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 A146357 Primes p such that continued fraction of (1+Sqrt[p])/2 has period 12 : primes in A146336. 1
 103, 127, 239, 263, 479, 887, 1567, 2711, 5743, 5903, 8311, 8447, 10567, 10847, 12391, 14783, 14831, 15887, 18191, 22343, 23447, 28151, 31391, 32359, 40087, 40343, 42703, 53407, 60103, 60623, 64231, 75431, 79943, 81559, 83663, 93503 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 LINKS MATHEMATICA \$MaxExtraPrecision = 4000; s = 10; aa = {}; Do[k = ContinuedFraction[(1 + Sqrt[Prime[n]])/2, 3000]; 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]] || k[[s + 4 m]] != k[[s + 5 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]] || k[[s + 4 m]] != k[[s + 5 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]] || k[[s + 4 m]] != k[[s + 5 m]]; AppendTo[aa, m]], {n, 1, 1495}]; bb = {}; Do[If[aa[[n]] == 12, AppendTo[bb, Prime[n]]], {n, 1, Length[aa]}]; bb (*Artur Jasinski*) Select[Prime[Range[10000]], Length[ContinuedFraction[(1+Sqrt[#])/2][[2]]] == 12&] (* Harvey P. Dale, May 18 2017 *) CROSSREFS Sequence in context: A139957 A077404 A139979 * A252269 A252262 A023080 Adjacent sequences:  A146354 A146355 A146356 * A146358 A146359 A146360 KEYWORD nonn AUTHOR Artur Jasinski, Oct 30 2008 EXTENSIONS Period length in definition corrected, 103 added, 607 and 2063 removed. - R. J. Mathar, Nov 08 2008 STATUS approved

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Last modified June 25 07:53 EDT 2019. Contains 324347 sequences. (Running on oeis4.)