[seqfan] Re: A "hard" quaternion-based sequence
Edwin Clark
eclark at math.usf.edu
Thu Jul 30 22:43:18 CEST 2009
On Thu, 30 Jul 2009, Benoît Jubin wrote:
> Are you considering only quaternions with integer coefficients,
> "Lipschitz integers", or quaternions with half-integer coefficients,
> "Hurwitz integers" (see
> http://en.wikipedia.org/wiki/Hurwitz_quaternion)?
> What are the corresponding sequences when you replace triplets by
> couples and/or Lipschitz integers by Gaussian integers?
>
If one takes the Lipschitz integers, then Creighton's problem becomes (in
general): find the number a(n) of sequences of length n:
(x[1],x[2],..,x[n]) such that x[i] is in the quaternion group Q =
{1,-1,i,-i,j,-j,k,-k} and the product x[1]x[2]...x[n] = 1. In this case
a little thought shows that a(n) = 8^(n-1)---the first n-1 elements
can be arbitrary and the n-th term is the inverse of the product
of the first n-1.
I seem to recall that given a finite group G there is a name for sequences
of elements of the group whose product is 1, but I cannot recall what it
is. In any case it seems that for a group of order k the sequence would be
simply a(n) = k^(n-1).
> Benoit
>
> 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
>> 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
>>
>>
>>
>> _______________________________________________
>>
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>
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