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
Keywords
Examples
a(3) = 3 because 3! divides sigma(3!) = 12.
Links
- Robert Israel, Table of n, a(n) for n = 1..1100
Programs
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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
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Mathematica
Table[k = 1; While[! Divisible[DivisorSigma[1, k!], n!], k++]; k, {n, 58}] (* Michael De Vlieger, Aug 08 2016 *)
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PARI
a(n) = {my(k = 1); while(sigma(k!) % n! != 0, k++); k; }
Comments