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A062328
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Length of period of continued fraction expansion of square root of 3^n+1.
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1
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1, 0, 1, 4, 1, 26, 1, 56, 1, 44, 1, 264, 1, 814, 1, 136, 1, 3730, 1, 20968, 1, 2448, 1, 287980, 1, 397238, 1, 2678, 1, 670896, 1, 8110044, 1, 20696, 1, 1066520, 1, 366601254, 1, 277444, 1, 5903828476, 1, 7701738148, 1, 8208058, 1, 30287795640, 1, 253244432640, 1, 11656644672, 1, 2376211301858, 1, 590009437260, 1
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OFFSET
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0,4
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COMMENTS
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a(n) = 1 iff n is even. In this case, 3^n + 1 = A002522(3^(n/2)) and the continued fraction expansion of sqrt(3^n+1) is {3^(n/2); 2*3^(n/2), 2*3^(n/2), 2*3^(n/2), 2*3^(n/2), ...}. - Bernard Schott, Sep 25 2019
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LINKS
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FORMULA
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EXAMPLE
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The period of sqrt(244) contains 26 terms: [1, 1, 1, 1, 1, 2, 1, 5, 1, 1, 9, 1, 6, 1, 9, 1, 1, 5, 1, 2, 1, 1, 1, 1, 1, 30], so a(5) = 26.
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MAPLE
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with(numtheory): [seq(nops(cfrac(sqrt(3^k+1), 'periodic', 'quotients')[2]), k=2..18)];
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MATHEMATICA
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Table[Length[Last[ContinuedFraction[Sqrt[3^w+1]]]], {w, 1, 40}] (* corrected by Harvey P. Dale, Dec 05 2014 *)
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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