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

0,2

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, a(-n) = -a(n-1).

Links

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

Index entries for linear recurrences with constant coefficients, signature (-48,-48,-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(4*k)^2.

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

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

From Amiram Eldar, Nov 25 2025: (Start)

a(n) = 9*A156093(n).

Sum_{n>=1} (-1)^n/a(n) = 35/9 - 7*phi/3, where phi is the golden ratio (A001622). (End)

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} ] ]
LinearRecurrence[{-48, -48, -1}, {0, -9, 432}, 16] (* Amiram Eldar, Nov 25 2025 *)

Crossrefs

Cf. A000032, A000045, A001622, A156086, A156087, A156093.

Sequence in context: A081481 A160376 A271652 * A092813 A242280 A266294

Adjacent sequences: A156089 A156090 A156091 * A156093 A156094 A156095

Keyword

sign,easy

Author

Stuart Clary, Feb 04 2009

Status

approved