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a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 2, a(1) = 1, a(2) = 2, a(3) = 1.
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%I #6 Aug 27 2021 21:23:09

%S 2,1,2,1,4,7,10,15,26,43,68,109,178,289,466,753,1220,1975,3194,5167,

%T 8362,13531,21892,35421,57314,92737,150050,242785,392836,635623,

%U 1028458,1664079,2692538,4356619,7049156,11405773,18454930,29860705,48315634,78176337

%N a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 2, a(1) = 1, 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="/A295686/b295686.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) = 2, a(1) = 1, a(2) = 2, a(3) = 1.

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

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

%Y Cf. A001622, A000045.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Nov 29 2017