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a(n) = a(n-1) + 2*a(n-2) - a(n-3), where a(0) = 2, a(1) = 4, a(2) = 7.
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%I #10 Apr 07 2020 23:05:34

%S 2,4,7,13,23,42,75,136,244,441,793,1431,2576,4645,8366,15080,27167,

%T 48961,88215,158970,286439,516164,930072,1675961,3019941,5441791,

%U 9805712,17669353,31838986,57371980,103380599,186285573,335674791,604865338,1089929347

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

%C Conjecture: a(n) is the number of letters (0's and 1's) in the n-th iteration of the mapping 00->0010, 1->10, starting with 00; see A287931.

%H Clark Kimberling, <a href="/A287381/b287381.txt">Table of n, a(n) for n = 0..2000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1, 2, -1).

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

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

%t LinearRecurrence[{1, 2, -1}, {2, 4, 7}, 40]

%Y Cf. A287931, A078038 (signed version).

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Jun 07 2017