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A146360
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Primes p such that continued fraction of (1+sqrt(p))/2 has period 15: primes in A146338.
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36
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193, 281, 1861, 1933, 2089, 2141, 2437, 2741, 2837, 3037, 3121, 3413, 4001, 4637, 4877, 5821, 6653, 7673, 8117, 10069, 10273, 10457, 11197, 11549, 11821, 12409
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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MAPLE
| A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146360 := proc(n) RETURN(isprime(n) and A146326(n) = 15) ; end: for n from 2 to 30000 do if isA146360(n) then printf("%d, \n", n) ; fi; od: [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 06 2009]
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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]] == 15, AppendTo[bb, Prime[n]]], {n, 1, Length[aa]}]; bb (*Artur Jasinski*)
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CROSSREFS
| Cf. A000290, A050950-A050969, A078370, A146326-A146345, A146348-A146360.
Sequence in context: A020352 A146338 A201858 * A050964 A015988 A142743
Adjacent sequences: A146357 A146358 A146359 * A146361 A146362 A146363
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KEYWORD
| nonn
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AUTHOR
| Artur Jasinski (grafix(AT)csl.pl), Oct 30 2008
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EXTENSIONS
| 8539 removed - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 06 2009
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