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%I #29 Apr 15 2019 09:30:28
%S 1,67,4421,291719,19249033,1270144459,83810285261,5530208682767,
%T 364909962777361,24078527334623059,1588817894122344533,
%U 104837902484740116119,6917712746098725319321,456464203340031130959067,30119719707695955917979101,1987445036504593059455661599
%N Chebyshev sequence with Diophantine property.
%C One fourth of bisection (even part) of A041024.
%C (4*a(n))^2 - 17*A078988(n)^2= -1 (Pell -1 equation, see A077232-3).
%H Indranil Ghosh, <a href="/A078989/b078989.txt">Table of n, a(n) for n = 0..548</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 (66, -1).
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%F G.f.: (1 + x)/(1 - 66*x + x^2).
%F a(n) = 66*a(n-1) - a(n-2) for n>=1, a(-1)=-1, a(0)=1.
%F a(n) = S(2*n, 2*sqrt(17)) = -i*((-1)^n)*T(2*n+1, 4*i)/4 = S(n, 66) + S(n-1, 66) with i^2=-1 and S(n, x), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120.
%F a(n) = A041024(2*n)/4.
%F a(n) = (1/4)*sinh((2*n + 1)*arcsinh(4)). - _Bruno Berselli_, Apr 03 2018
%e (x,y) = (4,1), (268,65), (17684,4289), ... give the positive integer solutions to x^2 - 17*y^2 =-1.
%t LinearRecurrence[{66, -1}, {1, 67}, 20] (* _Bruno Berselli_, Apr 03 2018 *)
%o (PARI) x='x+O('x^99); Vec((1+x)/(1-66*x+x^2)) \\ _Altug Alkan_, Apr 05 2018
%o (GAP) a:=[1,67];; for n in [3..20] do a[n]:=66*a[n-1]-a[n-2]; od; a; # _Muniru A Asiru_, Apr 05 2018
%Y Cf. A097316 for S(n, 66).
%Y Cf. A041024.
%Y Cf. similar sequences of the type (1/k)*sinh((2*n+1)*arcsinh(k)) listed in A097775.
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Jan 10 2003
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