# [seqfan] Re: A "hard" quaternion-based sequence

Creighton Kenneth Dement creighton.k.dement at mail.uni-oldenburg.de
Fri Jul 31 02:21:17 CEST 2009

```In my last reply I wrote:

> Here is a reformulation of the problem without refering to "stages":
>
> To calculate A(n) (offset 1), count the number of ways to place
> quaternions {1,-1,i,-i,j,-j,k,-k} on the spaces of an nx3 grid such that
>
> 1. the product of the 3 quaternions in each grid row is equal to 1 (taken
> from left to right)
>
> and
>
> 2. a*b*c = 1 where a, b, c are the products of the n quaternions in the
> first, second and third columns respectively (taken from top to bottom).

Now that I think about it, I should not have used the wording "to place"
because it might cause confusion.

To calculate A(n) (offset 1), count the number of "valid quaternion
configurations" on an nx3 grid where a valid configuration is one in
which:

1. a single quaternion from the set {1,-1,i,-i,j,-j,k,-k} is on each space
of the grid.

2. the product of the 3 quaternions in each grid row is equal to 1 (taken
from left to right)

and

3. a*b*c = 1 where a, b, c are the products of the n quaternions in the
first, second and
third columns respectively (taken from top to bottom).

```