A156093 One ninth of the alternating sum of the squares of the first n Fibonacci numbers with index divisible by 4.

0, -1, 48, -2256, 105985, -4979040, 233908896, -10988739073, 516236827536, -24252142155120, 1139334444463105, -53524466747610816, 2514510602693245248, -118128473859834915841, 5549523760809547799280, -260709488284188911650320

Offset: 0

Author

Stuart Clary, Feb 04 2009

Keywords

• sign
• easy

Comments

Natural bilateral extension (brackets mark index 0): ..., -105985, 2256, -48, 1, 0, [0], -1, 48, -2256, 105985, -4979040, ... This is (-A156093)-reversed followed by A156093. That is, A156093(-n) = -A156093(n-1).

Links

• Index entries for linear recurrences with constant coefficients, signature (-48,-48,-1).

Crossrefs

Cf. A156086, A156087, A156092.

Programs

Mathematica

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k (1/9) Fibonacci[4k]^2, {k, 1, n} ], Sum[ -(-1)^k (1/9) Fibonacci[-4k]^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) = (1/9) sum_{k=1..n} (-1)^k F(4k)^2.

Closed form: a(n) = (-1)^n (L(8n+4) - 7)/315.

Factored closed form: a(n) = (-1)^n F(4n) F(4n+4)/63.

Recurrence: a(n) + 47 a(n-1) + a(n-2) = (-1)^n.

Recurrence: a(n) + 48 a(n-1) + 48 a(n-2) + a(n-3) = 0.

G.f.: A(x) = -x/(1 + 48 x + 48 x^2 + x^3) = -x/((1 + x)(1 + 47 x + x^2)).