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a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 0, a(1) = 2, a(2) = 0, a(3) = 1.
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%I #6 Aug 27 2021 21:15:21

%S 0,2,0,1,3,5,6,10,18,29,45,73,120,194,312,505,819,1325,2142,3466,5610,

%T 9077,14685,23761,38448,62210,100656,162865,263523,426389,689910,

%U 1116298,1806210,2922509,4728717,7651225,12379944,20031170,32411112,52442281,84853395

%N a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 0, a(1) = 2, 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="/A295682/b295682.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, 0, 1, 1)

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

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

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

%Y Cf. A001622, A000045.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Nov 29 2017