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%I #36 Sep 05 2022 09:10:10
%S 23,96,219,392,615,888,1211,1584,2007,2480,3003,3576,4199,4872,5595,
%T 6368,7191,8064,8987,9960,10983,12056,13179,14352,15575,16848,18171,
%U 19544,20967,22440,23963,25536,27159,28832,30555,32328,34151,36024
%N a(n) = 25*n^2 - 2*n.
%C The identity (1250*n^2 - 100*n + 1)^2 - (25*n^2 - 2*n)*(250*n - 10)^2 = 1 can be written as A154374(n)^2 - a(n)*A154378(n)^2 = 1 (see also the second comment in A154374). - _Vincenzo Librandi_, Jan 30 2012
%C The continued fraction expansion of sqrt(a(n)) is [5n-1; {1, 3, 1, 10n-2}]. - _Magus K. Chu_, Sep 04 2022
%H Vincenzo Librandi, <a href="/A154376/b154376.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 From _Vincenzo Librandi_, Jan 30 2012: (Start)
%F G.f.: x*(23 + 27*x)/(1-x)^3.
%F a(n) = 3*a(n-1) -3*a(n-2) +a(n-3). (End)
%F E.g.f.: (25*x^2 + 23*x)*exp(x). - _G. C. Greubel_, Sep 15 2016
%t LinearRecurrence[{3, -3, 1}, {23, 96, 219}, 50] (* _Vincenzo Librandi_, Jan 30 2012 *)
%o (PARI) a(n)=25*n^2-2*n \\ _Charles R Greathouse IV_, Dec 26 2011
%Y Cf. A154374, A154378.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Jan 08 2009
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