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A247968
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a(n) = least k such that (k!*e^k)/(sqrt(2*Pi)*k^(k+1/2)) - 1 < 1/2^n.
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2
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1, 1, 1, 2, 3, 6, 11, 22, 43, 86, 171, 342, 683, 1366, 2731, 5462
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OFFSET
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1,4
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COMMENTS
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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
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REFERENCES
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Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 18 (Stirling's formula).
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LINKS
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Table of n, a(n) for n=1..16.
Eric Weisstein's World of Mathematics, Stirling's Approximation.
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MATHEMATICA
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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}]] (* A247968 *)
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CROSSREFS
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Cf. A005578.
Sequence in context: A318123 A226594 A043327 * A005578 A058050 A026418
Adjacent sequences: A247965 A247966 A247967 * A247969 A247970 A247971
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KEYWORD
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nonn,more
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AUTHOR
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Clark Kimberling, Sep 28 2014
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EXTENSIONS
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Name corrected by David A. Corneth, Mar 06 2019
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STATUS
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approved
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