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Least k such that k, k+n and k+2n have the same number of divisors.
3

%I #10 Aug 20 2017 17:26:21

%S 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,

%T 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,

%U 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

%N Least k such that k, k+n and k+2n have the same number of divisors.

%C Third row of A113465.

%H Harvey P. Dale, <a href="/A113467/b113467.txt">Table of n, a(n) for n = 1..1000</a>

%e a(7) = 8 because 8, 15 and 22 each have 4 divisors.

%t 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 *)

%o (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

%Y Cf. A005238, A113458, A113465.

%K easy,nonn

%O 1,1

%A _David Wasserman_, Jan 08 2006