%I #20 Sep 08 2022 08:45:42
%S 10662963,52387849,125674659,230523393,366934051,534906633,734441139,
%T 965537569,1228195923,1522416201,1848198403,2205542529,2594448579,
%U 3014916553,3466946451,3950538273,4465692019,5012407689,5590685283
%N a(n) = 15780962*n^2 - 5618000*n + 500001.
%C The identity (15780962*n^2 - 5618000*n + 500001)^2 - (2809*n^2 - 1000*n + 89)*(297754*n - 53000)^2 = 1 can be written as a(n)^2 -A157760(n)*A157761(n)^2 = 1.
%C This is the case s=53 and r=500 of the identity (2*(s^2*n-r)^2+1)^2 - (((s^2*n-r)^2+1)/s^2)*(2*s*(s^2*n-r))^2 = 1, where ((s^2*n-r)^2+1)/s^2 is an integer if r^2 == -1 (mod s^2). Therefore, for s=53, nonnegative r values are: 500, 2309, 3309, 5118, 6118, 7927, 8927, 10736, 11736, ... - _Bruno Berselli_, Apr 24 2018
%H Vincenzo Librandi, <a href="/A157762/b157762.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F G.f: x*(10662963 + 20398960*x + 500001*x^2)/(1 - x)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%t LinearRecurrence[{3, -3, 1}, {10662963, 52387849, 125674659}, 30]
%o (Magma) I:=[10662963, 52387849, 125674659]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..40]];
%o (PARI) a(n) = 15780962*n^2 - 5618000*n + 500001;
%Y Cf. A157760, A157761.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 06 2009