A156090 Alternating sum of the squares of the first n Fibonacci numbers with index divisible by 3.

0, -4, 60, -1096, 19640, -352460, 6324596, -113490320, 2036501104, -36543529620, 655747031980, -11766903046104, 211148507797800, -3788906237314396, 67989163763861220, -1220016041512187680, 21892299583455516896, -392841376460687116580

Offset

0,2

Comments

Natural bilateral extension (brackets mark index 0): ..., -19640, 1096, -60, 4, 0, [0], -4, 60, -1096, 19640, -352460, ... This is (-A156090)-reversed followed by A156090. That is, A156090(-n) = -A156090(n-1).

Links

Muniru A Asiru, Table of n, a(n) for n = 0..200

Index entries for linear recurrences with constant coefficients, signature (-16,34,-16,-1).

Formula

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

a(n) = Sum_{k=1..n} (-1)^k*F(3*k)^2.

Closed form: a(n) = (-1)^n*F(6*n+3)/10 - (2*n + 1)/5.

Recurrence: a(n) + 17*a(n-1) - 17*a(n-2) - a(n-3) = -8.

Recurrence: a(n) + 16*a(n-1) - 34*a(n-2) + 16*a(n-3) + a(n-4) = 0.

G.f.: A(x) = -(4*x + 4*x^2)/(1 + 16*x - 34*x^2 + 16*x^3 + x^4) = -4*x(1 + x)/((1 - x)^2 (1 + 18*x + x^2)).

Lim_{n -> inf} a(n)/a(n-1) = -(9 + sqrt(80)). - A.H.M. Smeets, Sep 11 2018

Maple

with(combinat, fibonacci): a:=n->add((-1)^k*fibonacci(3*k)^2, k=1..n): seq(a(n), n=0..20); # Muniru A Asiru, Sep 12 2018

Mathematica

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[3 k]^2, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[-3 k]^2, {k, 1, -n - 1} ] ]
LinearRecurrence[{-16, 34, -16, -1}, {0, -4, 60, -1096}, 30] (* Harvey P. Dale, Oct 24 2016 *)

Prog

(GAP) a:=[0, -4, 60, -1096];; for n in [5..20] do a[n]:=-16*a[n-1]+34*a[n-2]-16*a[n-3]-a[n-4]; od; a; # Muniru A Asiru, Sep 12 2018
(Magma) [(-1)^n*Fibonacci(6*n+3)/10 - (2*n + 1)/5: n in [0..20]]; // Vincenzo Librandi, Sep 12 2018

Crossrefs

Cf. A156084, A156085, A156091.

Sequence in context: A013482 A123480 A227528 * A181418 A208890 A370498

Adjacent sequences: A156087 A156088 A156089 * A156091 A156092 A156093

Keyword

sign,easy

Author

Stuart Clary, Feb 04 2009

Status

approved