A099919 a(n) = F(3) + F(6) + F(9) + ... + F(3n), F(n) = Fibonacci numbers A000045.
0, 2, 10, 44, 188, 798, 3382, 14328, 60696, 257114, 1089154, 4613732, 19544084, 82790070, 350704366, 1485607536, 6293134512, 26658145586, 112925716858, 478361013020, 2026369768940, 8583840088782, 36361730124070, 154030760585064, 652484772464328, 2763969850442378
Offset: 0
References
- A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 25.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Project Euler, Problem 2: Even Fibonacci Numbers.
- Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).
Crossrefs
Programs
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Magma
[(Fibonacci(3*n+2) - 1)/2: n in [0..30]]; // G. C. Greubel, Jan 17 2018
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Mathematica
CoefficientList[Series[2 x/((1 - x) (1 - 4 x - x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 15 2014 *) LinearRecurrence[{5, -3, -1}, {0, 2, 10}, 30] (* G. C. Greubel, Jan 17 2018 *) Accumulate[Fibonacci[3Range[0, 19]]] (* Alonso del Arte, Dec 23 2018 *)
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PARI
a(n) = sum(i=1, n, fibonacci(3*i)); \\ Michel Marcus, Mar 15 2014
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PARI
a(n) = fibonacci(3*n+2)\2 \\ Charles R Greathouse IV, Jun 11 2015
Formula
a(n) = (Fibonacci(3*n + 2) - 1)/2 = (A015448(n+1)-1)/2.
G.f.: 2*x/((1 - x)*(1 - 4*x - x^2)).
a(n) = (F(3n + 2) - 1)/2 = 2 * A049652(n).
a(n) = Sum_{0 <= j <= i <= n} binomial(i, j)*F(i + j). - Benoit Cloitre, May 21 2005
From Gary Detlefs, Dec 08 2010: (Start)
a(n) = 4*a(n - 1) + a(n - 2) + 2, n > 1.
a(n) = 5*a(n - 1) - 3*a(n - 2) - a(n - 3), n > 2.
a(n) = (Fibonacci(3*n + 3) + Fibonacci(3*n) - 2)/4. (End)
a(n) = (-10 + (5 - 3*sqrt(5))*(2 - sqrt(5))^n + (2 + sqrt(5))^n*(5 + 3*sqrt(5)))/20. - Colin Barker, Nov 26 2016
E.g.f.: exp(x)*(exp(x)*(5*cosh(sqrt(5)*x) + 3*sqrt(5)*sinh(sqrt(5)*x)) - 5)/10. - Stefano Spezia, Jun 03 2024
Extensions
a(0) = 0 prepended by Joerg Arndt, Mar 13 2014
Comments