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 A247968 a(n) = least k such that (k!*e^k)/(sqrt(2*Pi)*k^(k+1/2)) - 1 < 1/2^n. 2
 1, 1, 1, 2, 3, 6, 11, 22, 43, 86, 171, 342, 683, 1366, 2731, 5462 (list; graph; refs; listen; history; text; internal format)
 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/3). - Charlie Neder, Mar 06 2019 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, s[#] - 1 < 1/2^n &, 1] Flatten[Table[f[n], {n, 1, z}]]  (* A247968 *) CROSSREFS Cf. A005578. Sequence in context: A318123 A226594 A043327 * A005578 A058050 A026418 Adjacent sequences:  A247965 A247966 A247967 * A247969 A247970 A247971 KEYWORD nonn,more AUTHOR Clark Kimberling, Sep 28 2014 EXTENSIONS Name corrected by David A. Corneth, Mar 06 2019 STATUS approved

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Last modified June 29 14:46 EDT 2022. Contains 354913 sequences. (Running on oeis4.)