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%I #54 Jan 24 2023 14:37:48
%S 1,1,1,2,3,6,11,22,43,86,171,342,683,1366,2731,5462,10923,21846,43691,
%T 87382,174763,349526,699051,1398102,2796203,5592406,11184811,22369622
%N a(n) = least k such that (k!*e^k)/(sqrt(2*Pi)*k^(k+1/2)) - 1 < 1/2^n.
%C Is a(n) = A005578(n-2) for n >= 2?
%C A proof would likely follow from applying Stirling's formula to k!. - _R. J. Mathar_, Oct 07 2014
%C 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]
%D Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 18 (Stirling's formula).
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/StirlingsApproximation.html">Stirling's Approximation</a>.
%t z = 100; s[n_] := s[n] = (n!*E^n)/(Sqrt[2*Pi]*n^(n + 1/2));
%t N[Table[s[n], {n, 1, z}], 10]
%t f[n_] := f[n] = Select[Range[6000], s[#] - 1 < 1/2^n &, 1]
%t Flatten[Table[f[n], {n, 1, z}]]
%t (* alternate program *)
%t 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 *)
%Y Cf. A005578.
%K nonn,more
%O 1,4
%A _Clark Kimberling_, Sep 28 2014
%E Name corrected by _David A. Corneth_, Mar 06 2019
%E a(17)-a(28) from _J.W.L. (Jan) Eerland_, Jan 04 2023
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