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a(n) = 441*n^2 - 488*n + 135.
3

%I #27 Sep 23 2022 14:42:23

%S 88,923,2640,5239,8720,13083,18328,24455,31464,39355,48128,57783,

%T 68320,79739,92040,105223,119288,134235,150064,166775,184368,202843,

%U 222200,242439,263560,285563,308448,332215,356864,382395,408808,436103,464280

%N a(n) = 441*n^2 - 488*n + 135.

%C The identity (388962*n^2 - 430416*n + 119071)^2 - (441*n^2 - 488*n + 135)*(18522*n - 10248)^2 = 1 can be written as A157732(n)^2 - a(n)*A157731(n)^2 = 1.

%C 441*a(n) + 1 is a square. - _Bruno Berselli_, Apr 23 2018

%C The continued fraction expansion of sqrt(a(n)) is [21n-12; {2, 1, 1, 1, 2, 41n-24}]. - _Magus K. Chu_, Sep 23 2022

%H Vincenzo Librandi, <a href="/A157730/b157730.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*(88 + 659*x + 135*x^2)/(1 - x)^3.

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

%F a(n) = (9*n - 5)*(49*n - 27). - _Bruno Berselli_, Apr 23 2018

%t LinearRecurrence[{3, -3, 1}, {88, 923, 2640}, 40]

%o (Magma) I:=[88, 923, 2640]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]];

%o (PARI) a(n) = 441*n^2 - 488*n + 135.

%Y Cf. A157731, A157732.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Mar 05 2009