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Search: id:A146338
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| A146338 |
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Numbers k such that continued fraction of (1+Sqrt[k])/2 has period 15 |
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+0 3
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| 193, 281, 481, 1417, 1861, 1933, 2089, 2141, 2197, 2437, 2741, 2837, 3037, 3065, 3121, 3413, 3589, 3625, 3785, 3925, 3977
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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Something is wrong here! - N. J. A. Sloane (njas(AT)research.att.com), Oct 31 2008
For primes in this sequence see A146360.
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EXAMPLE
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a(1) = 193 because continued fraction of (1+Sqrt[193])/2 = 7, 2, 4, 6, 1, 2, 1, 1, 1, 1, 2, 1, 6, 4, 2, 13, 2, 4, 6, 1, 2, 1, 1, 1, 1, 2, 1, 6, 4, 2, 13...
has period (2, 4, 6, 1, 2, 1, 1, 1, 1, 2, 1, 6, 4, 2, 13) length 15
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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: isA146338 := proc(n) RETURN(A146326(n) = 15) ; end: for n from 2 to 4000 do if isA146338(n) then printf("%d, \n", 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]] == 14, 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: A142925 A060333 A020352 this_sequence A146360 A050964 A015988
Adjacent sequences: A146335 A146336 A146337 this_sequence A146339 A146340 A146341
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KEYWORD
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more,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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Extended beyond 3 terms by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 06 2009
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