%I #23 Sep 08 2022 08:46:00
%S 1,14,25,155,274,1691,2989,18446,32605,201215,355666,2194919,3879721,
%T 23942894,42321265,261176915,461654194,2849003171,5035874869,
%U 31077857966,54932969365,339007434455,599226788146,3698003921039,6536561700241,40339035696974
%N y-values in the solution to 13*x^2 - 12 = y^2.
%C When are both n+1 and 13*n+1 perfect squares? This problem gives the equation 13*x^2-12=y^2.
%H Vincenzo Librandi, <a href="/A199405/b199405.txt">Table of n, a(n) for n = 1..200</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,11,0,-1).
%F a(n+4) = 11*a(n+2)-a(n) with a(1)=1, a(2)=14, a(3)=25, a(4)=155.
%F G.f.: x*(1+x)*(1+13*x+x^2)/((1+3*x-x^2)*(1-3*x-x^2)). - _Bruno Berselli_, Nov 08 2011
%F a(n) = 2^(-1-n)*(2*(3-sqrt(13))^n+(-3-sqrt(13))^n*(-3+sqrt(13))-3*(-3+sqrt(13))^n-sqrt(13)*(-3+sqrt(13))^n+2*(3+sqrt(13))^n). - _Colin Barker_, Mar 27 2016
%t LinearRecurrence[{0, 11, 0, -1}, {1, 14, 25, 155}, 50] (* _Bruno Berselli_, Nov 08 2011 *)
%o (Magma) m:=27; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(1+x)*(1+13*x+x^2)/((1+3*x-x^2)*(1-3*x-x^2)))); // _Bruno Berselli_, Nov 08 2011
%o (PARI) Vec(x*(1+x)*(1+13*x+x^2)/((1+3*x-x^2)*(1-3*x-x^2)) + O(x^50)) \\ _Colin Barker_, Mar 27 2016
%Y Cf. A199404.
%K nonn,easy
%O 1,2
%A _Sture Sjöstedt_, Nov 05 2011
%E More terms from _Bruno Berselli_, Nov 08 2011