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a(n) = a(n-1) + a(n-2) for n >= 3, where a(0) = 2, a(1) = 4, a(2) = 7.
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%I #15 Apr 07 2020 20:50:07

%S 2,4,7,11,18,29,47,76,123,199,322,521,843,1364,2207,3571,5778,9349,

%T 15127,24476,39603,64079,103682,167761,271443,439204,710647,1149851,

%U 1860498,3010349,4870847,7881196,12752043,20633239,33385282,54018521,87403803,141422324

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

%C Empirically, a(n) is the number of letters (0's and 1's) in the n-th iterate of the mapping 00->1000, 10->010, starting with 00; see A288216.

%H Clark Kimberling, <a href="/A288219/b288219.txt">Table of n, a(n) for n = 0..2000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1, 1).

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

%F a(n) = L(n+2) for n >=1, where L = A000032 (Lucas numbers).

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

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

%Y Cf. A288216, A163695, A080023, A000204.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Jun 19 2017