A099920 a(n) = (n+1)*F(n), F(n) = Fibonacci numbers A000045.
0, 2, 3, 8, 15, 30, 56, 104, 189, 340, 605, 1068, 1872, 3262, 5655, 9760, 16779, 28746, 49096, 83620, 142065, 240812, 407353, 687768, 1159200, 1950650, 3277611, 5499704, 9216519, 15426870, 25793240, 43080608, 71884197, 119835652
Offset: 0
References
- A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 35.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
- S. Klavzar, On median nature and enumerative properties of Fibonacci-like cubes, Discr. Math. 299 (2005), 145-153.
- Franck Ramaharo, A one-variable bracket polynomial for some Turk's head knots, arXiv:1807.05256 [math.CO], 2018.
- Steven Finch, Variance of longest run duration in a random bitstring, arXiv:2005.12185 [math.CO], 2020.
- Tamás Szakács, Convolution of second order linear recursive sequences. II. Commun. Math. 25, No. 2, 137-148 (2017), remark 2.
- Tamás Szakács, Linear recursive sequences and factorials, Ph. D. Thesis, Univ. Debrecen (Hungary, 2024). See p. 35.
- Eric Weisstein's World of Mathematics, Edge Count.
- Eric Weisstein's World of Mathematics, Lucas Cube Graph.
- Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1)
Programs
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Haskell
a099920 n = a099920_list !! n a099920_list = zipWith (*) [1..] a000045_list -- Reinhard Zumkeller, Oct 07 2012
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Magma
[(n+1)*Fibonacci(n): n in [0..60]]; // Vincenzo Librandi, Apr 23 2011
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Mathematica
Table[(n + 1) Fibonacci[n], {n, 0, 40}] (* Harvey P. Dale, Jan 18 2012 *) LinearRecurrence[{2, 1, -2, -1}, {0, 2, 3, 8}, 40] (* Harvey P. Dale, Jan 18 2012 *) CoefficientList[Series[(2 - x) x/(-1 + x + x^2)^2, {x, 0, 20}], x] (* Eric W. Weisstein, Jul 28 2023 *)
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PARI
a(n)=(n+1)*fibonacci(n) \\ Charles R Greathouse IV, Jun 11 2015
Formula
G.f.: x*(2-x)/(1-x-x^2)^2;
a(n) = Sum_{k=0..n} F(n-k)*(L(k-1) + 0^k).
a(n) = Sum_{k=0..n+1} F(n-k)*binomial(n-k+1, k)*binomial(1, (k+1)/2)*(1-(-1)^k)/2.
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4); a(0)=0, a(1)=2, a(2)=3, a(3)=8. - Harvey P. Dale, Jan 18 2012
a(n) = a(n-1) + a(n-2) + A000032(n-1) (Lucas numbers). - Bob Selcoe, Aug 19 2015
Extensions
Entry revised by N. J. A. Sloane, Jan 23 2006. The offset changed, so some of the formulas may now be slightly off.
Comments