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 A031740 Numbers k such that the least term in the periodic part of the continued fraction for sqrt(k) is 62. 1
 962, 3846, 8652, 15380, 24030, 34602, 47096, 61512, 77850, 96110, 116292, 138396, 162422, 188370, 216240, 246032, 277746, 311382, 346940, 384420, 423822, 465146, 508392, 553560, 600650, 649662, 700596, 753452, 808230, 864930, 923552, 984096 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS a(n) = 961n^2 + n for n < 65, but a(65) = 3940288. - Charles R Greathouse IV, Aug 04 2017 LINKS Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 FORMULA Conjecture: a(n) = n*(961*n+1). a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). G.f.: 2*x*(481+480*x)/(1-x)^3. - Colin Barker, Sep 30 2012 Conjecture is false. m*(961*m+1) for m >= 1 is a subsequence (see comment in A031712) and 3940288 is a term not of the form m*(961*m+1). - Chai Wah Wu, Jun 19 2016 MATHEMATICA Select[Range[10^6], !IntegerQ[Sqrt[#]]&&Min[ContinuedFraction[Sqrt[#]][[2]]] == 62 &] (* Vincenzo Librandi, Jun 20 2016 *) CROSSREFS Sequence in context: A098207 A304315 A158414 * A031529 A347884 A031709 Adjacent sequences: A031737 A031738 A031739 * A031741 A031742 A031743 KEYWORD nonn AUTHOR David W. Wilson STATUS approved

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Last modified October 3 18:47 EDT 2023. Contains 365870 sequences. (Running on oeis4.)