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 A295854 a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4), where a(0) = -2, a(1) = -2, a(2) = 2, a(3) = 1. 1

%I #9 Aug 27 2021 21:15:48

%S -2,-2,2,1,15,18,57,79,184,271,551,838,1581,2451,4416,6931,12115,

%T 19174,32825,52255,88152,140919,235215,377158,624661,1003867,1653104,

%U 2661067,4363323,7032582,11494209,18543175,30233992,48809935,79437143,128312614,208536189

%N a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4), where a(0) = -2, a(1) = -2, a(2) = 2, a(3) = 1.

%C a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth rate of the Fibonacci numbers (A000045).

%H Clark Kimberling, <a href="/A295854/b295854.txt">Table of n, a(n) for n = 0..2000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1, 3, -2, -2)

%F a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = -2, a(1) = -2, a(2) = 2, a(3) = 1.

%F G.f.: (-2 + 10 x^2 + x^3)/(1 - x - 3 x^2 + 2 x^3 + 2 x^4).

%t LinearRecurrence[{1, 3, -2, -2}, {-2, -2, 2, 1}, 100]

%t CoefficientList[Series[(x^3+10*x^2-2)/(2*x^4+2*x^3-3*x^2-x+1),{x,0,40}],x] (* _Harvey P. Dale_, Mar 05 2018 *)

%Y Cf. A001622, A000045.

%K easy,sign

%O 0,1

%A _Clark Kimberling_, Dec 01 2017

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Last modified August 9 11:50 EDT 2024. Contains 375042 sequences. (Running on oeis4.)