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A275369
Least k such that n! divides sigma(k!) (k > 0).
1
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
1,2
LINKS
EXAMPLE
a(3) = 3 because 3! divides sigma(3!) = 12.
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 *)
PROG
(PARI) a(n) = {my(k = 1); while(sigma(k!) % n! != 0, k++); k; }
CROSSREFS
Sequence in context: A117772 A026924 A240313 * A377984 A062130 A151970
KEYWORD
nonn
AUTHOR
Altug Alkan, Jul 29 2016
STATUS
approved