascheroni constant.

Equals e^(-zeta'(2)/zeta(2)).

Equals exp((Sum_{k>=2} log(k)/(k^2))*(6/(Pi^2))).

Equals (Product_{k>=2} k^(1/(k^2)))^(6/(Pi^2)).

Equals exp(A306016). - Hugo Pfoertner, Feb 24 2025