A350922 a(0) = 2, a(1) = 5, and a(n) = 7*a(n-1) - a(n-2) - 4 for n >= 2.

2, 5, 29, 194, 1325, 9077, 62210, 426389, 2922509, 20031170, 137295677, 941038565, 6449974274, 44208781349, 303011495165, 2076871684802, 14235090298445, 97568760404309, 668746232531714, 4583654867317685, 31416837838692077, 215334210003526850, 1475922632185995869

Offset: 0

Author

Max Alekseyev, Jan 22 2022

Keywords

• nonn
• easy

Comments

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.

From William P. Orrick, Dec 20 2023: (Start)

Every term is a Markov number (see A002559) and, for n > 1, corresponds to a node of the Markov tree A368546 whose sibling and ancestors are all odd-indexed Fibonacci numbers. For n > 1, a(n) is the label of the node obtained from the root by going left n - 2 times and then right. Its Farey index, described in the comments to A368546, is 2 / (2*n - 1).

For instance, a(3) = 194 comes from going left once from the root node of the Markov tree and then right, which corresponds to the sequence of Markov numbers 5, 13, 194. The corresponding sequence of Farey indices is 1/2, 1/3, 2/5. The sibling of the final node corresponds to Markov number 34 and Farey index 1/4. (End)

Links

• Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Crossrefs

Cf. A049684, A064170, A350916, A002559, A000045.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350919, A350920, A350921, A350923, A350924, A350925, A350926.

Programs

Mathematica

CoefficientList[Series[(2 - x)*(1 - 5*x)/((1 - x)*(1 - 7*x + x^2)), {x, 0, 22}],x] (* James C. McMahon, Dec 22 2023 *)

Formula

G.f.: (2 - x)*(1 - 5*x)/((1 - x)*(1 - 7*x + x^2)). - Stefano Spezia, Jan 22 2022

a(n) = 3*A049684(n) + 2 = 3*A064170(n+2) - 1. - Hugo Pfoertner, Jan 22 2022

a(n) = 3*A000045(2*n - 1) * A000045(2*n + 1) - 1 = A000045(2*n - 1)^2 + A000045(2*n + 1)^2. - William P. Orrick, Jan 08 2023