

A305315


a(n) = sqrt(5*b(n)^2  4), with b(n) = A134493(n) = Fibonacci(6*n+1), n >= 0.


2



1, 29, 521, 9349, 167761, 3010349, 54018521, 969323029, 17393796001, 312119004989, 5600748293801, 100501350283429, 1803423556807921, 32361122672259149, 580696784543856761, 10420180999117162549, 186982561199565069121, 3355265920593054081629, 60207804009475408400201, 1080385206249964297121989
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OFFSET

0,2


COMMENTS

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 CayleyHamilton theorem the powers of the automorphic matrix A are: A^n = matrix([S(n)  9*S(n1), 20*S(n1)], [4*S(n1), S(n)  9*S(n1)]) with the Chebyshev polynomials S(n1) = S(n1, 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*k1) = A134497(k1) 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 oddindexed 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 oddindexed Fibonacci numbers in the Markoff sequence A002559 are given in A158381. The conjecture in this case is that the oddindexed Fibonacci numbers appear as largest numbers only in the ordered Markov triples with x = 1. See, e.g., the Aigner reference for the general FrobeniusMarkoff conjecture.


REFERENCES

Aigner, Martin. Markov's theorem and 100 years of the uniqueness conjecture. A mathematical journey from irrational numbers to perfect matchings. Springer, 2013.


LINKS



FORMULA

a(n) = sqrt(5*(F(6*n+1))^2  4), with F(6*n+1) = A134493(n), n >= 0.
a(n) = S(n, 18) + 11*S(n1, 18), n >= 0, with the Chebyshev polynomials S(n, 18) = A049660(n+1) and S(1, 18) = 0.
a(n) = 18*a(n1)  a(n2), n >= 1, with a(0)=1 and a(1) = 11.
G.f.: (1 + 11*x)/(1  18*x + x^2).
a(n) = 2*sinh((6*n + 1)*arccsch(2)).  Peter Luschny, May 25 2022


EXAMPLE

The solutions of the first class of positive proper solutions [a1(n), b1(n)] of the Pell equation a^2  5*b^2 = 4 begin: [1, 1], [29, 13], [521, 233], [9349, 4181], [167761, 75025], [3010349, 1346269], [54018521, 24157817], ...
The solutions of the second class of positive proper solutions [a5(n), b5(n)] begin: [11, 5], [199, 89], [3571, 1597], [64079, 28657], [1149851, 514229], [20633239, 9227465], [370248451, 165580141], ...
The solutions of the class of improper positive solutions [a3(n), b3(n)] begin: [4, 2], [76, 34], [1364, 610], [24476, 10946], [439204, 196418], [7881196, 3524578], [141422324, 63245986], ...


MATHEMATICA

Select[LinearRecurrence[{1, 1}, {1, 3}, 115], Mod[#, 4] == 1 &]. (* Fred Patrick Doty, Aug 03 2020 *)


PROG

(PARI) x='x+O('x^99); Vec((1+11*x)/(118*x+x^2)) \\ Altug Alkan, Jul 11 2018


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



