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a(n) = a(n-1) + 3*a(n-2) - 2*a(n-3) - a(n-4) for n >= 4, where a(0) = 2, a(1) = 4, a(2) = 7, a(3) = 12.
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%I #13 Apr 07 2020 22:17:06

%S 2,4,7,12,22,40,75,139,262,489,922,1726,3252,6097,11479,21540,40531,

%T 76096,143130,268816,505483,949575,1785262,3354205,6305358,11847874,

%U 22270276,41848977,78658699,147817204,277825071,522110308,981292414,1844155992,3465987547

%N a(n) = a(n-1) + 3*a(n-2) - 2*a(n-3) - a(n-4) for n >= 4, where a(0) = 2, a(1) = 4, a(2) = 7, a(3) = 12.

%C Conjecture: a(n) is the number of letters (0's and 1's) in the n-th iteration of the mapping 00->0001, 1->00, starting with 00; see A288314.

%H Clark Kimberling, <a href="/A288317/b288317.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, -1).

%F a(n) = a(n-1) + 3*a(n-2) - 2*a(n-3) - a(n-4) for n >= 4, where a(0) = 2, a(1) = 4, a(2) = 7, a(3) = 12.

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

%t Join[{2}, LinearRecurrence[{1, 3, -2, -1}, {4, 7, 12, 22}, 40]]

%Y Cf. A288314.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Jun 09 2017