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

Author

David Wasserman, Jan 08 2006

Keywords

• easy
• nonn

Comments

Third row of A113465.

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

Crossrefs

Cf. A005238, A113458, A113465.

Programs

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 *)

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