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%I #26 Sep 08 2022 08:45:41
%S 19603,143649,381939,734473,1201251,1782273,2477539,3287049,4210803,
%T 5248801,6401043,7667529,9048259,10543233,12152451,13875913,15713619,
%U 17665569,19731763,21912201,24206883,26615809,29138979
%N a(n) = 57122*n^2 - 47320*n + 9801.
%C The identity (57122*n^2 - 47320*n+9801)^2 - (169*n^2 - 140*n + 29)*(4394*n - 1820)^2 = 1 can be written as a(n)^2 - A156639(n)*A156627(n)^2 = 1.
%C This is the case s=13 and r=70 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=13, nonnegative r values are: 70, 99, 239, 268, 408, 437, 577, 606, 746, 775, 915, 944, ... - _Bruno Berselli_, Apr 24 2018
%H Vincenzo Librandi, <a href="/A156721/b156721.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*(19603 + 84840*x + 9801*x^2)/(1 - x)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%t LinearRecurrence[{3, -3, 1}, {19603, 143649, 381939}, 40]
%o (Magma) I:=[19603, 143649, 381939]; [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)=57122*n^2-47320*n+9801 \\ _Charles R Greathouse IV_, Dec 23 2011
%Y Cf. A156627, A156639.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Feb 15 2009
%E Edited by _Charles R Greathouse IV_, Jul 25 2010
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