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A275476
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Least k such that n divides d(k!) where d = A000005 (k > 0).
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2
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1, 2, 6, 3, 6, 6, 15, 4, 10, 6, 12, 7, 28, 15, 6, 5, 18, 10, 20, 7, 15, 12, 24, 8, 21, 28, 10, 15, 60, 6, 63, 8, 12, 18, 28, 11, 76, 20, 28, 9, 81, 15, 46, 12, 10, 24, 48, 8, 27, 21, 18, 28, 110, 10, 24, 15, 20, 60, 120, 7, 243, 63, 15, 15, 28, 12, 68, 18, 24, 28
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OFFSET
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1,2
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LINKS
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EXAMPLE
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a(3) = 6 because A000005(6!) = 30 is divisible by 3.
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MATHEMATICA
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Table[k = 1; While[! Divisible[DivisorSigma[0, k!], n], k++]; k, {n, 120}] (* Michael De Vlieger, Aug 07 2016 *)
With[{c=Table[{k, DivisorSigma[0, k!]}, {k, 250}]}, Table[SelectFirst[ c, Divisible[ #[[2]], n]&], {n, 70}]][[All, 1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Sep 17 2019 *)
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PROG
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(PARI) a(n) = {my(k = 1); while(numdiv(k!) % n != 0, k++); k; }
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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