A091140 a(n) = 2*a(n-1) + 4*a(n-2) - 2*a(n-3) with initial terms 1, 3, 9.

1, 3, 9, 28, 86, 266, 820, 2532, 7812, 24112, 74408, 229640, 708688, 2187120, 6749712, 20830528, 64285664, 198394016, 612269632, 1889544000, 5831378496, 17996393728, 55539213440, 171401244800, 528966555904, 1632459664128, 5037983062272, 15547871669248

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 r(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^2-x-1) / (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}})[[1, 1]]
LinearRecurrence[{2, 4, -2}, {1, 3, 9}, 30] (* Harvey P. Dale, May 18 2021 *)

Prog

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

Crossrefs

Cf. s(n) = A091141(n), t(n) = A091142(n).

Sequence in context: A291257 A005354 A084084 * A052541 A024738 A263841

Adjacent sequences: A091137 A091138 A091139 * A091141 A091142 A091143

Keyword

nonn,easy

Author

Gary W. Adamson, Dec 21 2003

Status

approved