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A014334
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Exponential convolution of Fibonacci numbers with themselves.
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10
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0, 0, 2, 6, 22, 70, 230, 742, 2406, 7782, 25190, 81510, 263782, 853606, 2762342, 8939110, 28927590, 93611622, 302933606, 980313702, 3172361830, 10265978470, 33221404262, 107506722406, 347899061862, 1125825013350, 3643246274150, 11789792601702
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OFFSET
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0,3
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LINKS
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Charles R. Wall, Problem B-573, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 24, No. 2 (1986), p. 181; Solution to Problem B-573 by Bob Prielipp, ibid., Vol. 25, No. 2 (1987), p. 184.
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FORMULA
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a(n) = 3*a(n-1) + 2*a(n-2) - 4*a(n-3), a(0)=0, a(1)=0, a(2)=2.
a(n) = Sum_{k=0..n-1} 2^k*Fibonacci(k) for n > 0.
a(n) = (-2 + ((1+sqrt(5))^n + (1-sqrt(5))^n))/5. (End)
a(n) = Sum_{k=0..n} Fibonacci(k)*Fibonacci(n-k)*binomial(n, k)). - Benoit Cloitre, May 11 2005
G.f.: 2*x^2/((1-x)*(1-2*x-4*x^2)).
a(n)= Sum_{k=1..n-1} A103435(k). (End)
a(n) = (Sum_{k=0..n} Lucas(k)*Lucas(n-k)*binomial(n, k)) - 4)/5 (Wall, 1986). - Amiram Eldar, Jan 27 2022
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MATHEMATICA
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LinearRecurrence[{3, 2, -4}, {0, 0, 2}, 30] (* Harvey P. Dale, Oct 24 2015 *)
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PROG
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(PARI) a(n)=if(n<1, 0, sum(k=0, n-1, fibonacci(k)*2^k))
(SageMath) [(2^n*lucas_number2(n, 1, -1) -2)/5 for n in range(41)] # G. C. Greubel, Jan 06 2023
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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