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%I #22 Jul 16 2021 01:34:16
%S 962,3846,8652,15380,24030,34602,47096,61512,77850,96110,116292,
%T 138396,162422,188370,216240,246032,277746,311382,346940,384420,
%U 423822,465146,508392,553560,600650,649662,700596,753452,808230,864930,923552,984096
%N Numbers k such that the least term in the periodic part of the continued fraction for sqrt(k) is 62.
%C a(n) = 961n^2 + n for n < 65, but a(65) = 3940288. - _Charles R Greathouse IV_, Aug 04 2017
%H Charles R Greathouse IV, <a href="/A031740/b031740.txt">Table of n, a(n) for n = 1..10000</a>
%F 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
%F 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
%t Select[Range[10^6], !IntegerQ[Sqrt[#]]&&Min[ContinuedFraction[Sqrt[#]][[2]]] == 62 &] (* _Vincenzo Librandi_, Jun 20 2016 *)
%K nonn
%O 1,1
%A _David W. Wilson_
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