A098183 -id:A098183 - cp's OEIS frontend

A098182 a(n) = 3*a(n-1) - a(n-2) + a(n-3), a(0)=1,a(1)=1,a(2)=3.

1, 1, 3, 9, 25, 69, 191, 529, 1465, 4057, 11235, 31113, 86161, 238605, 660767, 1829857, 5067409, 14033137, 38861859, 107619849, 298030825, 825334485, 2285592479, 6329473777, 17528163337, 48540608713, 134423136579

Offset: 0

Paul Barry, Aug 30 2004

G. C. Greubel, Table of n, a(n) for n = 0..1000
Olivier Carton, Jean-Michel Couvreur, Martin Delacourt, and Nicolas Ollinger, Linear Recurrence Sequence Automata and the Addition of Abstract Numeration Systems, 15th Int'l Conf. Comb. Words (WORDS 2025) Lecture Notes Comp. Sci. (LNCS Vol. 15729) 70-82. See p. 71.
Index entries for linear recurrences with constant coefficients, signature (3,-1,1).

Cf. A098183, A098184.

CoefficientList[Series[(1 - x)^2/(1 - 3 x + x^2 - x^3), {x,0,50}], x] (* G. C. Greubel, Mar 03 2017 *)

my(x='x+O(x^50)); Vec((1-x)^2/(1-3*x+x^2-x^3)) \\ G. C. Greubel, Mar 03 2017

G.f. : (1-x)^2/(1-3*x+x^2-x^3).

a(n) = Sum_{k=0..floor(n/2)} binomial(n+k, 3*k) * 2^k.

A098184 a(n) = 3a(n-1)+a(n-2)+a(n-3), a(0)=1, a(1)=1, a(2)=5.

Original entry on oeis.org

1, 1, 5, 17, 57, 193, 653, 2209, 7473, 25281, 85525, 289329, 978793, 3311233, 11201821, 37895489, 128199521, 433695873, 1467182629, 4963443281, 16791208345, 56804250945, 192167404461, 650097672673, 2199264673425

Offset: 0

Paul Barry, Aug 30 2004

Even bisection of the tribonacci sequence A000213. - Oboifeng Dira, Aug 03 2016

Index entries for linear recurrences with constant coefficients, signature (3,1,1).

Cf. A000213, A098182, A098183.

LinearRecurrence[{3,1,1},{1,1,5},30] (* Harvey P. Dale, Nov 29 2011 *)
CoefficientList[Series[(1 - x)^2/(1 - 3 x - x^2 - x^3), {x, 0, 24}], x] (* Michael De Vlieger, Aug 03 2016 *)

a(n)=([0,1,0; 0,0,1; 1,1,3]^n*[1;1;5])[1,1] \\ Charles R Greathouse IV, Aug 03 2016

from sage.combinat.sloane_functions import recur_gen3
it = recur_gen3(1,1,1,3,1,1)
[next(it) for i in range(32)] # Zerinvary Lajos, Jun 24 2008

G.f.: (1-x)^2/(1-3*x-x^2-x^3).
a(n) = Sum_{k=0..floor(n/2)} binomial(n+k, 3*k)*4^k.
a(n)/a(n-1) tends to 3.38297576..., the square of the tribonacci constant A058265. - Gary W. Adamson, Feb 28 2006

cp's OEIS Frontend

A098182 a(n) = 3*a(n-1) - a(n-2) + a(n-3), a(0)=1,a(1)=1,a(2)=3.

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