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A146336
Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 12.
3
94, 103, 124, 127, 128, 158, 160, 177, 190, 204, 208, 209, 216, 236, 239, 247, 263, 295, 296, 302, 316, 332, 351, 364, 376, 384, 385, 415, 423, 426, 432, 460, 464, 479, 492, 535, 544, 585, 606, 608, 609, 636, 639, 666, 668, 684, 696, 706, 734, 736, 744, 750
OFFSET
1,1
COMMENTS
For primes in this sequence see A146357.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)
EXAMPLE
a(2) = 103 because continued fraction of (1+sqrt(103))/2 = 15, 1, 1, 2, 1, 6, 20, 6, 1, 2, 1, 1, 9, 1, 1, 2, 1, 6, 20, 6, 1, 2, 1, 1, 9, 1, 1, 2, 1, 6, 20, 6, 1, 2 ... has period (1, 1, 2, 1, 6, 20, 6, 1, 2, 1, 1, 9) length 12.
MATHEMATICA
cf12Q[n_]:=!OddQ[Sqrt[n]]&&Length[ContinuedFraction[(1+Sqrt[n])/2][[2]]]==12; Select[Range[800], cf12Q] (* Harvey P. Dale, Oct 15 2011 *)
KEYWORD
nonn
AUTHOR
Artur Jasinski, Oct 30 2008
STATUS
approved