A275369 Least k such that n! divides sigma(k!) (k > 0).

1, 3, 3, 5, 5, 8, 14, 19, 19, 23, 23, 30, 30, 31, 50, 50, 50, 50, 50, 51, 51, 64, 90, 90, 91, 91, 91, 91, 126, 130, 130, 130, 130, 130, 130, 130, 130, 130, 131, 132, 132, 132, 132, 132, 134, 134, 234, 234, 234, 234, 236, 236, 236, 236, 288, 288, 288, 288

Offset: 1

Author

Altug Alkan, Jul 29 2016

Keywords

• nonn

Examples

			a(3) = 3 because 3! divides sigma(3!) = 12.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..1100

Crossrefs

Cf. A000142, A062569.

Programs

Maple

N:= 500: # to get a(1) .. a(N)
k:= 1; skf:= 1;
for n from 1 to N do
  nf:= n!;
  while skf mod nf <> 0 do
    k:= k+1;
    skf:= numtheory:-sigma(k!);
  od:
  A[n]:= k;
od:
seq(A[i],i=1..N); # Robert Israel, Aug 09 2016

Mathematica

Table[k = 1; While[! Divisible[DivisorSigma[1, k!], n!], k++]; k, {n, 58}] (* Michael De Vlieger, Aug 08 2016 *)

PARI

a(n) = {my(k = 1); while(sigma(k!) % n! != 0, k++); k; }