A039960 For n >= 2, a(n) = largest value of k such that n^k is <= n! (a(0) = a(1) = 1 by convention).
1, 1, 1, 1, 2, 2, 3, 4, 5, 5, 6, 7, 8, 8, 9, 10, 11, 11, 12, 13, 14, 14, 15, 16, 17, 18, 18, 19, 20, 21, 21, 22, 23, 24, 25, 25, 26, 27, 28, 29, 29, 30, 31, 32, 33, 33, 34, 35, 36, 37, 37, 38, 39, 40, 41, 42, 42, 43, 44, 45, 46, 46, 47, 48, 49, 50, 50, 51, 52, 53, 54, 55, 55, 56, 57
Offset: 0
Examples
a(7)=4 because 7! = 5040, 7^4 = 2401 but 7^5 = 16807. a(6)=3 since 6^3.67195... = 720 = 6! and 6^3 <= 6! < 6^4, i.e., 216 <= 720 < 1296.
Links
- Danny Rorabaugh, Table of n, a(n) for n = 0..10000
Programs
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Magma
[1,1] cat [Floor(Log(Factorial(n))/Log(n)): n in [2..80]]; // Vincenzo Librandi, Apr 15 2015
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Mathematica
ds[x_, y_] :=y!-y^x; a[n_] :=Block[{m=1, s=ds[m, n]}, While[Sign[s]!=-1&&!Greater[m, 256], m++ ];m]; Table[a[n]-1, {n, 3, 200}] (* or *) Table[Count[Part[Sign[Table[Table[n!-n^j, {j, 1, 128}], {n, 1, 128}]], u], 1], {u, 1, 128}] (* Labos Elemer *) Join[{1,1},Table[Floor[Log[n,n!]],{n,2,80}]] (* Harvey P. Dale, Sep 24 2019 *) -
PARI
a(n)=if(n>3,lngamma(n+1)\log(n),1) \\ Charles R Greathouse IV, Sep 02 2015
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Sage
[1,1] + [floor(log(factorial(n))/log(n)) for n in range(2,75)] # Danny Rorabaugh, Apr 14 2015
Formula
a(n) = floor(log_n(n!)) for n > 1.
a(n) = A060151(n) - 1 for n > 1. - Henry Bottomley, Mar 08 2001
From Danny Rorabaugh, Apr 14 2015: (Start)
a(n) = log_n(A074182(n)) for n > 1.
From Robert Israel, Apr 14 2015: (Start)
n*(1-1/log(n)) + 1 > log(n!)/log(n) > n*(1-1/log(n)) for n >= 7.
Thus a(n) is either floor(n*(1-1/log(n))) or ceiling(n*(1-1/log(n))) for n >= 7 (and in fact this is the case for n >= 3). (End)
Extensions
Corrected and extended by Henry Bottomley, Mar 08 2001
Edited by N. J. A. Sloane, Sep 26 2008 at the suggestion of R. J. Mathar
Comments