A054856 Number of ways to tile a 4 X n region with 1 X 1, 2 X 2, 3 X 3 and 4 X 4 tiles.

1, 1, 5, 13, 40, 117, 348, 1029, 3049, 9028, 26738, 79183, 234502, 694476, 2056692, 6090891, 18038173, 53420041, 158203433, 468519406, 1387520047, 4109140098, 12169216863, 36039131181, 106729873498, 316080480394, 936072224321

Offset: 0

Author

Silvia Heubach (silvi(AT)cine.net), Apr 21 2000

Keywords

• nonn
• easy

Comments

It is easy to see that the g.f. for indecomposable tilings, i.e. those that cannot be split vertically into smaller tilings, is g=z+4*z^2+2*z^3+z^4+2*z^3/(1-z); then G.f.=1/(1-g). - Emeric Deutsch, Oct 16 2006

Examples

			a(2)=5 as there is one tiling of a 4 X 2 region with only 1 X 1 tiles, 3 tilings with exactly one 2 X 2 tile and 1 tiling with exactly two 2 X 2 tiles.
		

Links

• S. Heubach, Tiling an m-by-n area with squares of size up to k-by-k (m<=5), Congressus Numerantium 140 (1999), 43-64.
• Index entries for linear recurrences with constant coefficients, signature (2,3,0,-1,-1).

Crossrefs

Cf. A002478, A054857, A226547.

Column k=4 of A219924. - Alois P. Heinz, Dec 01 2012

Programs

Maple

a[0]:=1: a[1]:=1: a[2]:=5: a[3]:=13: a[4]:=40: for n from 5 to 26 do a[n]:=2*a[n-1]+3*a[n-2]-a[n-4]-a[n-5] od: seq(a[n],n=0..26); # Emeric Deutsch, Oct 16 2006

Mathematica

f[ A_ ] := Module[ {til = A, sum}, sum = 2* Apply[ Plus, Drop[ til, -4 ] ]; AppendTo[ til, A[ [ -1 ] ] + 4A[ [ -2 ] ] + 4A[ [ -3 ] ] + 3A[ [ -4 ] ] + sum ] ]; NumOfTilings[ n_ ] := Nest[ f, {1, 1, 5, 13}, n - 2 ]; NumOfTilings[ 30 ]

Formula

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

a(n) = 2*a(n-1)+3*a(n-2)-a(n-4)-a(n-5). G.f.=(1-z)/((1+z)*(1-3*z+z^4)). - Emeric Deutsch, Oct 16 2006