%I #28 Jan 23 2020 00:57:37
%S 1,532,282491,150002189,79650879868,42294467207719,22458282436418921,
%T 11925305679271239332,6332314857410591666371,
%U 3362447263979344903603669,1785453164858174733221881868,948072268092426803995915668239
%N Chebyshev polynomials S(n,531) + S(n-1,531) with Diophantine property.
%C (23*a(n))^2 - 533*b(n)^2 = -4 with b(n)=A098259(n) give all positive solutions of this Pell equation.
%H Muniru A Asiru, <a href="/A098258/b098258.txt">Table of n, a(n) for n = 0..300</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H Giovanni Lucca, <a href="http://forumgeom.fau.edu/FG2019volume19/FG201902index.html">Integer Sequences and Circle Chains Inside a Hyperbola</a>, Forum Geometricorum (2019) Vol. 19, 11-16.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (531, -1).
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%F a(n) = S(n, 531) + S(n-1, 531) = S(2*n, sqrt(533)), with S(n, x)=U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x)= 0 = U(-1, x). S(n, 531)=A098257(n).
%F a(n)= (-2/23)*i*((-1)^n)*T(2*n+1, 23*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
%F G.f.: (1+x)/(1-531*x+x^2).
%F a(n) = 531*a(n-1) - a(n-2), n > 1; a(0)=1, a(1)=532. - _Philippe Deléham_, Nov 18 2008
%e All positive solutions of Pell equation x^2 - 533*y^2 = -4 are (23=23*1,1), (12236=23*532,530), (6497293=23*282491,281429), (3450050347=23*150002189,149438269), ...
%t LinearRecurrence[{531,-1},{1,532},20] (* _Harvey P. Dale_, Oct 09 2018 *)
%o (GAP) a:=[1,532];; for n in [3..12] do a[n]:=531*a[n-1]-a[n-2]; od; Print(a); # _Muniru A Asiru_, Apr 29 2019
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Sep 10 2004
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