A156088 Alternating sum of the squares of the first n even-indexed Fibonacci numbers.

0, -1, 8, -56, 385, -2640, 18096, -124033, 850136, -5826920, 39938305, -273741216, 1876250208, -12860010241, 88143821480, -604146740120, 4140883359361, -28382036775408, 194533374068496, -1333351581704065, 9138927697859960

Offset: 0

Author

Stuart Clary, Feb 04 2009

Keywords

• sign
• easy

Comments

Apart from signs, same as A092521.

Natural bilateral extension (brackets mark index 0): ..., 2640, -385, 56, -8, 1, 0, [0], -1, 8, -56, 385, -2640, 18096, ... This is (-a(n))-reversed followed by a(n). That is, a(-n) = -a(n-1).

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (-8,-8,-1).

Crossrefs

Cf. A000032, A000045, A081079, A092521, A103434, A103433, A156089.

Programs

Magma

A156088:= func< n | (-1)^n*(Lucas(4*n+2)-3)/15 >; // G. C. Greubel, Jun 12 2025

Mathematica

a[n_]:= If[n >= 0, Sum[(-1)^k Fibonacci[2k]^2, {k,n}], Sum[ -(-1)^k Fibonacci[-2k]^2, {k,-n-1}]];
LinearRecurrence[{-8,-8,-1}, {0,-1,8}, 41] (* G. C. Greubel, Jun 12 2025 *)

SageMath

def A156088(n): return (-1)^n*(lucas_number2(4*n+2,1,-1) -3)//15 # G. C. Greubel, Jun 12 2025

Formula

Let F(n) be the n-th Fibonacci number, A000045(n), and L(n) be the n-th Lucas number, A000032(n), then: (Start)

a(n) = Sum_{k=1..n} (-1)^k F(2*k)^2.

Closed form: a(n) = (-1)^n * (L(4*n+2) - 3)/15.

Factored closed form: a(n) = (1/3) * (-1)^n * F(n)*L(n)*F(n+1)*L(n+1) = (1/3)*(-1)^n * F(2*n)*F(2*n+2).

Recurrence: a(n) + 8*a(n-1) + 8*a(n-2) + a(n-3) = 0.

G.f.: -x/(1 + 8*x + 8*x^2 + x^3) = -x/((1 + x)(1 + 7*x + x^2)). (End)

From G. C. Greubel, Jun 12 2025: (Start)

a(n) = (-1)^n*A081079(n)/15.

E.g.f.: (1/15)*( exp(-7*x/2)*( 3*cosh(p*x) - sqrt(5)*sinh(p*x) ) - 3*exp(-x) ), where p = 3*sqrt(5)/2. (End)