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Search: id:A146330
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| A146330 |
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Numbers k such that continued fraction of (1+Sqrt[k])/2 has period 5 |
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+0
3
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| 41, 149, 157, 181, 269, 397, 425, 493, 565, 697, 761, 941, 1013, 1037, 1325, 1565, 1781, 1825
(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 A146350.
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EXAMPLE
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a(1) = 41 because continued fraction of (1+Sqrt[41])/2 = 3, 1, 2, 2, 1, 5, 1, 2, 2, 1, 5, 1, 2, 2, 1, 5, 1, 2, ...
has period (1,2,2,1,5) length 5
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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: isA146330 := proc(n) RETURN(A146326(n) = 5) ; end: for n from 2 to 2000 do if isA146330(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]] == 5, 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: A142449 A044373 A044754 this_sequence A146350 A050954 A141957
Adjacent sequences: A146327 A146328 A146329 this_sequence A146331 A146332 A146333
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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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Removed 259 and 1026 - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 06 2009
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