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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, 259, 269, 397, 425, 493, 565, 697, 761, 941, 1013, 1026
(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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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, A146346-A146360.
Adjacent sequences: A146327 A146328 A146329 this_sequence A146331 A146332 A146333
Sequence in context: A142449 A044373 A044754 this_sequence A146350 A050954 A141957
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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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