A156087 One ninth of the sum of the squares of the first n Fibonacci numbers with index divisible by 4.

0, 1, 50, 2354, 110595, 5195620, 244083556, 11466731525, 538692298134, 25307071280790, 1188893657899015, 55852694849972936, 2623887764290829000, 123266872226818990089, 5790919106896201705210, 272049931151894661154810

Offset

0,3

Comments

Natural bilateral extension (brackets mark index 0): ..., -110595, -2354, -50, -1, 0, [0], 1, 50, 2354, 110595, 5195620, ... This is (-A156087)-reversed followed by A156087. That is, A156087(-n) = -A156087(n-1).

Links

Table of n, a(n) for n=0..15.

Formula

Let F(n) be the Fibonacci number A000045(n).

a(n) = sum_{k=1..n} F(4k)^2.

Closed form: a(n) = F(8n+4)/135 - (2n + 1)/45.

Recurrence: a(n) - 48 a(n-1) + 48 a(n-2) - a(n-3) = 2.

Recurrence: a(n) - 49 a(n-1) + 96 a(n-2) - 49 a(n-3) + a(n-4) = 0.

G.f.: A(x) = (x + x^2)/(1 - 49 x + 96 x^2 - 49 x^3 + x^4) = x (1 + x)/((1 - x)^2 (1 - 47 x + x^2)).

Mathematica

a[n_Integer] := If[ n >= 0, Sum[ (1/9) Fibonacci[4k]^2, {k, 1, n} ], -Sum[ (1/9) Fibonacci[-4k]^2, {k, 1, -n - 1} ] ]
Accumulate[Fibonacci[4Range[0, 20]]^2]/9 (* Harvey P. Dale, Sep 22 2011 *)

Crossrefs

Cf. A156086, A156092, A156093.

Sequence in context: A240191 A231696 A285877 * A162919 A163290 A163837

Adjacent sequences: A156084 A156085 A156086 * A156088 A156089 A156090

Keyword

nonn,easy

Author

Stuart Clary, Feb 04 2009

Status

approved