cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A377024 Decimal expansion of the constant F(2) related to asymptotic products of factorials.

Original entry on oeis.org

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

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Author

Bernd C. Kellner, Oct 13 2024

Keywords

Comments

The constants F(1) = A213080, F(2), ... occur in the context of asymptotic constants related to asymptotic products of factorials as well as of binomial and multinomial coefficients. Moreover, the sequence (F(k)){k >= 1} is strictly decreasing with limit 1. For example, for k >= 1 the asymptotic product Prod{v >= 1} (k*v)! has the asymptotic constant F(k)*A^k*(2*Pi)^(1/4), where A = A074962 denotes the Glaisher-Kinkelin constant. Let gamma = A001620 be Euler's constant and Gamma(x) be the gamma function.
For k >= 1, the constants F(k) can be computed by an explicit formula and a divergent series expansion, as follows. We have log(F(k)) = (1/(12*k))*(1-log(k)) + (k/4)*log(2*Pi) - ((k^2+1)/k)*log(A) - Sum_{v=1..k-1} (v/k)*log(Gamma(v/k)) = gamma/(12*k) - t*zeta(3)/(360*k^3) with some t in (0,1), respectively.
It follows that log(F(2)) = 1/24 + log(2*Pi)/4 + (5/24)*log(2) - (5/2)*log(A) = gamma/24 - t*zeta(3)/2880 with some t in (0,1), and so F(2) lies in the interval (1.023914..., 1.024342...) (see Kellner 2009 and 2024).

Examples

			1.02393741163711840157795078258621780080376098043644005129469909513476924124007...
		

Crossrefs

Programs

  • Maple
    exp(-1/6+5/2*Zeta(1, -1))*(2*Pi)^(1/4)*2^(5/24); evalf(%, 100);
  • Mathematica
    RealDigits[Exp[1/24] (2 Pi)^(1/4) 2^(5/24) / Glaisher^(5/2), 10, 100][[1]]
  • PARI
    default(realprecision, 100);
    exp(-1/6+5/2*zeta'(-1))*(2*Pi)^(1/4)*2^(5/24)
  • Sage
    import mpmath
    mpmath.mp.pretty = True; mpmath.mp.dps = 100
    mpmath.exp(-1/6+5/2*mpmath.zeta(-1, 1, 1))*(2*pi)^(1/4)*2^(5/24)
    

Formula

Equals exp(1/24)*(2*Pi)^(1/4)*2^(5/24)/A^(5/2) where A = A074962.
Equals exp(-1/6+(5/2)*zeta'(-1))*(2*Pi)^(1/4)*2^(5/24).