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 A305315 a(n) = sqrt(5*b(n)^2 - 4), with b(n) = A134493(n) = Fibonacci(6*n+1), n >= 0. 2

%I

%S 1,29,521,9349,167761,3010349,54018521,969323029,17393796001,

%T 312119004989,5600748293801,100501350283429,1803423556807921,

%U 32361122672259149,580696784543856761,10420180999117162549,186982561199565069121,3355265920593054081629,60207804009475408400201,1080385206249964297121989

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

%C 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.

%C 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.

%C 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).

%C 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.

%C 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.

%C Also Lucas numbers that are congruent to 1 mod 4. - _Fred Patrick Doty_, Aug 03 2020

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

%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (18,-1).

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

%F 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.

%F a(n) = 18*a(n-1) - a(n-2), n >= 1, with a(0)=1 and a(-1) = -11.

%F G.f.: (1 + 11*x)/(1 - 18*x + x^2).

%e 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], ...

%e 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], ...

%e 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], ...

%t Select[LinearRecurrence[{1, 1}, {1, 3}, 115], Mod[#, 4] == 1 &]. (* _Fred Patrick Doty_, Aug 03 2020 *)

%o (PARI) x='x+O('x^99); Vec((1+11*x)/(1-18*x+x^2)) \\ _Altug Alkan_, Jul 11 2018

%Y Cf. A000045, A002559, A007805, A049629, A049660, A134493, A134495, A134497, A158381, A305316.

%K nonn,easy

%O 0,2

%A _Wolfdieter Lang_, Jul 10 2018

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Last modified September 21 14:54 EDT 2020. Contains 337272 sequences. (Running on oeis4.)