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Search: id:A103227
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| A103227 |
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Least k such that the Gaussian integer (2n-1)+ki is squarefull. |
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+0
2
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| 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, 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, 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
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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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.
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LINKS
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T. D. Noe, Table of n, a(n) for n=1..1000
T. D. Noe, Plot of the Moebius function for Gaussian Integers
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EXAMPLE
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a(4)=1 because 7+i has the factor (2+i)^2, but 7+0i has no square factors because it is prime.
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MATHEMATICA
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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}]
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CROSSREFS
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Cf. A103226 (Moebius function for Gaussian integers).
Sequence in context: A117028 A138282 A155823 this_sequence A165415 A069199 A138339
Adjacent sequences: A103224 A103225 A103226 this_sequence A103228 A103229 A103230
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KEYWORD
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nonn
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AUTHOR
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T. D. Noe (noe(AT)sspectra.com), Jan 26 2005
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