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a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4), where a(0) = -1, a(1) = 0, a(2) = 0, a(3) = 1.
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%I #6 Aug 27 2021 21:02:46

%S -1,0,0,1,3,6,13,23,44,75,135,226,393,651,1108,1823,3059,5010,8325,

%T 13591,22428,36531,59983,97538,159569,259155,422820,686071,1117083,

%U 1811346,2944813,4772543,7750124,12555435,20371095,32992066,53494233,86617371,140373748

%N a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4), where a(0) = -1, a(1) = 0, a(2) = 0, 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="/A295730/b295730.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) = -1, a(1) = 0, a(2) = 0, a(3) = 1.

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

%t LinearRecurrence[{1, 3, -2, -2}, {-1, 0, 0, 1}, 100]

%Y Cf. A001622, A000045, A005672.

%K easy,sign

%O 0,5

%A _Clark Kimberling_, Nov 30 2017