A122252 Binet's factorial series. Numerators of the coefficients of a convergent series for the logarithm of the Gamma function.
1, 1, 59, 29, 533, 1577, 280361, 69311, 36226519, 7178335, 64766889203, 32128227179, 459253205417, 325788932161, 2311165698322609, 287144996287039, 1215091897184850539, 402833263943353393, 476099430416027805187, 236881416523193720213, 650730651653461090091101
Offset: 1
Examples
Rational sequence starts: 1/12, 1/12, 59/360, 29/60, 533/280, 1577/168, 280361/5040, ...
c(1) = Integral_{x=0..1} x*(x - 1/2) / 1 = Integral_{x=0..1} (x^2 - x/2) = (x^3/3 - x^2/4) | {x, 0, 1} = 1/12.
Links
- Robert G. Wilson v, Table of n, a(n) for n = 1..100
- J. P. M. Binet, Mémoire sur les intégrales définites Eulériennes et sur leur application à la théorie des suites ainsi qu'à l`évaluation des functions des grands nombres, Journal de l`École Polytechnique, XVI:123-343, July 1839.
- Ch. Hermite, Sur la function log Gamma(a) Journal für die reine und angewandte Mathematik, 115:201-208, 1895.
- G. Nemes, Generalization of Binet's Gamma function formulas, Integral Transforms and Special Functions, 24(8):595-606, 2013.
- Raphael Schumacher, Rapidly Convergent Summation Formulas involving Stirling Series, arXiv:1602.00336 [math.NT], 2016.
- P. Van Mieghem, Binet's factorial series and extensions to Laplace transforms, arXiv:2102.04891 [math.FA], 2021.
- Wikipedia, Stirling's Approximation
Programs
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Maple
r := n -> add((-1)^(n-j)*Stirling1(n,j)*j/((j+1)*(j+2)), j=1..n)/(2*n): a := n -> numer(r(n)); seq(a(n), n=1..21); # Peter Luschny, Sep 22 2021
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Mathematica
Rising[z_, n_Integer/;n>0] := z Rising[z + 1, n - 1]; Rising[z_, 0] := 1; c[n_Integer/;n>0] := Integrate[Rising[x, n] (x - 1/2), {x, 0, 1}] / n; Numerator@ Array[c, 19] (* updated by Robert G. Wilson v, Aug 15 2015 *) -
PARI
a(n) = numerator(sum(j=1, n, (-1)^(n-j)*stirling(n,j,1)*j/((j+1)*(j+2)))/(2*n)); \\ Michel Marcus, Sep 22 2021
Formula
a(n) = numerator(c(n)), where c(n) are given by Binet's formulas:
log Gamma z = (z - 1/2) log z - z + log(2*Pi)/2 + Sum_{n >= 1} c(n)/(z+1)^(n), where z^(n) is the rising factorial.
c(n) = (1/n)*Integral_{x=0..1} x^(n)*(x - 1/2).
a(n) = numerator((1/2n)*Sum_{j=1..n} (-1)^(n-j)*Stirling1(n,j)*j/((j+1)*(j+2))). - Peter Luschny, Sep 22 2021
Extensions
Edited by Peter Luschny, Sep 22 2021