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 A146332 Numbers k such that the continued fraction of (1+sqrt(k))/2 has period 7. 2
 89, 109, 113, 137, 373, 389, 509, 653, 685, 797, 853, 925, 949, 997, 1009 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS For primes in this sequence see A146352. LINKS EXAMPLE a(4) = 137 because continued fraction of (1+sqrt(137))/2 = 6, 2, 1, 5, 5, 1, 2, 11, 2, 1, 5, 5, 1, 2, 11, 2, 1, 5, 5, 1, 2, 11 ... has period (2, 1, 5, 5, 1, 2, 11) length 7. MAPLE A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146332 := proc(n) RETURN(A146326(n) = 7) ; end: for n from 2 to 1100 do if isA146332(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]] == 7, AppendTo[bb, n]], {n, 1, Length[aa]}]; bb CROSSREFS Cf. A000290, A078370, A146326-A146345, A146348-A146360. Sequence in context: A215165 A157764 A032691 * A146352 A050956 A121608 Adjacent sequences:  A146329 A146330 A146331 * A146333 A146334 A146335 KEYWORD nonn,more AUTHOR Artur Jasinski, Oct 30 2008 EXTENSIONS 997 added by R. J. Mathar, Sep 06 2009 STATUS approved

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Last modified July 22 14:53 EDT 2019. Contains 325224 sequences. (Running on oeis4.)