# [seqfan] Re: Tatami

zbi74583.boat at orange.zero.jp zbi74583.boat at orange.zero.jp
Sat Apr 30 04:10:52 CEST 2016

>     Hi,Seqfans
>
>     The case of 5 x n room, n is odd, number of tiling is represented as
follows
>
>     S(n)=S_1(n)+S_5(n)+S_3(n)
>
>     S_1(n)=2*(Sum_{0<=k<=[(n-1)/6]}
> ((n+3)/4-1/2*k)*((n-1)/4-1/2*k)!/(k!*((n-1)/4-3/2*k)!)
>     If n=1 Mod 4 then k is even else k is odd
>
>     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)!)
>     If n=1 Mod 4 then k is even else k is odd
>
>     S_3(n)=2*(Sum_{0<=k<=[(n-3)/6]} 2*((n-3)/4-1/2*k)!/(k!*((n-3)/4-3/2*k)!)
>     If n=1 Mod 4 then k is odd  else k is even
>
>     S(n) : 10,8,18,24,32,52
>     OFFSET  5
>
>     Numbers computed with the formula and numbers with blute force are the
same
>     So, it seem to be correct
>
>     Richard
>     Could you confirm them and compute more term?
>
>
>
>     Yasutoshi
>
>

Let us compute S_1(n) the case n=17

n=1  Mod 4  so  k is even
0<=k<=[(17-1)/6] , [m] means Floor(m)
So
k=0 or 2
S_1(n)=2*(5*4!/(0!*4!)+4*3!/(2!*1!)=34