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A287439 a(n) = 2*a(n-2) + 2*a(n-3) for n >= 3, where a(0) = 2, a(2) = 4, a(3) = 7. 3

%I #8 Apr 07 2020 20:51:09

%S 2,4,7,12,22,38,68,120,212,376,664,1176,2080,3680,6512,11520,20384,

%T 36064,63808,112896,199744,353408,625280,1106304,1957376,3463168,

%U 6127360,10841088,19181056,33936896,60044288,106235904,187962368,332560384,588396544

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

%C Conjecture: a(n) is the number of letters (0's and 1's) in the n-th iterate of the mapping 00->1000, 10->000, starting with 00; see A288729.

%H Clark Kimberling, <a href="/A287439/b287439.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 (0, 2, 2).

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

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

%t LinearRecurrence[{0, 2, 2}, {2, 4, 7}, 40]

%Y Cf. A287372.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Jun 17 2017

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