cp's OEIS Frontend

For positive n, a(n) equals (-1)^n times the permanent of the (4n) X (4n) tridiagonal matrix with sqrt(i)'s along the three central diagonals, where i is the imaginary unit. - John M. Campbell, Jul 12 2011

a(n) = 5^n*a(n; 3/5) = (16/5)^n*a(2*n; 3/4), and F(4*n) = 5^n*b(n; 3/5) = (16/5)^n*b(2*n; 3/4), where a(n; d) and b(n; d), n=0, 1, ..., d in C, denote the delta-Fibonacci numbers defined in comments to A014445. Two of these identities from the following relations follows: F(k+1)^n*a(n; F(k)/F(k+1)) = F(k*n+1) and F(k+1)^n*b(n; F(k)/F(k+1)) = F(k*n) (see also Witula's et al. papers). - Roman Witula, Jul 24 2012

Partial sums of A033888, i.e., a(n) = Sum_{k=0..n} Fibonacci(4*k). - Vladeta Jovovic, Jun 09 2002

From Paul Weisenhorn, May 17 2009: (Start)

a(n) is the solution of the 2 equations a(n)+1=A^2 and 5*a(n)+1=B^2

which are equivalent to the Pell equation (10*a(n)+3)^2-5*(A*B)^2=4.

(End)

Numbers a(n) such as a(n)+1 and 5*a(n)+1 are perfect squares. - Sture Sjöstedt, Nov 03 2011

Table starts

.1....2........4..........8............16...............32..................64

.1....5.......25........105...........441.............1869................7921

.1...13......169.......1573.........14641...........146410.............1464100

.1...34.....1156......20570........366025..........7320500...........146410000

.1...89.....7921.....269225.......9150625........398967250.........17394972100

.1..233....54289....3524125.....228765625......21860843125.......2089022169025

.1..610...372100...46131250....5719140625....1201888840625.....252579343635025

.1.1597..2550409..603865625..142978515625...66099082156250...30557658560100100

.1.4181.17480761.7904703125.3574462890625.3635200164062500.3696969485250250000

Table starts

.1.....5........25.........80.........256..........976.........3721

.1....13.......169.......1040........6400........53280.......443556

.1....34......1156......13600......160000......2920000.....53290000

.1....89......7921.....178000.....4000000....160564000...6445199524

.1...233.....54289....2330000...100000000...8830490000.779775536401

.1...610....372100...30500000..2500000000.485643650000

.1..1597...2550409..399250000.62500000000

.1..4181..17480761.5226250000

.1.10946.119814916

.1.28657

The 12-gonal square numbers k correspond to the nonnegative integer solutions of the Diophantine equation k = d*(5*d-4) = c^2, equivalent to (5*d-2)^2 - 5*c^2 = 4. Substituting x = 5*d-2 and y = c gives the Pell-Fermat's equation x^2 - 5*y^2 = 4.

The solutions x are in A342710, while corresponding solutions y that are also the indices c of the squares which are 12-gonal are in A033890.

The indices d of the corresponding 12-gonal which are squares are in A081068.

a(n) is the square of limit of (G(j+2n-1) + G(j-2n+1))/G(j) as j -> infinity, where G(n) is any sequence of the form G(n+1) = G(n) + G(n-1), with any starting values, including non-integer values. G(n) includes Lucas and Fibonacci. Compare with A005248 for even number offsets from j in any such G(n). - Richard R. Forberg, Nov 16 2014

a(n) = (t(i+6n+3) + t(i))/(t(i+4n+2) + t(i+2n+1)) + 3, where (t) is any sequence of the form t(n+2) = 2t(n+1) + 2t(n) - t(n-1) or of the form t(n+1) = 3t(n) - t(n-1) without regard to initial values as long as t(i+4n+2) + t(i+2n+1) != 0. - Klaus Purath, Jun 23 2024

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).