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

%I #12 Feb 19 2019 10:22:45

%S 4,3,2,1,8,13,16,25,46,75,116,187,308,499,802,1297,2104,3405,5504,

%T 8905,14414,23323,37732,61051,98788,159843,258626,418465,677096,

%U 1095565,1772656,2868217,4640878,7509099,12149972,19659067,31809044,51468115,83277154

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

%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="/A295673/b295673.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) = 4, a(1) = 3, a(2) = 2, a(3) = 1.

%F G.f.: (-4 + x + x^2 + 5 x^3)/(-1 + x + x^3 + x^4).

%t LinearRecurrence[{1, 0, 1, 1}, {4, 3, 2, 1}, 100]

%Y Cf. A001622, A000045.

%K easy,nonn

%O 0,1

%A _Clark Kimberling_, Nov 27 2017