OFFSET
0,3
REFERENCES
Ralph P. Grimaldi, Fibonacci and Catalan Numbers: An Introduction, (2012). See Exercise 10 at p. 110.
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).
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
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 *)
PROG
(PARI) a(n)=if(n<1, 0, sum(k=0, n-1, fibonacci(k)*2^k))
(Magma) [(2^n*Lucas(n) -2)/5: n in [0..40]]; // Vincenzo Librandi, Jul 15 2018
(SageMath) [(2^n*lucas_number2(n, 1, -1) -2)/5 for n in range(41)] # G. C. Greubel, Jan 06 2023
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved
