cp's OEIS Frontend

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.

From Jon E. Schoenfield, Jan 18 2019: (Start)

Previously, the Name had included the comment, "This sequence is such that 20*(a(n)^2) + 20*a(n) + 1 = j^2 = a square."

However, Anthony Hernandez observed that this statement is not true for all terms; e.g., at a(4)=16, 20*16^2 + 20*16 + 1 = 5441, a nonsquare.

It is true that 20*a(n)^2 + 20*a(n) + 1 = A305315(n/3)^2 when n == 0 (mod 3) and A305316((n-2)/3)^2 when n == 2 (mod 3); however, for n == 1 (mod 3) with n > 1, sqrt(20*a(n)^2 + 20*a(n) + 1) is a noninteger number whose fractional part apparently approaches 3 - sqrt(5) as n increases, and Andrey Zabolotskiy observes that round(sqrt(20*a(n)^2 + 20*a(n) + 1) + sqrt(5)) appears to be equal to A002878(n). (End)