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A094686
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A Fibonacci convolution.
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14
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1, 0, 1, 2, 2, 4, 7, 10, 17, 28, 44, 72, 117, 188, 305, 494, 798, 1292, 2091, 3382, 5473, 8856, 14328, 23184, 37513, 60696, 98209, 158906, 257114, 416020, 673135, 1089154, 1762289, 2851444, 4613732, 7465176, 12078909, 19544084, 31622993, 51167078
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OFFSET
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0,4
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COMMENTS
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Let (b(n)) be the p-INVERT of (1,1,0,0,0,0,0,0,...) using p(S) = 1 - S^2; then b(n) = a(n+1) for n >=0. See A292324. - Clark Kimberling, Sep 15 2017
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LINKS
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Elena Barcucci, Antonio Bernini, Stefano Bilotta, and Renzo Pinzani, Non-overlapping matrices, arXiv:1601.07723 [cs.DM], 2016 (see 1st column of Table 1 p. 8).
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FORMULA
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G.f.: 1/((1-x-x^2)*(1+x+x^2)).
a(n) = 2*sqrt(3)*Sum_{k=0..n} Fibonacci(k+1)*cos((4*(n-k)+1)*Pi/6)/3.
a(n) = a(n-2) + 2*a(n-3) + a(n-4).
a(n) = A005252(n) - (-cos((2*n+1)*Pi/3)/2 - sqrt(3)*sin((2*n+1)*Pi/3)/6 + sqrt(3)*cos(Pi*n/3+Pi/6)/6 + sin((2*n+1)*Pi/6)/2).
a(n) = Sum_{k=0..floor(n/2)} if(mod(n-k, 2)=0, binomial(n-k, k), 0).
a(n) = A093040(n-1) - Fibonacci(n). (End)
a(n) = Sum_{k=0..floor(n/2)} C(n-k, k)*(1+(-1)^(n-k))/2. - Paul Barry, Sep 09 2005
a(n) = Sum_{k=0..floor(n/2)} C(2*k, n-2*k).
a(n) = Sum_{k=0..floor(n/2)} C(n-k,k)*C(3*k,n-k)/C(3*k,k). (End)
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MATHEMATICA
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PROG
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(PARI) Vec(1/((1-x-x^2)*(1+x+x^2)) + O(x^50)) \\ Michel Marcus, Sep 27 2014
(Magma) [(Fibonacci(n+1) +((n+2) mod 3) -1)/2: n in [0..40]]; // G. C. Greubel, Feb 09 2023
(SageMath) [(fibonacci(n+1) + (n+2)%3 - 1)/2 for n in range(41)] # G. C. Greubel, Feb 09 2023
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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