A156084 Sum of the squares of the first n Fibonacci numbers with index divisible by 3.

0, 4, 68, 1224, 21960, 394060, 7071116, 126886032, 2276877456, 40856908180, 733147469780, 13155797547864, 236071208391768, 4236125953503964, 76014195954679580, 1364019401230728480, 24476335026198433056, 439210011070341066532

Offset: 0

Author

Stuart Clary, Feb 04 2009

Keywords

• nonn
• easy

Comments

Natural bilateral extension (brackets mark index 0): ..., -21960, -1224, -68, -4, 0, [0], 4, 68, 1224, 21960, 394060, ... This is (-A156084)-reversed followed by A156084. That is, A156084(-n) = -A156084(n-1).

Links

• Index entries for linear recurrences with constant coefficients, signature (17,17,-1).

Crossrefs

Partial sums of A014729.

Cf. A156085, A156090, A156091

Programs

Mathematica

a[n_Integer] := If[ n >= 0, Sum[ Fibonacci[3k]^2, {k, 1, n} ], -Sum[ Fibonacci[-3k]^2, {k, 1, -n - 1} ] ]

Formula

Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).

a(n) = sum_{k=1..n} F(3k)^2.

Closed form: a(n) = L(6n+3)/20 - (-1)^n/5.

Factored closed form: a(n) = (1/4) F(n) F(n+1) (L(n) - 1)(L(n) + 1)(L(2n+2) - 1) if n is even; a(n) = (1/4) F(n) F(n+1) (L(n+1) - 1)(L(n+1) + 1)(L(2n) - 1) if n is odd.

Recurrence: a(n) - 17 a(n-1) - 17 a(n-2) + a(n-3) = 0.

G.f.: A(x) = 4 x/(1 - 17 x - 17 x^2 + x^3) = 4 x/((1 + x)(1 - 18 x + x^2)).

a(n) = 4*A156085(n). - R. J. Mathar, Aug 06 2019