A247968 a(n) = least k such that (k!*e^k)/(sqrt(2*Pi)*k^(k+1/2)) - 1 < 1/2^n.

1, 1, 1, 2, 3, 6, 11, 22, 43, 86, 171, 342, 683, 1366, 2731, 5462, 10923, 21846, 43691, 87382, 174763, 349526, 699051, 1398102, 2796203, 5592406, 11184811, 22369622

Offset: 1

Author

Clark Kimberling, Sep 28 2014

Keywords

• nonn
• more

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.

Crossrefs

Cf. A005578.

Programs

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 *)

Extensions

Name corrected by David A. Corneth, Mar 06 2019

a(17)-a(28) from J.W.L. (Jan) Eerland, Jan 04 2023