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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.

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).

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).

For the explicit form of powers of the automorphic matrix A in terms of Chebyshev polynomials S(n, 18) see a comment in A305315.

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.

All terms are divisible by 8. - Alonso del Arte, Jul 27 2013

Conjecture: For n >= 2, the terms of this sequence are exactly those Fibonacci numbers which are the sum of the three numbers of a Pythagorean triple (checked up to F(80)). - Felix Huber, Nov 03 2023

Gives those numbers which are Fibonacci numbers in A103135.

Generally, for any sequence where a(0)= Fibonacci(p), a(1) = F(p+q) and Lucas(q)*a(1) +- a(0) = F(p+2q), then a(n) = L(q)*a(n-1) +- a(n-2) generates the following Fibonacci sequence: a(n) = F(q(n)+p). So for this sequence, a(n) = 18*a(n-1) - a(n-2) = F(6n+4): q=6, because 18 is the 6th Lucas number (L(0) = 2, L(1)=1); F(4)=3, F(10)=55 and F(16)=987 (F(0)=0 and F(1)=1). See Lucas sequence A000032. This is a special case where a(0) and a(1) are increasing Fibonacci numbers and Lucas(m)*a(1) +- a(0) is another Fibonacci. - Bob Selcoe, Jul 08 2013

a(n) = x + y where x and y are solutions to x^2 = 5*y^2 - 1. (See related sequences with formula below.) - Richard R. Forberg, Sep 05 2013

For positive n, a(n) equals (-1)^n times the permanent of the (6n)X(6n) tridiagonal matrix with ((-1)^(1/6))'s along the three central diagonals. - John M. Campbell, Jul 12 2011

a(n) = x + y where those two values are solutions to: x^2 = 5*y^2 + 1. (See related sequences with formula below). - Richard R. Forberg, Sep 05 2013

Define a sequence c(n) by c(0)=0, c(1)=1; thereafter c(n) = (c(n-2)*c(n-1)-1)/(c(n-2)+c(n-1)+2). Then it appears that (apart from signs), a(n) is the denominator of c(n). - Jonas Holmvall, Jun 21 2023

The o.g.f. of {F(m*n + 2)}_{n>=0}, for m = 1, 2, ..., is

G(m,x) = (1 + F(m - 2)*x) / (1 - L(m)*x + (-1)^m*x^2), with F = A000045 and F(-1) = 1, and L = A000032. - Wolfdieter Lang, Feb 06 2023

This sequence gives all solutions of one of two classes of positive proper solutions a(n) = a1(n) of the Pell equation a(n)^2 - 5*b(n) = -4 with b(n) = b1(n) = Fibonacci(6*n+1) = A134493(n). These solutions are obtained from the fundamental positive solution [1, 1] (to be read as a column vector) by application of positive powers of the automorphic matrix A = matrix([9, 20], [4, 9]) with determinant +1.

The other class of positive proper solutions is obtained similarly from the fundamental solution [11,5] and is given by [a5(n), b5(n)], with a5(n) = A305316(n) and b5(n) = A134497(n) = F(6*n+5), with the Fibonacci numbers F = A000045.

The remaining positive solutions are improper and are obtained by application of the same matrix A on the fundamental improper solution [4, 2]. They are given by [a3(n), b3(n)], with a3(n) = 4*A049629(n) and b3(n) = A134495(n) = 2*A007805(n).

Via the Cayley-Hamilton theorem the powers of the automorphic matrix A are: A^n = matrix([S(n) - 9*S(n-1), 20*S(n-1)], [4*S(n-1), S(n) - 9*S(n-1)]) with the Chebyshev polynomials S(n-1) = S(n-1, x=18) = A049660(n), n >= 0.

This shows that ordered Markoff (Markov) triples [1, y, m], with 1 <= y <= m, have for m from the union of sets {m1(k)}{k>=0} U {m5(k)}{k>=0) U {m3(k)}_{k>=0)}, with mj(k) = F(6*k+j), for j = 1, 5, and 3, the unique solutions yj(k) = (3*F(6*k+j) - aj(k))/2 < mj(k), namely y1(k) = F(6*k-1) = A134497(k-1) with F(-1) = 1, y5(k) = F(6*k+3) = A134495(k) and y3(k) = F(6*k+1) = A134493. The solutions with the + sign are excluded because they are > mj(k). This trisection of the odd-indexed Fibonacci numbers as m numbers shows again the well known fact that each of them appears as largest member in a Markoff triple if the smallest member is x = 1. The positions of the odd-indexed Fibonacci numbers in the Markoff sequence A002559 are given in A158381. The conjecture in this case is that the odd-indexed Fibonacci numbers appear as largest numbers only in the ordered Markov triples with x = 1. See, e.g., the Aigner reference for the general Frobenius-Markoff conjecture.

Also Lucas numbers that are congruent to 1 mod 4. - Fred Patrick Doty, Aug 03 2020