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%I #14 Feb 06 2017 09:44:57
%S 1,258,66563,17172996,4430566405,1143068959494,294907360983047,
%T 76084956064666632,19629623757323008009,5064366844433271399690,
%U 1306587016240026698112011,337094385823082454841499148
%N Chebyshev U(n,x) polynomial evaluated at x=129 = 3*43.
%C Used to form integer solutions of Pell equation a^2 - 65*b^2 =-1. See A097735 with A097736.
%H Indranil Ghosh, <a href="/A097734/b097734.txt">Table of n, a(n) for n = 0..413</a>
%H R. Flórez, R. A. Higuita, A. Mukherjee, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Mukherjee/mukh2.html">Alternating Sums in the Hosoya Polynomial Triangle</a>, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (258, -1).
%F a(n) = 2*129*a(n-1) - a(n-2), n>=1, a(0)=1, a(-1):=0.
%F a(n) = S(n, 2*129)= U(n, 129), Chebyshev's polynomials of the second kind. See A049310.
%F G.f.: 1/(1-258*x+x^2).
%F a(n)= sum((-1)^k*binomial(n-k, k)*258^(n-2*k), k=0..floor(n/2)), n>=0.
%F a(n) = ((129+16*sqrt(65))^(n+1) - (129-16*sqrt(65))^(n+1))/(32*sqrt(65)), n>=0.
%t LinearRecurrence[{258, -1},{1, 258},12] (* _Ray Chandler_, Aug 11 2015 *)
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Aug 31 2004
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