OFFSET
1,4
COMMENTS
Is a(n) = A005578(n-2) for n >= 2?
A proof would likely follow from applying Stirling's formula to k!. - R. J. Mathar, Oct 07 2014
a(n) is the least k such that the Stirling approximation to k! underestimates the real value by a factor of less than 1/2^k. The MathWorld link notes that replacing sqrt(2k) with sqrt(2k+1/3) in Stirling's approximation gives a much closer approximation of k!, which leads to the formula a(n) = ceiling(2^n/12). - Charlie Neder, Mar 06 2019 [corrected by Jon E. Schoenfield, Dec 18 2022]
REFERENCES
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 18 (Stirling's formula).
LINKS
Eric Weisstein's World of Mathematics, Stirling's Approximation.
MATHEMATICA
z = 100; s[n_] := s[n] = (n!*E^n)/(Sqrt[2*Pi]*n^(n + 1/2));
N[Table[s[n], {n, 1, z}], 10]
f[n_] := f[n] = Select[Range[6000], s[#] - 1 < 1/2^n &, 1]
Flatten[Table[f[n], {n, 1, z}]]
(* alternate program *)
Table[k=1; Monitor[Parallelize[While[True, If[((Factorial[k]*Exp[k])/(Sqrt[2*Pi]*k^(k+(1/2))))-1<1/2^n, Break[]]; k++]; k], k], {n, 1, 10}] (* J.W.L. (Jan) Eerland, Dec 08 2022 *)
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Clark Kimberling, Sep 28 2014
EXTENSIONS
Name corrected by David A. Corneth, Mar 06 2019
a(17)-a(28) from J.W.L. (Jan) Eerland, Jan 04 2023
STATUS
approved