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a(n) = 3*a(n-1) - 2*a(n-2) + 2*a(n-3) - 4*a(n-4) + 2*a(n-5), where a(0) = 2, a(1) =3, a(2) = 6, a(3)=13, a(4) = 29.
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%I #10 Apr 06 2020 19:50:34

%S 2,3,6,13,29,65,145,323,719,1599,3555,7903,17567,39047,86791,192911,

%T 428783,953055,2118351,4708447,10465439,23261471,51703135,114920255,

%U 255432575,567748479,1261931199,2804887039,6234405887,13857177215,30800266111,68459569919

%N a(n) = 3*a(n-1) - 2*a(n-2) + 2*a(n-3) - 4*a(n-4) + 2*a(n-5), where a(0) = 2, a(1) =3, a(2) = 6, a(3)=13, a(4) = 29.

%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->110, starting with 00; see A287125.

%H Clark Kimberling, <a href="/A287128/b287128.txt">Table of n, a(n) for n = 0..2000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3, -2, 2, -4, 2).

%F a(n) = 3*a(n-1) - 2*a(n-2) + 2*a(n-3) - 4*a(n-4) + 2*a(n-5), where a(0) = 2, a(1) =3, a(2) = 6, a(3)=13, a(4) = 29..

%F G.f.: (-2 + 3*x - x^2 + 3*x^3 - 4*x^4)/(-1 + 3*x - 2*x^2 + 2*x^3 - 4*x^4 + 2*x^5).

%t LinearRecurrence[{3, -2, 2, -4, 2}, {2, 3, 6, 13, 29}, 40]

%Y Cf. A287125.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Jun 06 2017