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Search: id:A067564
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| A067564 |
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Numbers n such that determinant[{{n, phi(n), sigma(n)}, {n+1, phi(n+1), sigma(n+1)}, {n+2, phi(n+2), sigma(n+2)}] is a perfect cube. |
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1
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OFFSET
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1,2
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COMMENT
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If n is a term of the sequence, then the parallelipiped formed by the vectors {n, phi(n), sigma(n)}, {n+1, phi(n+1), sigma(n+1)}, {n+2, phi(n+2), sigma(n+2) has the same volume as that of an integral cube.
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EXAMPLE
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For 217, the corresponding matrix is {{217,phi(217), sigma(217)},{218,phi(218), sigma(218)},{219,phi(219), sigma(219)} = {{217,180,256},{218,108,330},{219,144,296}}, whose determinant is 216 = 6^3. Therefore 217 is a term of the sequence.
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MATHEMATICA
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p[n_] := Det[{{n, EulerPhi[n], DivisorSigma[1, n]}, {n + 1, EulerPhi[n + 1], DivisorSigma[1, n + 1]}, {n + 2, EulerPhi[n + 2], DivisorSigma[1, n + 2]}}]; Do[If[IntegerQ[p[i]^(1/3)], Print[i]], {i, 1, 10^5}]
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CROSSREFS
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Sequence in context: A111832 A114563 A112311 this_sequence A165797 A011823 A122894
Adjacent sequences: A067561 A067562 A067563 this_sequence A067565 A067566 A067567
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KEYWORD
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nonn
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AUTHOR
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Joseph L. Pe (joseph_l_pe(AT)hotmail.com), Jan 29 2002
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