A226308 a(n) = a(n-1) + a(n-2) + 2*a(n-3) with a(0)=2, a(1)=1, a(2)=5.

2, 1, 5, 10, 17, 37, 74, 145, 293, 586, 1169, 2341, 4682, 9361, 18725, 37450, 74897, 149797, 299594, 599185, 1198373, 2396746, 4793489, 9586981, 19173962, 38347921, 76695845, 153391690, 306783377, 613566757, 1227133514, 2454267025, 4908534053, 9817068106, 19634136209

Offset: 0

Author

N. J. A. Sloane, Jun 07 2013

Keywords

• nonn
• easy

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• G. Cerda-Morales, A note on modified third-order jacobsthal numbers, arXiv:1905.00725 [math.CO], 2019.
• Charles K. Cook and Michael R. Bacon, Some identities for Jacobsthal and Jacobsthal-Lucas numbers satisfying higher order recurrence relations, Annales Mathematicae et Informaticae, 41 (2013) pp. 27-39.
• Evren Eyican Polatlı and Yüksel Soykan, On generalized third-order Jacobsthal numbers, Asian Res. J. of Math. (2021) Vol. 17, No. 2, 1-19, Article No. ARJOM.66022.
• Index entries for linear recurrences with constant coefficients, signature (1,1,2).

Crossrefs

Cf. A046636, A047853, A233328.

Programs

Maple

A226308 := n -> 1/7*(2^(n+3) + 6*cos(2*Pi*n/3) - 4*sqrt(3)*sin(2*Pi*n/3)):
seq(A226308(n), n = 0 .. 34); # Mélika Tebni, Mar 09 2024

Mathematica

CoefficientList[Series[-(2 x^2 - x + 2) / ((2 x - 1) (x^2 + x + 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 18 2013 *)
LinearRecurrence[{1,1,2},{2,1,5},40] (* Harvey P. Dale, Nov 03 2024 *)

PARI

a(n)=([0,1,0; 0,0,1; 2,1,1]^n*[2;1;5])[1,1] \\ Charles R Greathouse IV, Jul 19 2016

Python

a0, a1, a2 = 2, 1, 5
for n in range(77):
  a = a2 + a1 + 2*a0
  print(a0, end=', ')
  a0, a1, a2 = a1, a2, a # Alex Ratushnyak, Jun 08 2013

Formula

G.f.: -(2*x^2-x+2) / ((2*x-1)*(x^2 + x + 1)). - Colin Barker, Jun 08 2013

a(3*n) = A047853(n+1), a(3*n+1) = A233328(n), a(3*n+2) = A046636(n+1). - Philippe Deléham, Feb 24 2014

From Mélika Tebni, Mar 09 2024: (Start)

E.g.f.: (1/7)*(8*exp(2*x) + exp(-x/2)*(6*cos(sqrt(3)*x/2) - 4*sqrt(3)*sin(sqrt(3)*x/2))) (Charles K. Cook and Michael R. Bacon, 2013).

a(n) = (1/7)*(2^(n+3) + 6*cos(2*Pi*n/3) - 4*sqrt(3)*sin(2*Pi*n/3)). (End)

Extensions

Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021