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 A146329 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 4. 1
 6, 7, 8, 14, 20, 23, 24, 28, 32, 33, 34, 42, 47, 48, 52, 55, 60, 62, 69, 72, 75, 78, 79, 80, 95, 98, 110, 119, 120, 126, 133, 135, 136, 138, 140, 141, 142, 156, 167, 168, 174, 180, 189, 194, 205, 210, 213, 215, 219, 220, 222, 223, 224, 248, 252, 254, 272, 287, 288 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS For primes in this sequence see A146349. LINKS EXAMPLE a(2) = 7 because continued fraction of (1 + sqrt(7))/2 = 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, ... has period (1,1,4,1) length 4. MAPLE isA146329 := proc(n) RETURN(A146326(n) = 4) ; end: for n from 2 to 400 do if isA146329(n) then printf("%d, ", n) ; fi; od: # R. J. Mathar, Sep 06 2009 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]] == 4, AppendTo[bb, n]], {n, 1, Length[aa]}]; bb (* Artur Jasinski *) cf4Q[n_]:=Module[{s=(1+Sqrt[n])/2}, If[IntegerQ[s], 1, Length[ ContinuedFraction[ s][[2]]]]==4]; Select[Range[300], cf4Q] (* Harvey P. Dale, Dec 14 2017 *) CROSSREFS Cf. A000290, A078370, A146326-A146345, A146348-A146360. Sequence in context: A118733 A047275 A047590 * A272344 A133893 A101647 Adjacent sequences:  A146326 A146327 A146328 * A146330 A146331 A146332 KEYWORD nonn AUTHOR Artur Jasinski, Oct 30 2008 EXTENSIONS 39, 68, 150, 155, etc. removed by R. J. Mathar, Sep 06 2009 STATUS approved

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Last modified July 15 14:07 EDT 2019. Contains 325030 sequences. (Running on oeis4.)