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
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).
Programs
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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
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Magma
[(-1)^n*Fibonacci(6*n+3)/10 - (2*n + 1)/5: n in [0..20]]; // Vincenzo Librandi, Sep 12 2018
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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
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Mathematica
a[n_]:= 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 *) -
SageMath
def A156090(n): return ((-1)^n*fibonacci(6*n+3) -2*(2*n+1))//10 # G. C. Greubel, Jun 12 2025
Formula
Let F(n) be the n-th Fibonacci number, A000045(n), then: (Start)
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.: -4*x*(1 + x)/(1 + 16*x - 34*x^2 + 16*x^3 + x^4) = -4*x(1 + x)/((1 - x)^2*(1 + 18*x + x^2)). (End)
Limit_{n -> oo} a(n)/a(n-1) = -(9 + sqrt(80)). - A.H.M. Smeets, Sep 11 2018
E.g.f.: (-1/(5*sqrt(5)))*( exp(-9*x)*(2*sinh(p*x) - sqrt(5)*cosh(p*x)) + sqrt(5)*(1 + 2*x)*exp(x) ), where p = 4*sqrt(5). - G. C. Greubel, Jun 12 2025
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