A120940 Alternating sum of the Fibonacci numbers multiplied by their (combinatorial) indices.
0, 1, 3, 6, 14, 26, 52, 95, 177, 318, 572, 1012, 1784, 3117, 5423, 9382, 16170, 27758, 47500, 81035, 137885, 234046, 396408, 670056, 1130544, 1904281, 3202587, 5378310, 9020102, 15109058, 25279012, 42248567, 70537929, 117657342, 196076468, 326485852
Offset: 0
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- M. M. Herreshoff, A Combinatorial proof of the summation from k = 0 to n of k times f sub k, Presented at The Twelfth International Conference on Fibonacci Numbers and Their Applications.
- Index entries for linear recurrences with constant coefficients, signature (1,3,-1,-3,-1).
Programs
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Mathematica
CoefficientList[Series[(2*z^2 + z)/((z + 1)*(z^2 + z - 1)^2), {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 08 2011 *) LinearRecurrence[{1,3,-1,-3,-1},{0,1,3,6,14},40] (* Harvey P. Dale, Apr 21 2018 *) -
PARI
concat(0, Vec(x*(1 + 2*x) / ((1 + x)*(1 - x - x^2)^2) + O(x^40))) \\ Colin Barker, Apr 03 2019
Formula
a(n) = Sum_{k=0..n} (-1)^(n-k)*k*f(k) also, when n >= 3, a(n) = nf(n-1) + f(n-3) + (-1)^n where f(n) = F(n+1).
a(n) = (-1)^n+A000045(n)-A001629(n+2)+3*A001629(n+1). - R. J. Mathar, Jul 11 2011 [Corrected by Alan Michael Gómez Calderón, Jul 23 2025]
From Colin Barker, Apr 03 2019: (Start)
G.f.: x*(1 + 2*x) / ((1 + x)*(1 - x - x^2)^2).
a(n) = a(n-1) + 3*a(n-2) - a(n-3) - 3*a(n-4) - a(n-5) for n>4.
(End)