A119283 Alternating sum of the squares of the first n Fibonacci numbers.

0, -1, 0, -4, 5, -20, 44, -125, 316, -840, 2185, -5736, 15000, -39289, 102840, -269260, 704909, -1845500, 4831556, -12649205, 33116020, -86698896, 226980625, -594243024, 1555748400, -4073002225, 10663258224, -27916772500, 73087059221, -191344405220, 500946156380, -1311494063981, 3433536035500, -8989114042584, 23533806092185, -61612304234040

Offset: 0

Author

Stuart Clary, May 13 2006

Keywords

• sign
• easy

Comments

Natural bilateral extension (brackets mark index 0): ..., 840, -316, 125, -44, 20, -5, 4, 0, 1, 0, [0], -1, 0, -4, 5, -20, 44, -125, 316, -840, 2185, ... This is (-A119283)-reversed followed by A119283.

Examples

			The first few squares of Fibonacci numbers are: 0, 1, 1, 4, 9, 25, 64, 169.
a(0) = 0.
a(1) = 0 - 1 = -1.
a(2) = 0 - 1 + 1 = 0.
a(3) = 0 - 1 + 1 - 4 = -4.
a(4) = 0 - 1 + 1 - 4 + 9 = 5.
		

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (-1,4,-1,-1)

Crossrefs

Cf. A001654, A119282, A119284, A119285, A119286, A119287, A128696, A128698.

Programs

Mathematica

altFiboSqSum[n_Integer] := If[n >= 0, Sum[(-1)^k Fibonacci[k]^2, {k, n}], Sum[-(-1)^k Fibonacci[-k]^2, {k, 1, -n - 1}]]; altFiboSqSum[Range[0, 39]] (* Clary *)
Accumulate[Table[(-1)^n Fibonacci[n]^2, {n, 0, 39}]] (* Alonso del Arte, Apr 25 2017 *)
Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[0,40]]^2,{1,-1},{2,-1,2}],2]] (* Harvey P. Dale, Jul 29 2024 *)

PARI

concat(0, Vec(-x*(1 + x) / ((1 - x)^2*(1 + 3*x + x^2)) + O(x^40))) \\ Colin Barker, Apr 25 2017

Formula

Let F(n) be the Fibonacci number A000045(n).

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

Closed form: a(n) = (-1)^n F(2n+1)/5 - (2 n + 1)/5.

Recurrence: a(n) + a(n-1) - 4 a(n-2) + a(n-3) + a(n-4) = 0.

G.f.: A(x) = (-x - x^2)/(1 + x - 4 x^2 + x^3 + x^4) = -x(1 + x)/((1 - x)^2 (1 + 3 x + x^2)).

a(n) = ( -1-2*n+(-1)^n*( A001906(n+1)-A001906(n) ))/5. - R. J. Mathar, Nov 16 2007

a(n) = (2^(-1-n)*(-5*2^(1+n)+5*(-3+sqrt(5))^n - sqrt(5)*(-3+sqrt(5))^n + (-3-sqrt(5))^n*(5+sqrt(5)) - 5*2^(2+n)*n)) / 25. - Colin Barker, Apr 25 2017