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A156090
Alternating sum of the squares of the first n Fibonacci numbers with index divisible by 3.
4
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).
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
KEYWORD
sign,easy
AUTHOR
Stuart Clary, Feb 04 2009
STATUS
approved