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Least primitive number k such that 1/k is in the Cantor set and the fraction 1/k has period n in base 3.
1

%I #3 Mar 30 2012 17:22:56

%S 1,4,13,10,121,28,1093,82,757,244,88573,730,797161,2188,59293,6562,

%T 64570081,1036,581130733,2362,4785157,177148,47071589413,84253,

%U 3501192601,1594324,387440173,4782970,34315188682441,66124

%N Least primitive number k such that 1/k is in the Cantor set and the fraction 1/k has period n in base 3.

%C Primitive means that 3 does not divide k. The term a(n) is the first term in row n of A173800. For n=p^k, with p prime and k>0, it appears that a(n) = Phi(n,3), the n-th cyclotomic polynomial evaluated at 3.

%H T. D. Noe, <a href="/A175174/b175174.txt">Table of n, a(n) for n=1..200</a>

%K nonn

%O 1,2

%A _T. D. Noe_, Feb 28 2010