A065027 a(n) is the smallest k > 0 such that n^k < k!.
1, 2, 4, 7, 9, 12, 14, 17, 20, 22, 25, 28, 30, 33, 36, 38, 41, 44, 47, 49, 52, 55, 57, 60, 63, 65, 68, 71, 73, 76, 79, 82, 84, 87, 90, 92, 95, 98, 101, 103, 106, 109, 111, 114, 117, 119, 122, 125, 128, 130, 133, 136, 138, 141, 144, 147, 149, 152, 155, 157, 160, 163
Offset: 0
Examples
2^3 > 3! but 2^4 < 4!, so a(2)=4.
Links
- Harry J. Smith and Ely Golden, Table of n, a(n) for n = 0..10000 (first 1000 terms from Harry J. Smith)
- Robert Israel, Plot of a(n) - n e + log(sqrt(2 Pi n)) for 1 <= n <= 20000
- Tomas Schonbek, POLYA008: First n for which m^n < n! [From Nikos Apostolakis, Feb 17 2009]
Programs
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Maple
m:= 1: for n from 1 to 100 do while n^m >= m! do m:=m+1 od: A[n]:= m; od: seq(A[n],n=0..100); # Robert Israel, Oct 28 2016
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Mathematica
Table[Length[Select[Table[m^n/n!,{n,1,180}],#>=1&]]+1,{m,1,61}] sm0[n_]:=Module[{m=1},While[n^m>=m!,m++];m]; Array[sm0,70] (* Harvey P. Dale, Jan 24 2018 *) -
PARI
{ m=1; for (n=1, 1000, until (n^m < m!, m++); write("b065027.txt", n, " ", m) ) } \\ Harry J. Smith, Oct 03 2009
Formula
It appears that L(n) < a(n) - n e + log(sqrt(2 Pi n)) < 1/2, where L(n) = -1/2 + o(1), and L(n) > -0.53 for all n. - Robert Israel, Oct 28 2016 (In other words, a(n) - n e + log(sqrt(2 Pi n)) < 1/2 for all n, and there is some function L(n) = -1/2 + o(1) such that -0.53 < L(n) < a(n) - n e + log(sqrt(2 Pi n)) for all n. - Charles R Greathouse IV, Nov 04 2016)
Extensions
More terms from Robert G. Wilson v, Dec 05 2001
a(0)=1 prepended by Alois P. Heinz, Dec 21 2019
Comments