A113467 Least k such that k, k+n and k+2n have the same number of divisors.
33, 3, 119, 3, 77, 5, 8, 3, 77, 3, 35, 5, 8, 3, 187, 6, 21, 5, 8, 3, 145, 33, 39, 5, 8, 39, 8, 3, 33, 7, 15, 12, 189, 3, 28, 7, 21, 3, 55, 3, 33, 5, 8, 66, 209, 69, 35, 5, 8, 3, 115, 39, 141, 5, 51, 6, 8, 27, 15, 7, 21, 66, 95, 3, 40, 5, 27, 3, 8, 15, 35, 7, 69, 55, 287, 6, 65, 11, 8, 3, 24
Offset: 1
Examples
a(7) = 8 because 8, 15 and 22 each have 4 divisors.
Links
- Harvey P. Dale, Table of n, a(n) for n = 1..1000
Programs
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Mathematica
snd[n_]:=Module[{k=1},While[Length[Union[DivisorSigma[0,{k,k+n,k+2n}]]]>1, k++];k]; Array[snd,90] (* Harvey P. Dale, Aug 20 2017 *)
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PARI
a(n) = {k = 1; until ((numdiv(k) == numdiv(k+n)) && (numdiv(k) == numdiv(k+2*n)), k++); return (k);} \\ Michel Marcus, Jun 16 2013
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