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Least k > 1 such that k^n divides (k-1)!.
2

%I #20 Apr 29 2023 00:06:07

%S 2,6,12,12,12,24,24,36,40,45,48,60,60,60,72,72,72,80,80,90,90,120,120,

%T 120,120,120,120,120,144,144,144,144,144,144,144,180,180,180,180,180,

%U 180,180,180,180,240,240,240,240,240,240,240,240,240,240,240,240,240

%N Least k > 1 such that k^n divides (k-1)!.

%C Is 45 the only odd number in this sequence? - _Derek Orr_, Apr 16 2015

%H Michael De Vlieger, <a href="/A092427/b092427.txt">Table of n, a(n) for n = 0..1000</a>

%F Does lim_{n->oo} a(n)/n exist?

%t Table[k = 2; While[Mod[(k - 1)!, k^n] != 0, k++]; k, {n, 0, 60}] (* _Michael De Vlieger_, Apr 16 2015 *)

%o (PARI) a(n)=if(n<0,0,k=2;while((k-1)!%(k^n)>0,k++);k)

%Y A061770 gives values of n such that a(n) > a(n-1).

%K nonn

%O 0,1

%A _Benoit Cloitre_, Mar 22 2004

%E Name edited by _Derek Orr_, Apr 16 2015