

A305316


a(n) = sqrt(5*b(n)^2  4) with b(n) = Fibonacci(6*n+5) = A134497(n).


2



11, 199, 3571, 64079, 1149851, 20633239, 370248451, 6643838879, 119218851371, 2139295485799, 38388099893011, 688846502588399, 12360848946698171, 221806434537978679, 3980154972736918051, 71420983074726546239, 1281597540372340914251, 22997334743627409910279, 412670427844921037470771, 7405070366464951264563599, 132878596168524201724674011
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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 oddindexed Fibonacci number appears as largest member in Markoff (Markov) triples with smallest member 1 see also A305315.


LINKS

Table of n, a(n) for n=0..20.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (18,1).


FORMULA

a(n) = sqrt(5*(F(6*n+5))^2  4), with F(6*n+5) = A134497(n), n >= 0.
a(n) = 11*S(n, 18) + 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(1) = 1 and a(0) = 11.
G.f.: (11 + x)/(1  18*x + x^2).


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)/(118*x+x^2)) \\ Altug Alkan, Jul 11 2018
(MAGMA) I:=[11, 199]; [n le 2 select I[n] else 18*Self(n1)Self(n2): n in [1..25]]; // Vincenzo Librandi, Jul 22 2018


CROSSREFS

Cf. A000045, A007805, A049629, A049660, A134493, A134495, A134497, A305315.
Sequence in context: A140534 A112733 A088295 * A268959 A213531 A064748
Adjacent sequences: A305313 A305314 A305315 * A305317 A305318 A305319


KEYWORD

nonn,easy


AUTHOR

Wolfdieter Lang, Jul 10 2018


STATUS

approved



