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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
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: A304315 A393861 A158414 * A031529 A347884 A031709
KEYWORD
nonn
STATUS
approved