A337929 Numbers w such that (F(2*n-1)^2, -F(2*n)^2, w) are primitive solutions of the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 1, where F(n) is the n-th Fibonacci number (A000045).

1, 11, 79, 545, 3739, 25631, 175681, 1204139, 8253295, 56568929, 387729211, 2657535551, 18215019649, 124847601995, 855718194319, 5865179758241, 40200540113371, 275538601035359, 1888569667134145, 12944449068903659, 88722573815191471, 608113567637436641

Offset: 1

Author

XU Pingya, Sep 30 2020

Keywords

• nonn
• easy

Examples

			2*(F(3)^2)^3 + 2*(-F(4)^2)^3 + 11^3 = 2*4^3 + 2*(-9)^3 + 11^3 = 1, 11 is a term.
		

Links

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

Crossrefs

Cf. A000045, A000578, A003482, A056573, A337928.

Programs

Mathematica

Table[(2*Fibonacci[2n]^6 - 2*Fibonacci[2n-1]^6 + 1)^(1/3), {n, 22}]
LinearRecurrence[{8,-8,1},{1,11,79},30] (* Harvey P. Dale, Aug 23 2021 *)

Formula

a(n) = (2*F(2*n)^6 - 2*F(2*n-1)^6 + 1)^(1/3).

From Colin Barker, Oct 01 2020: (Start)

G.f.: x*(1 + 3*x - x^2) / ((1 - x)*(1 - 7*x + x^2)).

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3) for n>3.

(End)

a(n) = 2*A003482(n) + 1. - Hugo Pfoertner, Oct 01 2020

a(n) = A033888(n) - A064170(n+2). - Flávio V. Fernandes, Jan 10 2021