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A103227 Least k such that the Gaussian integer (2n-1)+ki is squarefull.

%I A103227 #7 Mar 30 2012 17:22:35
%S A103227 7,4,10,1,0,2,9,5,6,8,3,7,0,0,2,8,6,5,9,2,1,1,0,4,0,7,4,10,1,12,2,0,5,
%T A103227 6,8,3,11,0,11,3,0,6,5,6,2,10,1,10,4,0,7,4,10,1,12,2,9,5,0,8,0,3,0,11,
%U A103227 3,8,6,0,4,2,12,1,10,0,7,7,0,10,1,12,2,9,5,6,0,0,11,0,11,3,5,6,5,9,0,6,1,10
%N A103227 Least k such that the Gaussian integer (2n-1)+ki is squarefull.
%C A103227 We treat only the odd case 2n-1 because for the even case we can always take k=0; that is, for n>0, 2n always has a factor of (1+i)^2. Using quadratic residues mod 25, it can be proved that 0<=a(n)<=12 for all n. The plot shows the squarefull Gaussian integers as black squares.
%H A103227 T. D. Noe, <a href="/A103227/b103227.txt">Table of n, a(n) for n=1..1000</a>
%H A103227 T. D. Noe, <a href="http://www.sspectra.com/math/ComplexMoebius.gif">Plot of the Moebius function for Gaussian Integers</a>
%e A103227 a(4)=1 because 7+i has the factor (2+i)^2, but 7+0i has no square factors because it is prime.
%t A103227 moebius[z_] := Module[{f, mu}, If[z==0, mu=0, If[Abs[z]==1, mu=1, f=FactorInteger[z, GaussianIntegers->True]; If[Abs[f[[1, 1]]]==1, f=Drop[f, 1]]; mu=1; Do[If[f[[i, 2]]==1, mu=-mu, mu=0], {i, Length[f]}]]]; mu]; Table[k=0; While[z=n+k*I; moebiusMuZ[z]!=0, k++ ]; k, {n, 1, 250, 2}]
%Y A103227 Cf. A103226 (Moebius function for Gaussian integers).
%K A103227 nonn
%O A103227 1,1
%A A103227 _T. D. Noe_, Jan 26 2005