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%I #22 Apr 10 2013 22:21:01
%S 0,0,2,2,4,2,3,3,3,3,5,3,12,4,4,4,16,5,9,3,5,6,14,3,4,4,4,4,18,4,30,4,
%T 6,9,4,4,36,6,8,4,40,5,21,5,5,10,23,4,7,7,10,7,52,8,9,6,13,7,15,4,8,
%U 14,5,5,5,6,18,8,17,5,7,5,72,21,5,14,9,8,23,5
%N The least number k that maximizes k! mod n.
%H T. D. Noe, <a href="/A221861/b221861.txt">Table of n, a(n) for n = 1..10000</a>
%e For n=11, we see that the factorial of 5 (120), modulo 11 is 10, which is the highest possible value, so the 11th term is 5.
%o (Ruby) (1..100).map{|n|(0..n).max_by{|x|[(1..x).inject(1,:*)%n,-x]}}
%Y Cf. A062170.
%K nonn
%O 1,3
%A _Aaron Weiner_, Apr 10 2013
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