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A305315
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a(n) = sqrt(5*b(n)^2 - 4), with b(n) = A134493(n) = Fibonacci(6*n+1), n >= 0.
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2
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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
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0,2
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COMMENTS
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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.
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REFERENCES
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Aigner, Martin. Markov's theorem and 100 years of the uniqueness conjecture. A mathematical journey from irrational numbers to perfect matchings. Springer, 2013.
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LINKS
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FORMULA
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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(n-1, 18), n >= 0, with the Chebyshev polynomials S(n, 18) = A049660(n+1) and S(-1, 18) = 0.
a(n) = 18*a(n-1) - a(n-2), 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
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EXAMPLE
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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], ...
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MATHEMATICA
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Select[LinearRecurrence[{1, 1}, {1, 3}, 115], Mod[#, 4] == 1 &]. (* Fred Patrick Doty, Aug 03 2020 *)
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PROG
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(PARI) x='x+O('x^99); Vec((1+11*x)/(1-18*x+x^2)) \\ Altug Alkan, Jul 11 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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