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n^k - n! where n^k > n! >= n^(k-1).
4

%I #11 Apr 14 2015 16:46:56

%S 2,3,40,5,576,11767,221824,168561,6371200,174442081,4680778752,

%T 4377478573,202076363776,7342081491375,260552186822656,

%U 226934809133761,14420591159943168,677361585374052121,30335097991823360000

%N n^k - n! where n^k > n! >= n^(k-1).

%H Danny Rorabaugh, <a href="/A111683/b111683.txt">Table of n, a(n) for n = 2..447</a>

%F a(n) = n^(1+floor(log_n(n!))) - n! = n^A060151(n) - A000142(n). - _Danny Rorabaugh_, Apr 14 2015

%e a(5) = 125 - 120 = 5, because 125 > 120 >= 25.

%t For[n = 2, n < 20, n++, k := 0; While[n^k <= n!, k++ ]; Print[n^k - n! ]] (* _Stefan Steinerberger_, Jan 26 2006 *)

%o (Sage) [n^(1+floor(log(factorial(n))/log(n))) - factorial(n) for n in range(2,21)] # _Danny Rorabaugh_, Apr 14 2015

%K nonn,easy

%O 2,1

%A _Amarnath Murthy_, Aug 16 2005

%E More terms from _Stefan Steinerberger_, Jan 26 2006