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Least m>0 such that (1+r)^m >= n!, where r = (1+sqrt(5))/2, the golden ratio.
2

%I #7 Jul 25 2012 12:49:59

%S 1,1,2,4,5,7,9,12,14,16,19,21,24,27,29,32,35,38,41,44,48,51,54,57,61,

%T 64,68,71,75,78,82,85,89,93,96,100,104,107,111,115,119,123,127,131,

%U 135,139,143,147,151,155,159,163,167,171,175

%N Least m>0 such that (1+r)^m >= n!, where r = (1+sqrt(5))/2, the golden ratio.

%H Clark Kimberling, <a href="/A214051/b214051.txt">Table of n, a(n) for n = 1..1000</a>

%p A214051 := proc(n)

%p local r,m;

%p r := (1+sqrt(5))/2 ;

%p for m from 1 do

%p if floor((1+r)^m) >= n! then

%p return m ;

%p end if;

%p end do:

%p end proc: # _R. J. Mathar_, Jul 23 2012

%t r=GoldenRatio;

%t Table[m=1;While[n!>(1+r)^m,m++];m,{n,1,100}]

%Y Cf. A214050.

%K nonn,easy

%O 1,3

%A _Clark Kimberling_, Jul 18 2012