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Least k such that n! divides d(k!) (d = A000005, k > 0).
1

%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