[seqfan] Re: A "hard" quaternion-based sequence
benoit.jubin at gmail.com
Thu Jul 30 16:40:26 CEST 2009
Are you considering only quaternions with integer coefficients,
"Lipschitz integers", or quaternions with half-integer coefficients,
"Hurwitz integers" (see
What are the corresponding sequences when you replace triplets by
couples and/or Lipschitz integers by Gaussian integers?
On Thu, Jul 30, 2009 at 4:00 PM, Creighton Kenneth
Dement<creighton.k.dement at mail.uni-oldenburg.de> wrote:
> 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:
> Start with a triplet of quaternions [a0, a1, a2]
> 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
> Stage 2:
> This stage consists of finding the number of pairs (A,B) in TxT
> such that
> (A*B)*(A*B)*(A*B) = 1
> For ex.
> A = ['i, 'i, -1]
> B = [-'i, 'j, 'k]
> We have
> A*A*A = 'i * 'i * (-1) = 1
> B*B*B = -'i * 'j * 'k = 1
> (A*B)*(A*B)*(A*B) = ee * 'k * (-'k) = 1
> |T2| = 2176
> Stage 3:
> Find the number of triplets (A,B,C) in TxTxT
> such that
> (A*B*C)*(A*B*C)*(A*B*C) = 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?
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