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

Creighton Kenneth Dement creighton.k.dement at mail.uni-oldenburg.de
Thu Jul 30 16:00:12 CEST 2009

```Dear Seqfans,

Stage 1, stage 2, ... below give the terms of the sequence. I can't find
the first 5 terms (i.e. first 5 stages) so it would be nice to have some
help or new ideas. Of course, instead of considering triplets in stage 1,
it is clear that quadruplets, etc. could also be considered resulting in a
triangle of such sequences.

Stage 1:
Find the number of quaternions so that
a0*a1*a2 = 1 (writing "1" for the unit quaternion)

Examples are
['i, 'i, -1]
['i, 'j, -'k]
...

Let T (stage 1) be the set of all such quaternion triplets.
My count gives |T| = 64. This can be taken as the first element of the
sequence.

Stage 2:

This stage consists of finding the number of pairs (A,B) in TxT
such that

(A[0]*B[0])*(A[1]*B[1])*(A[2]*B[2]) = 1

For ex.
A = ['i, 'i, -1]
B = [-'i, 'j, 'k]

We have
A[0]*A[1]*A[2] = 'i * 'i * (-1) = 1
B[0]*B[1]*B[2] = -'i * 'j * 'k = 1
(A[0]*B[0])*(A[1]*B[1])*(A[2]*B[2]) = ee * 'k * (-'k) = 1

|T2| = 2176

Stage 3:
Find the number of triplets (A,B,C) in TxTxT
such that

(A[0]*B[0]*C[0])*(A[1]*B[1]*C[1])*(A[2]*B[2]*C[2]) = 1

The pattern for defining higher stages should be now be obvious.

This seems to be a difficult sequence to calculate- at least
using my initial brute force methods.

For quaterions I get (64, 2176, ...)
For floretions I get (1024, ...)

Is anyone able to extend or independently verify these sequences?

Sincerely,
Creighton

```