A091141 a(n) = 2*a(n-1) + 4*a(n-2) - 2*a(n-3) with initial terms 1, 4, 13.

1, 4, 13, 40, 124, 382, 1180, 3640, 11236, 34672, 107008, 330232, 1019152, 3145216, 9706576, 29955712, 92447296, 285304288, 880486336, 2717295232, 8385927232, 25880062720, 79869243904, 246486884224, 760690618624, 2347590286336, 7244969278720, 22358918465536

Offset

1,2

Comments

One of 3 related sequences generated from finite difference operations. Let r(1)=s(1)=t(1)=1. Given r(n), s(n) and t(n), let f(x) = r(n) x^2 + s(n) x + t(n) and let r(n+1), s(n+1) and t(n+1) be the 0th, 1st and 2nd differences of f(x) at x=1. I.e., r(n+1) = f(1) = r(n)+s(n)+t(n), s(n+1) = f(2)-f(1) = 3r(n)+s(n) and t(n+1) = f(3)-2f(2)+f(1) = 2r(n). This sequence gives s(n).

Links

Colin Barker, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (2,4,-2).

Formula

Let v(n) be the column vector with elements r(n), s(n), t(n); then v(n) = [1 1 1 / 3 1 0 / 2 0 0] v(n-1).

The limit as n->infinity of a(n+1)/a(n) is the largest root of x^3 - 2x^2 - 4x + 2 = 0, which is about 3.086130197651494.

G.f.: x*(x+1)^2 / (2*x^3-4*x^2-2*x+1). - Colin Barker, May 21 2015

Mathematica

a[n_] := (MatrixPower[{{1, 1, 1}, {3, 1, 0}, {2, 0, 0}}, n-1].{{1}, {1}, {1}})[[2, 1]]
LinearRecurrence[{2, 4, -2}, {1, 4, 13}, 30] (* Harvey P. Dale, Jun 19 2018 *)

Prog

(PARI) Vec(x*(x+1)^2/(2*x^3-4*x^2-2*x+1) + O(x^100)) \\ Colin Barker, May 21 2015

Crossrefs

Cf. r(n) = A091140(n), t(n) = A091142(n).

Sequence in context: A003462 A076040 A261547 * A098183 A171556 A227747

Adjacent sequences: A091138 A091139 A091140 * A091142 A091143 A091144

Keyword

nonn,easy

Author

Gary W. Adamson, Dec 21 2003

Status

approved