A360467 a(n) = Fibonacci(4*n+2) + 3*Fibonacci(2*n+1)^2.

4, 20, 130, 884, 6052, 41474, 284260, 1948340, 13354114, 91530452, 627359044, 4299982850, 29472520900, 202007663444, 1384581123202, 9490060198964, 65045840269540, 445830821687810, 3055769911545124, 20944558559128052, 143556140002351234, 983948421457330580

Offset: 0

Author

Alexander M. Domashenko, Feb 08 2023

Keywords

• nonn

Comments

Values of x + 3*y in solutions of x^2 = 5*y^2 - 4*y in positive integers. In the solutions, the values of x and y are given by Fibonacci(4*n + 2) and Fibonacci(2*n + 1)^2 respectively.

The above Diophantine equation arises out of the following problem regarding the subdivision of a square into four triangles of integer area. For n >= 1, the sequence gives the areas of the squares in the solutions (see illustration in Links). Two lines are drawn from a corner of a square to points on the opposing sides. A third line is added between the two points so that the square is divided into four triangles. The area of each triangle is required to be an integer and those of the right triangles to form an arithmetic progression with difference 1. The smallest right triangle by area is the one formed by the third line. In the solutions, the area of the inner triangle is given by Fibonacci(4*n + 2) and the total area of the three right triangles is 3*Fibonacci(2*n + 1)^2. The area of the square is then equal to a(n).

Examples

			a(2) = F(4*2+2) + 3*F(2*2 +1)^2 = F(10) + 3*F(5)^2 = 55 + 3*5^2 = 130.
a(4) = F(4*4+2) + 3*F(2*4 +1)^2 = F(18) + 3*F(9)^2 = 2584+ 3*34^2 = 6052.
G.f. = 4 + 20*x + 130*x^2 + 884*x^3 + 6052*x^4 + ... - _Michael Somos_, Mar 02 2023
		

Links

• Nicolay Avilov, Problem 2450. The sixth parallelogram (in Russian),
• Nicolay Avilov, Problem 2447. Four triangles in a parallelogram (in Russian),
• Nicolay Avilov, Problem 2442. Herringbone in a parallelogram (in Russian).
• Alexander M. Domashenko, Illustration of square divided into four triangles.
• Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Crossrefs

Cf. A000045, A033890, A064170, A081068, A295683.

Programs

Maple

a := proc(n) option remember; if n < 3 then return [4, 20, 130][n + 1] fi;
a(n-3) - 8 * (a(n-2) - a(n-1)) end: seq(a(n), n = 0..22); # Peter Luschny, Feb 17 2023

Mathematica

LinearRecurrence[{8, -8, 1}, {4, 20, 130}, 22] (* Amiram Eldar, Feb 17 2023 *)
a[ n_] := 2 * Fibonacci[2*n+1] * Fibonacci[2*n+3]; (* Michael Somos, Mar 02 2023 *)

PARI

Vec(2*(2 - 6*x + x^2)/((1 - x)*(1 - 7*x + x^2)) + O(x^25)) \\ Andrew Howroyd, Feb 16 2023

SageMath

print([2*(lucas_number2(n+1, 7, 1) + 3) // 5 for n in range(23)]) # Peter Luschny, Feb 17 2023

Formula

a(n) = A033890(n) + 3*A081068(n)^2.

a(n) = Fibonacci(2*n+1)*(Fibonacci(2*n) + Fibonacci(2*n+2) + 3*Fibonacci(2*n+1)).

a(n) = 2*A064170(n+3).

G.f.: 2*(2 - 6*x + x^2)/((1 - x)*(1 - 7*x + x^2)). - Andrew Howroyd, Feb 16 2023

a(n) = a(n-3) - 8 * (a(n-2) - a(n-1)) for n >= 3. - Peter Luschny, Feb 17 2023

a(n) = a(-2-n) = 2*F(2*n+1) * F(2*n+3) = A295683(4*(n+1)) for all n in Z. - Michael Somos, Mar 02 2023