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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

%I

%S 1,2,6,18,56,172,532,1640,5064,15624,48224,148816,459280,1417376,

%T 4374240,13499424,41661056,128571328,396788032,1224539264,3779088000,

%U 11662756992,35992787456,111078426880,342802489600,1057933111808,3264919328256,10075966124544

%N a(n) = 2*a(n-1) + 4*a(n-2) - 2*a(n-3) with initial terms 1, 2, 6.

%C 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).

%H Colin Barker, <a href="/A091142/b091142.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,4,-2).

%F 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).

%F 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.

%F G.f.: -x*(2*x^2-1) / (2*x^3-4*x^2-2*x+1). - _Colin Barker_, May 21 2015

%t a[n_] := (MatrixPower[{{1, 1, 1}, {3, 1, 0}, {2, 0, 0}}, n-1].{{1}, {1}, {1}})[[3, 1]]

%o (PARI) Vec(-x*(2*x^2-1)/(2*x^3-4*x^2-2*x+1) + O(x^100)) \\ _Colin Barker_, May 21 2015

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

%K nonn,easy

%O 1,2

%A _Gary W. Adamson_, Dec 21 2003

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Last modified July 31 00:50 EDT 2021. Contains 346365 sequences. (Running on oeis4.)