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a(n) = 2*a(n-1) - a(n-3) for n >= 5, a(0) = 2, a(1) = 4, a(2) = 7, a(3) = 12, a(4) = 22.
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%I #14 Apr 07 2020 21:27:19

%S 2,4,7,12,22,37,62,102,167,272,442,717,1162,1882,3047,4932,7982,12917,

%T 20902,33822,54727,88552,143282,231837,375122,606962,982087,1589052,

%U 2571142,4160197,6731342,10891542,17622887,28514432,46137322,74651757,120789082

%N a(n) = 2*a(n-1) - a(n-3) for n >= 5, a(0) = 2, a(1) = 4, a(2) = 7, a(3) = 12, a(4) = 22.

%C Conjecture: a(n) is the number of letters (0's and 1's) in the n-th iterate of the mapping 00->0010, 01->011, 10->000, starting with 00; see A289104.

%H Clark Kimberling, <a href="/A289107/b289107.txt">Table of n, a(n) for n = 0..3000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2, 0, -1).

%F a(n) = 2*a(n-1) - a(n-3) for n >= 5, a(0) = 2, a(1) = 4, a(2) = 7, a(3) = 12, a(4) = 22.

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

%F a(n) = -3 + 2^(-1-n)*sqrt(5)*(-(1-sqrt(5))^(1 + n) + (1+sqrt(5))^(1+n)) for n>1. - _Colin Barker_, Jun 28 2017

%t Join[{2, 4}, LinearRecurrence[{2, 0, -1}, {7, 12, 22}, 40]]

%Y Cf. A289104.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Jun 28 2017