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%I #43 Sep 08 2022 08:46:13
%S 3,36,393,4287,46764,510117,5564523,60699636,662131473,7222746567,
%T 78788080764,859446141837,9375119479443,102266868132036,
%U 1115560429972953,12168897861570447,132742316047301964,1447996578658751157,15795220049198960763,172299423962529817236,1879498443538629028833,20502183454962389499927
%N Numbers n such that 13*n^2 + 52 is a square.
%C The limit of a(n)/a(n-1) approaches (11+sqrt(117))/2 as n -> infinity.
%C The continued fraction [a(n); a(n), a(n), ...] = ((3+sqrt(13))/2)^(2*n-1).
%C Equivalently, numbers n such that (n^2+4)/13 is a square.
%C Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(13)). - _Greg Dresden_, Jul 22 2019
%C As 13*n^2 + 52 = 13 * (n^2 + 4), n == 3 (mod 13) or n == 10 (mod 13) alternately. - _Bernard Schott_, Jul 23 2019
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11,-1).
%F G.f.: 3*x*(1+x)/(1-11*x+x^2).
%F a(n) = 11*a(n-1) - a(n-2); a(0) = 3, a(1) = 36.
%F a(n) = floor(((3+sqrt(13))/2)^(2*n+1)+((3+sqrt(13))/2)^(1-2*n)).
%F a(n) = 3*A097783(n-1). - _R. J. Mathar_, Jun 07 2016
%t Table[Floor[((3 + Sqrt[13])/2)^(2*n + 1) + ((3 + Sqrt[13])/2)^(1 - 2 n)], {n, 21}] (* _Michael De Vlieger_, Jun 20 2015 *)
%t LinearRecurrence[{11, -1}, {3, 36}, 25] (* _Vincenzo Librandi_, Jul 23 2019 *)
%o (PARI) for(n=1,20,q=((3+sqrt(13))/2)^(2*n-1);print1(contfrac(q)[1],", "))
%o (Magma) I:=[3,36]; [n le 2 select I[n] else 11*Self(n-1)-Self(n-2): n in [1..30]]; // _Vincenzo Librandi_, Jul 23 2019
%Y Cf. A002878, A077444.
%K nonn,easy
%O 1,1
%A _Derek Orr_, Jun 18 2015