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%I #5 Aug 21 2013 05:12:43
%S 1,2,3,4,5,4,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,
%T 27,26,29,30,31,32,33,34,35,36,37,38,39,40,41,40,43,44,43,46,47,48,49,
%U 50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,64,67,68,69,70,71,72
%N Least k such that n! divides (n+k)!/(n+1)!.
%C Motivated by 6!=10!/7!.
%C It appears that for most n, a(n) = n. The sequence of n for which a(n) != n begins: 6, 28, 42, 45, 66, 77, 91, 110, 126 ... that is probably A120624. - _Michel Marcus_, Aug 21 2013
%o (PARI) for (n=1,100,k=1;while((n+k)!/(n+1)!%n!,k++);print1(k,","))
%Y Cf. A000142.
%K nonn
%O 1,2
%A _Jaume Oliver Lafont_, Mar 17 2009