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

%I #18 Jun 19 2018 14:43:40

%S 1,4,13,40,124,382,1180,3640,11236,34672,107008,330232,1019152,

%T 3145216,9706576,29955712,92447296,285304288,880486336,2717295232,

%U 8385927232,25880062720,79869243904,246486884224,760690618624,2347590286336,7244969278720,22358918465536

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

%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 s(n).

%H Colin Barker, <a href="/A091141/b091141.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*(x+1)^2 / (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}})[[2, 1]]

%t LinearRecurrence[{2,4,-2},{1,4,13},30] (* _Harvey P. Dale_, Jun 19 2018 *)

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

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

%K nonn,easy

%O 1,2

%A _Gary W. Adamson_, Dec 21 2003

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