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A091142 a(n) = 2*a(n-1) + 4*a(n-2) - 2*a(n-3) with initial terms 1, 2, 6. 4
1, 2, 6, 18, 56, 172, 532, 1640, 5064, 15624, 48224, 148816, 459280, 1417376, 4374240, 13499424, 41661056, 128571328, 396788032, 1224539264, 3779088000, 11662756992, 35992787456, 111078426880, 342802489600, 1057933111808, 3264919328256, 10075966124544 (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 t(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*(2*x^2-1) / (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}})[[3, 1]]

PROG

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

CROSSREFS

Cf. r(n) = A091140(n), s(n) = A091141(n).

Sequence in context: A291730 A002999 A291228 * A275857 A111961 A190861

Adjacent sequences:  A091139 A091140 A091141 * A091143 A091144 A091145

KEYWORD

nonn,easy

AUTHOR

Gary W. Adamson, Dec 21 2003

STATUS

approved

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Last modified June 14 10:04 EDT 2021. Contains 345025 sequences. (Running on oeis4.)