A156092 Alternating sum of the squares of the first n Fibonacci numbers with index divisible by 4.

0, -9, 432, -20304, 953865, -44811360, 2105180064, -98898651657, 4646131447824, -218269279396080, 10254010000167945, -481720200728497344, 22630595424239207232, -1063156264738514242569, 49945713847285930193520, -2346385394557700204852880

Offset: 0

Author

Stuart Clary, Feb 04 2009

Keywords

• sign
• easy

Comments

Natural bilateral extension (brackets mark index 0): ..., -953865, 20304, -432, 9, 0, [0], -9, 432, -20304, 953865, -44811360, ... This is (-A156092)-reversed followed by A156092. That is, A156092(-n) = -A156092(n-1).

Crossrefs

Cf. A156086, A156087, A156093.

Programs

Mathematica

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

Closed form: a(n) = (-1)^n (L(8n+4) - 7)/35.

Factored closed form: a(n) = (-1)^n F(4n) F(4n+4)/7.

Recurrence: a(n) + 47 a(n-1) + a(n-2) = (-1)^n 9.

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

G.f.: A(x) = -9 x/(1 + 48 x + 48 x^2 + x^3) = -9 x/((1 + x)(1 + 47 x + x^2)).