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Search: id:A146332
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| A146332 |
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Numbers k such that continued fraction of (1+Sqrt[k])/2 has period 7 |
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+0
3
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| 89, 109, 113, 137, 373, 389, 509, 653, 685, 797, 853, 925, 949, 997, 1009
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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For primes in this sequence see A146352.
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EXAMPLE
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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
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MAPLE
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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: [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 06 2009]
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MATHEMATICA
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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 (*Artur Jasinski*)
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CROSSREFS
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A000290, A078370, A146326-A146345, A146348-A146360.
Sequence in context: A033254 A157764 A032691 this_sequence A146352 A050956 A121608
Adjacent sequences: A146329 A146330 A146331 this_sequence A146333 A146334 A146335
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KEYWORD
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nonn
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AUTHOR
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Artur Jasinski (grafix(AT)csl.pl), Oct 30 2008
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EXTENSIONS
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997 added by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 06 2009
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