%I #12 Aug 23 2016 11:27:50
%S 1,2,6,8,20,20,28,36,36,56,90,90,120,126,126,126,172,172,342,342,374,
%T 380,464,464,464,464,464,464,464,464,847,847,1056,1105,1105,1105,1105,
%U 1105,1330,1330,1728,1728,1728,1728,1728,2410,2856,2856,2856,2856
%N Least k such that n! divides d(k!) (d = A000005, k > 0).
%C From _Michael De Vlieger_, Aug 10 2016: (Start)
%C Values in a(n): 1, 2, 6, 8, 20, 28, 36, 56, 90, 120, 126, 172, 342, 374, 380, 464, 847, 1056, 1105, 1330, 1728, 2410, 2856, ...
%C First position of values in a(n): 1, 2, 3, 4, 5, 7, 8, 10, 11, 13, 14, 17, 19, 21, 22, 23, 31, 33, 34, 39, 41, 46, 47, ...
%C (End)
%H Chai Wah Wu, <a href="/A275826/b275826.txt">Table of n, a(n) for n = 1..500</a>
%e a(4) = 8 because A000005(8!) = 96 is divisible by 4! = 24.
%t Table[k = 1; While[! Divisible[DivisorSigma[0, k!], n!], k++]; k, {n, 50}] (* _Michael De Vlieger_, Aug 10 2016 *)
%o (PARI) a(n) = {my(k=1); while(numdiv(k!) % n!, k++); k; }
%Y Cf. A000005, A027423, A275476.
%K nonn
%O 1,2
%A _Altug Alkan_, Aug 10 2016