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Search: id:A146333
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| A146333 |
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Numbers k such that continued fraction of (1+Sqrt[k])/2 has period 8 |
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3
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| 31, 40, 46, 71, 76, 88, 91, 92, 96, 104, 108, 152, 153, 155, 176, 188, 192, 200, 206, 207, 234, 238, 261, 266, 276, 279, 280, 282, 320, 328, 335, 336, 348, 366, 378, 383, 386, 392, 408, 414, 450, 476, 477, 480, 488, 501, 503, 504, 505, 540, 542, 555, 558, 581
(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 A146353.
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EXAMPLE
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a(1) = 31 because continued fraction of (1+Sqrt[31])/2 = 3, 3, 1, 1, 10, 1, 1, 3, 5, 3, 1, 1, 10, 1, 1, 3, 5, 3, 1, 1, 10, 1, ...
has period (3, 1, 1, 10, 1, 1, 3, 5) length 8
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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: isA146333 := proc(n) RETURN(A146326(n) = 8) ; end: for n from 2 to 700 do if isA146333(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]] == 8, 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: A039378 A043201 A043981 this_sequence A109645 A156298 A099180
Adjacent sequences: A146330 A146331 A146332 this_sequence A146334 A146335 A146336
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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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155 and 279 etc. added, 311 etc. removed by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 06 2009
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