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%I #18 Sep 08 2022 08:46:21
%S 11,199,3571,64079,1149851,20633239,370248451,6643838879,119218851371,
%T 2139295485799,38388099893011,688846502588399,12360848946698171,
%U 221806434537978679,3980154972736918051,71420983074726546239,1281597540372340914251,22997334743627409910279,412670427844921037470771,7405070366464951264563599,132878596168524201724674011
%N a(n) = sqrt(5*b(n)^2 - 4) with b(n) = Fibonacci(6*n+5) = A134497(n).
%C This sequence gives all solutions of one of two classes of positive proper solutions a(n) = a5(n) of the Pell equation a(n)^2 - 5*b(n) = -4 with b(n) = b5(n) = F(6*n+5) = A134497(n), with the Fibonacci numbers F = A000045. These solutions are obtained from the fundamental positive solution [11, 5] (to be read as a column vector) by application of positive powers of the automorphic matrix A = matrix([9, 20], [4, 9]) of determinant +1.
%C The other class of positive proper solutions is obtained similarly from the fundamental solution [1,1] and is given by [a1(n), b1(n)], with a1(n) = A305315(n) and b1(n) = A134493(n) = F(6*n+1).
%C The remaining positive solutions are improper and are obtained by application of positive powers of the same matrix A on the fundamental improper solution [4, 2]. They are given by [a3(n), b3(n)], with a3(n) = F(6*n+3) = 4* A049629(n) and b3(n) = A134495(n) = 2*A007805(n).
%C For the explicit form of powers of the automorphic matrix A in terms of Chebyshev polynomials S(n, 18) see a comment in A305315.
%C The relation to a proof using this Pell equation of the well known fact that each odd-indexed Fibonacci number appears as largest member in Markoff (Markov) triples with smallest member 1 see also A305315.
%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 (18,-1).
%F a(n) = sqrt(5*(F(6*n+5))^2 - 4), with F(6*n+5) = A134497(n), n >= 0.
%F a(n) = 11*S(n, 18) + S(n-1, 18), n >= 0, with the Chebyshev polynomials S(n, 18) = A049660(n+1) and S(-1, 18) = 0.
%F a(n) = 18*a(n-1) - a(n-2), n >= 1, with a(-1) = -1 and a(0) = 11.
%F G.f.: (11 + x)/(1 - 18*x + x^2).
%e See A305315 for the three classes of solutions of this Pell equation
%t f[n_] := Sqrt[5 Fibonacci[6 n + 5]^2 - 4]; Array[f, 17, 0] (* or *)
%t CoefficientList[ Series[(x + 11)/(x^2 - 18x + 1), {x, 0, 18}], x] (* or *)
%t LinearRecurrence[{18, -1}, {11, 199}, 18] (* _Robert G. Wilson v_, Jul 21 2018 *)
%o (PARI) x='x+O('x^99); Vec((11+x)/(1-18*x+x^2)) \\ _Altug Alkan_, Jul 11 2018
%o (Magma) I:=[11, 199]; [n le 2 select I[n] else 18*Self(n-1)-Self(n-2): n in [1..25]]; // _Vincenzo Librandi_, Jul 22 2018
%Y Cf. A000045, A007805, A049629, A049660, A134493, A134495, A134497, A305315.
%K nonn,easy
%O 0,1
%A _Wolfdieter Lang_, Jul 10 2018
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