A100334 An inverse Catalan transform of Fibonacci(2n).
0, 1, 2, 2, 0, -5, -13, -21, -21, 0, 55, 144, 233, 233, 0, -610, -1597, -2584, -2584, 0, 6765, 17711, 28657, 28657, 0, -75025, -196418, -317811, -317811, 0, 832040, 2178309, 3524578, 3524578, 0, -9227465, -24157817, -39088169, -39088169, 0, 102334155, 267914296, 433494437, 433494437, 0, -1134903170
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (3,-4,2,-1).
Programs
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Magma
I:=[0,1,2,2]; [n le 4 select I[n] else 3*Self(n-1) -4*Self(n-2) +2*Self(n-3) -Self(n-4): n in [1..41]]; // G. C. Greubel, Jan 30 2023
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Mathematica
Table[FullSimplify[GoldenRatio^n*Sqrt[2/5 + 2*Sqrt[5]/25]*Sin[Pi*n/5 + Pi/5] - (1/GoldenRatio)^n*Sqrt[2/5 - 2*Sqrt[5]/25]*Sin[2*Pi*n/5 + 2*Pi/5]], {n, 0, 41}] (* Arkadiusz Wesolowski, Oct 26 2012 *) LinearRecurrence[{3,-4,2,-1}, {0,1,2,2}, 41] (* G. C. Greubel, Jan 30 2023 *) -
SageMath
def A100334(n): return sum((-1)^k*binomial(n-k,k)*fibonacci(2*n-2*k) for k in range(1+(n//2))) [A100334(n) for n in range(41)] # G. C. Greubel, Jan 30 2023
Formula
G.f.: x*(1-x)/(1-3*x+4*x^2-2*x^3+x^4).
a(n) = (phi)^n*sqrt(2/5+2*sqrt(5)/25)*sin(Pi*(n+1)/5) -(1/phi)^n*sqrt(2/5-2*sqrt(5)/25)*sin(2*Pi*(n+1)/5), where phi=(1+sqrt(5))/2;
a(n) = Sum_{k=0..floor(n/2)} (C(n-k, k)*(-1)^k*Sum_{j=0..n-k} C(n-k, j)*F(j));
a(n) = Sum_{k=0..floor(n/2)} C(n-k, k)*(-1)^k*Fibonacci(2n-2k).
a(n) = 3*a(n-1)-4*a(n-2)+2*a(n-3)-a(n-4). - Paul Curtz, May 13 2008
a(5*n) = -F(-5*n), a(5*n+1) = -F(-5*n-2), a(5*n+2) = a(5*n+3) = F(-5*n-3), a(5*n+4) = 0. - Ehren Metcalfe, Apr 04 2019