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a(n) = 2*a(n-1) - a(n-2) + a(n-3) - a(n-4), where a(0) = 2, a(1) = 4, a(2) = 5, a(3) = 6.
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%I #12 Apr 06 2020 22:12:17

%S 2,4,5,6,9,13,18,26,38,55,80,117,171,250,366,536,785,1150,1685,2469,

%T 3618,5302,7770,11387,16688,24457,35843,52530,76986,112828,165357,

%U 242342,355169,520525,762866,1118034,1638558,2401423,3519456,5158013,7559435,11078890

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

%C Conjecture: a(n) is the number of letters (0's and 1's) in the n-th iteration of the mapping 00->0101, 1->100, starting with 00; see A288426.

%H Clark Kimberling, <a href="/A288429/b288429.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2, -1, 1, -1).

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

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

%F a(n) = a(n-1) + a(n-3) - 1, for n > 2. - _Greg Dresden_, Feb 09 2020

%t LinearRecurrence[{2, -1, 1, -1}, {2, 4, 5, 6}, 40]

%Y Cf. A288426.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Jun 11 2017