%I #33 Jul 14 2023 05:20:43
%S 1,198,39203,7761996,1536836005,304285766994,60247045028807,
%T 11928610629936792,2361804657682456009,467625393610496352990,
%U 92587466130220595436011,18331850668390067399977188,3629613844875103124600047213,718645209434602028603409370986
%N Chebyshev U(n,x) polynomial evaluated at x=99 gives 2*7^2+1.
%C Used to form integer solutions of Pell equation a^2 - 50*b^2 =-1. See A097732 with A097733.
%H Indranil Ghosh, <a href="/A097731/b097731.txt">Table of n, a(n) for n = 0..434</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 (198,-1).
%F a(n) = 2*99*a(n-1) - a(n-2), n>=1, a(0)=1, a(-1):=0.
%F a(n) = S(n, 2*99)= U(n, 99), Chebyshev's polynomials of the second kind. See A049310.
%F G.f.: 1/(1-198*x+x^2).
%F a(n) = sum((-1)^k*binomial(n-k, k)*198^(n-2*k), k=0..floor(n/2)), n>=0.
%F a(n) = ((99+70*sqrt(2))^(n+1) - (99-70*sqrt(2))^(n+1))/(140*sqrt(2)), n>=0.
%F a(n) = Pell(6*n + 6)/Pell(6). Sum_{n >= 0} 1/( 14*a(n) + 1/(14*a(n)) ) = 1/14. - _Peter Bala_, Mar 25 2015
%F a(n) = A002965(12*(n+1))/70. - _Gerry Martens_, Jul 14 2023
%p with(combinat): seq(fibonacci(6*n+6,2)/70, n=0..12); # _Zerinvary Lajos_, Apr 21 2008
%t LinearRecurrence[{198, -1},{1, 198},12] (* _Ray Chandler_, Aug 11 2015 *)
%Y Cf. A002965.
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Aug 31 2004
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