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a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 1, a(1) = 1, a(2) = 3, a(3) = -3.
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%I #6 Nov 28 2017 10:30:27

%S 1,1,3,-3,-1,3,3,-1,1,7,9,9,17,33,51,77,127,211,339,543,881,1431,2313,

%T 3737,6049,9793,15843,25629,41471,67107,108579,175679,284257,459943,

%U 744201,1204137,1948337,3152481,5100819,8253293,13354111,21607411,34961523

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

%C Lim_{n->inf} 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="/A295676/b295676.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) = 1, a(1) = 1, a(2) = 3, a(3) = -3.

%F G.f.: (-1 - 2 x^2 + 7 x^3)/(-1 + x + x^3 + x^4).

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

%Y Cf. A001622, A000045.

%K easy,sign

%O 0,3

%A _Clark Kimberling_, Nov 27 2017