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%I #47 Oct 31 2024 13:39:06
%S 3,2,19,54,107,178,267,374,499,642,803,982,1179,1394,1627,1878,2147,
%T 2434,2739,3062,3403,3762,4139,4534,4947,5378,5827,6294,6779,7282,
%U 7803,8342,8899,9474,10067,10678,11307,11954,12619,13302,14003,14722,15459,16214,16987
%N a(n) = 9*n^2 - 10*n + 3.
%C The identity (81*n^2 + 72*n + 17)^2 - (9*n^2 + 8*n + 2)*(27*n + 12)^2 = 1 can be written as A154295(n+1)^2 - a(n+1)*A154266(n)^2 = 1. - _Vincenzo Librandi_, Feb 03 2012
%C For n >= 1, the continued fraction expansion of sqrt(a(n)) is [3n-2; {2, 1, 3n-3, 1, 2, 6n-4}]. For n=1, this collapses to [1; {2}]. - _Magus K. Chu_, Sep 05 2022
%H Vincenzo Librandi, <a href="/A154262/b154262.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 From _Vincenzo Librandi_, Feb 02 2012: (Start)
%F G.f.: (3 - 7*x + 22*x^2)/(1-x)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
%F E.g.f.: (3 - x + 9*x^2)*exp(x). - _Elmo R. Oliveira_, Oct 31 2024
%t LinearRecurrence[{3, -3, 1}, {2, 19, 54}, 50] (* _Vincenzo Librandi_, Feb 02 2012 *)
%t Table[9n^2-10n+3,{n,0,50}] (* _Harvey P. Dale_, Feb 11 2023 *)
%o (PARI) a(n)=9*n^2-10*n+3 \\ _Charles R Greathouse IV_, Dec 27 2011
%o (Magma) I:=[2, 19, 54]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // _Vincenzo Librandi_, Feb 02 2012
%Y Cf. A154266, A154295.
%K nonn,easy
%O 0,1
%A _Vincenzo Librandi_, Jan 06 2009
%E Edited by _Charles R Greathouse IV_, Jul 25 2010