A156093 - OEIS

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,3

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, 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) = (1/9)*Sum_{k=1..n} (-1)^k*F(4*k)^2.

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

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

From Amiram Eldar, Nov 25 2025: (Start)