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A065027
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a(n) is the smallest k > 0 such that n^k < k!.
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8
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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
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history;
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OFFSET
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0,2
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COMMENTS
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Differences are 1, 2 or 3 (see A065067). The limit as n -> infinity of a(n)/n is e. - Robert G. Wilson v, Dec 05 2001. [Apparently the Schonbek link contains a proof of the first assertion.]
a(10) = 25, a(100) = 269, a(1000) = 2714, a(10000) = 27177, a(10^5) = 271822, see A085830.
a(n) = least k such that geometric mean of {1,1/2,...,1/k} <= 1/n. - Clark Kimberling, Jul 11 2013
Let b(n) be the largest k such that n^k > k!. Then b(n) = a(n)-1. - Joseph Damico, Jun 30 2019
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LINKS
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FORMULA
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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)
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EXAMPLE
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2^3 > 3! but 2^4 < 4!, so a(2)=4.
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MAPLE
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m:= 1:
for n from 1 to 100 do
while n^m >= m! do m:=m+1 od:
A[n]:= m;
od:
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MATHEMATICA
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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 *)
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PROG
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(PARI) { m=1; for (n=1, 1000, until (n^m < m!, m++); write("b065027.txt", n, " ", m) ) } \\ Harry J. Smith, Oct 03 2009
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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