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A091140 a(n) = 2*a(n-1) + 4*a(n-2) - 2*a(n-3) with initial terms 1, 3, 9. 4
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 (list; graph; refs; listen; history; text; internal format)
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]]

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

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Last modified October 13 20:38 EDT 2019. Contains 327981 sequences. (Running on oeis4.)