OFFSET
0,1
COMMENTS
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.
LINKS
FORMULA
EXAMPLE
See A305315 for the three classes of solutions of this Pell equation
MATHEMATICA
f[n_] := Sqrt[5 Fibonacci[6 n + 5]^2 - 4]; Array[f, 17, 0] (* or *)
CoefficientList[ Series[(x + 11)/(x^2 - 18x + 1), {x, 0, 18}], x] (* or *)
LinearRecurrence[{18, -1}, {11, 199}, 18] (* Robert G. Wilson v, Jul 21 2018 *)
PROG
(PARI) x='x+O('x^99); Vec((11+x)/(1-18*x+x^2)) \\ Altug Alkan, Jul 11 2018
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Jul 10 2018
STATUS
approved