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%I #17 Oct 24 2015 04:03:23
%S 2,1,2,2,4,3,6,4,6,5,10,6,12,7,10,8,16,9,18,10,14,11,22,12,24,13,18,
%T 14,28,15,30,16,22,17,28,18,36,19,26,20,40,21,42,22,30,23,46,24,42,25,
%U 34,26,52,27,44,28,38,29,58,30,60,31,42,32,52,33,66,34,46,35
%N Least number k such that (n!-k)/(n-k) is an integer.
%C It is conjectured that a(2n) = n.
%C The largest k such that (n!-k)/(n-k) is an integer is n!
%C a(2n+1) <= 2n for all n >= 1.
%C Does a(n) = A060681(n) for n>1? - _Michel Marcus_, Oct 24 2015
%e 6-1 is not divisible by 6!-1, 6-2 is not divisible by 6!-2, but 6-3 is divisible by 6!-3. So a(6) = 3.
%o (PARI) a(n)=for(k=1, n!, if(k!=n, s = (n!-k)/(n-k); if(floor(s)==s, return(k));));
%o n=1;while(n<100,print(a(n));n+=1) \\ corrected by _Michel Marcus_, Oct 24 2015
%K nonn
%O 1,1
%A _Derek Orr_, May 21 2014