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a(n) = 2*a(n-1) + a(n-2) - a(n-3), where a(0) = 2, a(1) = 3, a(2) = 6.
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%I #10 Apr 07 2020 21:15:44

%S 2,3,6,13,29,65,146,328,737,1656,3721,8361,18787,42214,94854,213135,

%T 478910,1076101,2417977,5433145,12208166,27431500,61638021,138499376,

%U 311205273,699271901,1571249699,3530566026,7933109850,17825536027,40053615878,89999657933

%N a(n) = 2*a(n-1) + a(n-2) - a(n-3), where a(0) = 2, a(1) = 3, a(2) = 6.

%C Conjecture: a(n) is the number of letters (0's and 1's) in the n-th iteration of the mapping 00->001, 1->011, starting with 00; see A286939.

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

%F a(n) = 2*a(n-1) + a(n-2) - a(n-3), where a(0) = 2, a(1) = 3, a(2) = 6.

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

%t LinearRecurrence[{2, 1, -1}, {2, 3, 6}, 40]

%Y Cf. A286059.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Jun 05 2017