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A146336
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Numbers k such that continued fraction of (1+sqrt(k))/2 has period 12.
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3
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94, 103, 124, 127, 128, 158, 160, 177, 190, 204, 208, 209, 216, 236, 239, 247, 263, 295, 296, 302, 316, 332, 351, 364, 376, 384, 385, 415, 423, 426, 432, 460, 464, 479, 492, 535, 544, 585, 606, 608, 609, 636, 639, 666, 668, 684, 696, 706, 734, 736, 744, 750
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| For primes in this sequence see A146357.
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LINKS
| Harvey P. Dale, Table of n, a(n) for n = 1..1000
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EXAMPLE
| a(2) = 103 because continued fraction of (1+Sqrt[103])/2 = 15, 1, 1, 2, 1, 6, 20, 6, 1, 2, 1, 1, 9, 1, 1, 2, 1, 6, 20, 6, 1, 2, 1, 1, 9, 1, 1, 2, 1, 6, 20, 6, 1, 2 ...
has period (1, 1, 2, 1, 6, 20, 6, 1, 2, 1, 1, 9) length 12
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MATHEMATICA
| 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]] == 11, AppendTo[bb, n]], {n, 1, Length[aa]}]; bb (*Artur Jasinski*)
cf12Q[n_]:=!OddQ[Sqrt[n]]&&Length[ContinuedFraction[(1+Sqrt[n])/2][[2]]]==12; Select[Range[800], cf12Q] (* From Harvey P. Dale, Oct 15 2011 *)
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CROSSREFS
| A000290, A078370, A146326-A146345, A146348-A146360.
Sequence in context: A118684 A020444 A045301 * A044974 A020355 A039551
Adjacent sequences: A146333 A146334 A146335 * A146337 A146338 A146339
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KEYWORD
| nonn
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AUTHOR
| Artur Jasinski (grafix(AT)csl.pl), Oct 30 2008
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