# [seqfan] Re: A193376 Tabl = 20 existing sequences

israel at math.ubc.ca israel at math.ubc.ca
Mon Jul 25 19:18:16 CEST 2011

```Yes, with z+1 tiles on an n x 1 grid (with n >= z), either there is a tile
(of any of the k colours) on the first spot, followed by any configuration
on the remaining (n-z) x 1 grid, or the first spot is vacant, followed by
any configuration on the remaining (n-1) x 1. So T(n,k) = T(n-1,k) +
k*T(n-z,k), with T(n,k) = 1 for n=0,1,...,z-1. The solution is T(n,k) =
sum_r r^(-n-1)/(1 + z k r^(z-1)) where the sum is over the roots of the
polynomial k x^z + x - 1.

Robert Israel                                israel at math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel
University of British Columbia            Vancouver, BC, Canada

On Jul 25 2011, Ron Hardin wrote:

>experimentally zX1 tiles give a table with the corresponding
>a(n)=a(n-1)+k*a(n-z) column recurrences, taking a quick spot preview.
>
> rhhardin at mindspring.com
>rhhardin at att.net (either)
>
>
>
>----- Original Message ----
>> From: Ron Hardin <rhhardin at att.net>
>> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
>> Sent: Mon, July 25, 2011 7:33:50 AM
>> Subject: [seqfan] Re: A193376 Tabl = 20 existing sequences
>>
>> The same problem with 3X1 tiles apparently gives a n-1 n-3 recurrence
>> (b-file
>
>> still in progress), needs a formula  too:
>
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```