A014334 Exponential convolution of Fibonacci numbers with themselves.
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
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- C. A. Church and Marjorie Bicknell, Exponential generating functions for Fibonacci identities, Fibonacci Quarterly 11, no. 3 (1973), 275-281.
- Sergio Falcon, Half self-convolution of the k-Fibonacci sequence, Notes on Number Theory and Discrete Mathematics, Vol. 26, No. 3 (2020), pp. 96-106.
- Helmut Prodinger, Convolution identities for Tribonacci numbers via the diagonal of a bivariate generating function, arXiv:1910.08323 [math.NT], 2019.
- 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.
- Index entries for linear recurrences with constant coefficients, signature (3,2,-4).
Programs
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Magma
[(2^n*Lucas(n) -2)/5: n in [0..40]]; // Vincenzo Librandi, Jul 15 2018
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Mathematica
LinearRecurrence[{3,2,-4}, {0,0,2}, 30] (* Harvey P. Dale, Oct 24 2015 *) Table[(2^n LucasL[n] -2)/5, {n,0,100}] (* Vladimir Reshetnikov, May 18 2016 *) -
PARI
a(n)=if(n<1,0,sum(k=0,n-1,fibonacci(k)*2^k))
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SageMath
[(2^n*lucas_number2(n,1,-1) -2)/5 for n in range(41)] # G. C. Greubel, Jan 06 2023
Formula
From Benoit Cloitre, May 29 2003: (Start)
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
From R. J. Mathar, Sep 29 2010: (Start)
a(n) = 2*A014335(n).
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) = (2^n*A000032(n) - 2)/5. - Vladimir Reshetnikov, May 18 2016
E.g.f.: 2*(cosh(sqrt(5)*x)-1)*exp(x)/5. - Ilya Gutkovskiy, May 18 2016
a(n) = ((Sum_{k=0..n} Lucas(k)*Lucas(n-k)*binomial(n, k)) - 4)/5 (Wall, 1986). - Amiram Eldar, Jan 27 2022