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%I #65 Mar 02 2023 22:42:09
%S 4,20,130,884,6052,41474,284260,1948340,13354114,91530452,627359044,
%T 4299982850,29472520900,202007663444,1384581123202,9490060198964,
%U 65045840269540,445830821687810,3055769911545124,20944558559128052,143556140002351234,983948421457330580
%N a(n) = Fibonacci(4*n+2) + 3*Fibonacci(2*n+1)^2.
%C 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.
%C 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).
%H Nicolay Avilov, <a href="https://www.diofant.ru/problem/4244/">Problem 2450. The sixth parallelogram</a> (in Russian),
%H Nicolay Avilov, <a href="https://www.diofant.ru/problem/4241/">Problem 2447. Four triangles in a parallelogram</a> (in Russian),
%H Nicolay Avilov, <a href="https://www.diofant.ru/problem/4236/">Problem 2442. Herringbone in a parallelogram</a> (in Russian).
%H Alexander M. Domashenko, <a href="/A360467/a360467.jpg">Illustration of square divided into four triangles</a>.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (8,-8,1).
%F a(n) = A033890(n) + 3*A081068(n)^2.
%F a(n) = Fibonacci(2*n+1)*(Fibonacci(2*n) + Fibonacci(2*n+2) + 3*Fibonacci(2*n+1)).
%F a(n) = 2*A064170(n+3).
%F G.f.: 2*(2 - 6*x + x^2)/((1 - x)*(1 - 7*x + x^2)). - _Andrew Howroyd_, Feb 16 2023
%F a(n) = a(n-3) - 8 * (a(n-2) - a(n-1)) for n >= 3. - _Peter Luschny_, Feb 17 2023
%F 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
%e a(2) = F(4*2+2) + 3*F(2*2 +1)^2 = F(10) + 3*F(5)^2 = 55 + 3*5^2 = 130.
%e a(4) = F(4*4+2) + 3*F(2*4 +1)^2 = F(18) + 3*F(9)^2 = 2584+ 3*34^2 = 6052.
%e G.f. = 4 + 20*x + 130*x^2 + 884*x^3 + 6052*x^4 + ... - _Michael Somos_, Mar 02 2023
%p a := proc(n) option remember; if n < 3 then return [4, 20, 130][n + 1] fi;
%p a(n-3) - 8 * (a(n-2) - a(n-1)) end: seq(a(n), n = 0..22); # _Peter Luschny_, Feb 17 2023
%t LinearRecurrence[{8, -8, 1}, {4, 20, 130}, 22] (* _Amiram Eldar_, Feb 17 2023 *)
%t a[ n_] := 2 * Fibonacci[2*n+1] * Fibonacci[2*n+3]; (* _Michael Somos_, Mar 02 2023 *)
%o (PARI) Vec(2*(2 - 6*x + x^2)/((1 - x)*(1 - 7*x + x^2)) + O(x^25)) \\ _Andrew Howroyd_, Feb 16 2023
%o (SageMath)
%o print([2*(lucas_number2(n+1, 7, 1) + 3) // 5 for n in range(23)]) # _Peter Luschny_, Feb 17 2023
%Y Cf. A000045, A033890, A064170, A081068, A295683.
%K nonn
%O 0,1
%A _Alexander M. Domashenko_, Feb 08 2023
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