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a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4), where a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 2.
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%I #6 Aug 27 2021 21:19:55

%S 0,0,1,2,5,9,18,31,57,96,169,281,482,795,1341,2200,3669,5997,9922,

%T 16175,26609,43296,70929,115249,188226,305523,497845,807464,1313501,

%U 2129157,3459042,5604583,9096393,14733744,23895673,38694953,62721698,101547723,164531565

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

%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="/A295724/b295724.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) = 0, a(1) = 0, a(2) = 1, a(3) = 2.

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

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

%Y Cf. A001622, A000045, A005672.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Nov 29 2017