%I #29 Sep 08 2022 08:45:40

%S 99,649,3699,9249,17299,27849,40899,56449,74499,95049,118099,143649,

%T 171699,202249,235299,270849,308899,349449,392499,438049,486099,

%U 536649,589699,645249,703299,763849,826899,892449,960499,1031049

%N a(n) = 1250*n^2 - 700*n + 99.

%C The identity (1250*n^2 - 700*n + 99)^2 - (25*n^2 - 14*n + 2)*(250*n - 70)^2 = 1 can be written as a(n)^2 - A154357(n)*A154361(n)^2 = 1. See also the third comment in A154357.

%H Vincenzo Librandi, <a href="/A154359/b154359.txt">Table of n, a(n) for n = 0..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.: (99 + 352*x + 2049*x^2)/(1-x)^3. - _Bruno Berselli_, Dec 13 2011

%F a(n) = 3*a(n-1) -3*a(n-2) +a(n-3). - _Vincenzo Librandi_, Feb 21 2012

%F E.g.f.: (99 + 550*x + 1250*x^2)*exp(x). - _G. C. Greubel_, Sep 15 2016

%t LinearRecurrence[{3, -3, 1}, {99, 649, 3699}, 50] (* _Vincenzo Librandi_, Feb 21 2012 *)

%o (PARI) for(n=0, 40, print1(1250*n^2-700*n+99", ")); \\ _Vincenzo Librandi_, Feb 21 2012

%o (Magma) [1250*n^2-700*n+99: n in [0..40]]; // _Bruno Berselli_, Sep 15 2016

%K nonn,easy

%O 0,1

%A _Vincenzo Librandi_, Jan 08 2009

%E Edited by _Charles R Greathouse IV_, Jul 29 2010

%E Librandi's comment rewritten by _Bruno Berselli_, Dec 13 2011