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 A146340 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 17. 3
 521, 617, 709, 1433, 1597, 2549, 2909, 2965, 3161, 3581, 3821, 4013, 4285, 4649, 5501, 5585, 5693, 5813, 6197, 6409, 7825, 7853, 8093, 8125, 8573, 8917, 9281, 9665, 9677, 9925, 10265, 10597, 10973, 11273, 12085, 12805, 13061, 13109, 13613, 13957, 14677 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS For primes in this sequence see A146362. LINKS Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..150 from Harvey P. Dale) EXAMPLE a(1) = 521 because continued fraction of (1+sqrt(521))/2 = 11, 1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21, 1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21, 1, 10, 2, 5, ... has period (1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21) length 17. MAPLE A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146340 := proc(n) RETURN(A146326(n) = 17) ; end: for n from 2 do if isA146340(n) then printf("%d, \n", n) ; fi; od: # R. J. Mathar, Sep 06 2009 MATHEMATICA cf17Q[n_]:=Module[{s=(1+Sqrt[n])/2}, If[IntegerQ[s], 1, Length[ ContinuedFraction[ s][[2]]]]==17]; Select[Range[5000], cf17Q] (* Harvey P. Dale, Dec 20 2017 *) CROSSREFS Cf. A000290, A078370, A146326-A146345, A146348-A146360. Sequence in context: A291998 A300395 A139663 * A146362 A050966 A113158 Adjacent sequences:  A146337 A146338 A146339 * A146341 A146342 A146343 KEYWORD more,nonn AUTHOR Artur Jasinski, Oct 30 2008 EXTENSIONS 998 and 1006 removed, sequence extended by R. J. Mathar, Sep 06 2009 More terms from Harvey P. Dale, Dec 20 2017 STATUS approved

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Last modified September 24 12:36 EDT 2021. Contains 347642 sequences. (Running on oeis4.)