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%I #7 Sep 23 2018 16:01:09
%S 1,1,2,2,3,3,4,5,5,6,7,8,8,9,10,10,11,12,13,13,14,15,15,16,17,18,18,
%T 19,20,21,21,22,23,23,24,25,26,26,27,28,29,29,30,31,32,32,33,34,35,36,
%U 36,37,38,39,39,40,41,42,42,43,44,45,45,46
%N Least m such that m!*5^m >= n!.
%H Clark Kimberling, <a href="/A213856/b213856.txt">Table of n, a(n) for n = 1..1000</a>
%e a(6) = 3 because (5^2)*2! < 6! <= (5^3)*3!.
%t Table[m=1; While[n!>(5^m)*m!, m++]; m, {n,1,100}]
%t Module[{nn=70,m},mt=Table[{m,m! 5^m},{m,nn}];Table[SelectFirst[mt,#[[2]] >= n!&],{n,nn}]][[All,1]] (* _Harvey P. Dale_, Sep 23 2018 *)
%Y Cf. A213855.
%K nonn,easy
%O 1,3
%A _Clark Kimberling_, Jul 17 2012
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