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 A146360 Primes p such that continued fraction of (1 + sqrt(p))/2 has period 15: primes in A146338. 36
 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; text; internal format)
 OFFSET 1,1 LINKS 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: # R. J. Mathar, Sep 06 2009 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 *) Select[Prime[Range[1500]], Length[ContinuedFraction[(Sqrt[#]+1)/2][[2]]] == 15&] (* Harvey P. Dale, Aug 16 2014 *) CROSSREFS Cf. A000290, A050950-A050969, A078370, A146326-A146345, A146348-A146360. Sequence in context: A020352 A146338 A201858 * A050964 A015988 A260539 Adjacent sequences:  A146357 A146358 A146359 * A146361 A146362 A146363 KEYWORD nonn AUTHOR Artur Jasinski, Oct 30 2008 EXTENSIONS 8539 removed by R. J. Mathar, Sep 06 2009 STATUS approved

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Last modified July 23 00:43 EDT 2019. Contains 325228 sequences. (Running on oeis4.)