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

0,2

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

Table of n, a(n) for n=0..17.

Index entries for linear recurrences with constant coefficients, signature (17,17,-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

Mathematica

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

Crossrefs

Partial sums of A014729.

Cf. A156085, A156090, A156091

Sequence in context: A339787 A323276 A141032 * A362730 A000658 A351027

Adjacent sequences: A156081 A156082 A156083 * A156085 A156086 A156087

Keyword

nonn,easy

Author

Stuart Clary, Feb 04 2009

Status

approved