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Search: id:A146334
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| A146334 |
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Numbers k such that continued fraction of (1+Sqrt[k])/2 has period 10 |
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+0 2
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| 43, 67, 116, 129, 134, 161, 162, 184, 218, 242, 243, 246, 270, 274, 297, 301, 314, 338, 339, 345, 354, 356, 407, 411, 451, 452, 459, 465, 475, 498, 515, 517, 532, 534, 561, 563, 590, 591, 595, 597, 603, 611, 638, 648, 657, 665, 669, 671, 690, 705, 715
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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For primes in this sequence see A146355.
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EXAMPLE
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a(1) = 43 because continued fraction of (1+Sqrt[43])/2 = 3, 1, 3, 1, 1, 12, 1, 1, 3, 1, 5, 1, 3, 1, 1, 12, 1, 1, 3, 1, 5, 1, 3, 1, 1, 12, 1, 1, 3, 1, ...
has period (1, 3, 1, 1, 12, 1, 1, 3, 1, 5) length 10
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MAPLE
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A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146334 := proc(n) RETURN(A146326(n) = 10) ; end: for n from 2 to 715 do if isA146334(n) then printf("%d, ", n) ; fi; od: [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 06 2009]
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MATHEMATICA
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s = 10; aa = {}; Do[k = ContinuedFraction[(1 + Sqrt[n])/2, 1000]; If[Length[k] < 190, AppendTo[aa, 0], m = 1; While[k[[s ]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]], m++ ]; s = s + 1; While[k[[s ]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]], m++ ]; s = s + 1; While[k[[s ]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]], m++ ]; AppendTo[aa, m]], {n, 1, 500}]; bb = {}; Do[If[aa[[n]] == 10, AppendTo[bb, n]], {n, 1, Length[aa]}]; bb (*Artur Jasinski*)
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CROSSREFS
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A000290, A078370, A146326-A146345, A146348-A146360.
Sequence in context: A020349 A050959 A139917 this_sequence A039385 A043208 A043988
Adjacent sequences: A146331 A146332 A146333 this_sequence A146335 A146336 A146337
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KEYWORD
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nonn
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AUTHOR
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Artur Jasinski (grafix(AT)csl.pl), Oct 30 2008
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EXTENSIONS
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284 removed by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 06 2009
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