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The number of ways to tile (with squares and rectangles) a 2 X (n+2) strip with the upper left and upper right squares removed.
2

%I #22 Jul 16 2018 09:13:09

%S 2,4,15,46,150,480,1545,4964,15958,51292,164871,529946,1703418,

%T 5475328,17599457,56570280,181834970,584475732,1878691887,6038716422,

%U 19410365422,62391120800,200545011401,644615789580,2072001259342,6660074556204,21407609138375

%N The number of ways to tile (with squares and rectangles) a 2 X (n+2) strip with the upper left and upper right squares removed.

%C Each number in the sequence is the partial sum of A033505 (n starts at 0, each number add one if n is even). We can also find the recursion relation a(n) = 2*a(n-1) + 4*a(n-2) - a(n-4) for the sequence, which can be proved by induction.

%H Colin Barker, <a href="/A316726/b316726.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,4,0,-1).

%F a(n) = 2*a(n-1) + 4*a(n-2) - a(n-4) for n>=4.

%F G.f.: (2 - x^2) / ((1 + x)*(1 - 3*x - x^2 + x^3)). - _Colin Barker_, Jul 12 2018

%e For n=4, a(4) = 150 = 2*a(3) + 4*a(2) - a(0).

%t CoefficientList[ Series[(-x^2 + 1)/(x^4 - 4x^2 - 2x + 1), {x, 0, 27}], x] (* or *) LinearRecurrence[{2, 4, 0, -1}, {2, 4, 15, 46}, 27] (* _Robert G. Wilson v_, Jul 15 2018 *)

%o (PARI) Vec((2 - x^2) / ((1 + x)*(1 - 3*x - x^2 + x^3)) + O(x^30)) \\ _Colin Barker_, Jul 12 2018

%Y Cf. A033505.

%K nonn,easy

%O 0,1

%A _Zijing Wu_, Jul 11 2018

%E More terms from _Colin Barker_, Jul 12 2018