A281791 Ways to tile a 5 X (2n+1) floor with tatami mats, including one monomer.

3, 18, 10, 8, 18, 24, 32, 52, 68, 100, 142, 196, 280, 388, 542, 756, 1046, 1452, 2006, 2768, 3816, 5248, 7212, 9896, 13562, 18568, 25392, 34692, 47354, 64580, 88002, 119824, 163034, 221672, 301200, 409004, 555060, 752844, 1020550, 1382732, 1872520, 2534596, 3429206, 4637556, 6269070

Offset: 0

Author

Yasutoshi Kohmoto, Jan 30 2017

Keywords

• nonn
• easy

Comments

Apart from a single 1 X 1 monomer, the area is tiled with 2 X 1 mats. No four mats are permitted to meet at a point.

Examples

			For n=0, the 5X1 floor allows the monomer to be placed at one of the two ends or in the middle: a(n=0)=3.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (0,2,2,-1,-2,-1).

Crossrefs

Cf. A271786 [3X(2n+1) floor]. 2nd column of A272474.

Programs

PARI

s1(n)=my(s); forstep(k=(n%4!=1),(n-1)\6,2, s+=((n+3)/4-k/2)*((n-1)/4-k/2)!/(k!*((n-1)/4-3*k/2)!)); 2*s
s3(n)=my(s); forstep(k=(n%4==1),(n-3)\6,2, s+=((n-3)/4-k/2)!/(k!*((n-3)/4-3*k/2)!)); 2*s
s5(n)=my(s); forstep(k=(n%4!=1),(n-5)\6,2, s+=((n+7)/4-k/2)*((n-5)/4-k/2)!/(k!*((n-5)/4-3*k/2)!)); 2*s
a(n)=s1(n) + s3(n) + s5(n) \\ Charles R Greathouse IV, Feb 20 2017

Formula

a(n) = S_1(2n+1) + S_5(2n+1) + S_3(2n+1) for n>1 where

S_1(n) = 2* Sum_{k= 0<=k<=[(n-1)/6]} ((n+3)/4-1/2*k) *((n-1)/4-1/2*k)!/(k!*((n-1)/4-3/2*k)!). The sum is over even k if n==1 (mod 4), else over odd k.

S_5(n) = 2* Sum_{0<=k<=[(n-5)/6]} ((n+7)/4-1/2*k) *((n-5)/4-1/2*k)!/(k!*((n-5)/4-3/2*k)!). The sum is over even k if n==1 (mod 4) else over odd k.

S_3(n) = 2* Sum_{0<=k<=[(n-3)/6]} 2*((n-3)/4-1/2*k)!/(k!*((n-3)/4-3/2*k)!). The sum is over odd k if n==1 (mod 4), else over even k.

Where [m] is floor(m).

G.f. x +14*x^3 +2*x*(1 +2*x^2 +3*x^4 -2*x^6 -4*x^8 -2*x^10)/ (1-x^4-x^6)^2. (Includes zeros for even floor widths).- R. J. Mathar, Apr 10 2017

a(n) = 2*(A228577(n-1)+A228577(n+1))+4*(A182097(n-2)+A182097(n-1)), n>1. - R. J. Mathar, Apr 10 2017