Pure quaternions and 2nd order recurrence relations

[Creighton Dement]

> Dear Seqfans,
> 
> In a previous post (see below) I mention that, without a minimum
> amount of theory to back it up, I see no direct way to conclude, for
> ex., that a floretion sequence given by FAMP as (1, 2, 3, 4, 5, 6,
> 7...)  is the sequence of naturals - regardless of how many terms are
> calculated.
> 
> There now appears to be a proof that if F = .25('ii' + 'jj' + 'kk' +
> e) and x is any floretion, then tesseq[Fx] really is a power sequence
> (as opposed to a sequence that just looks a lot like it might be one).
> 
> The results are given under:
> http://www.crowdog.de/Flointseq.pdf
> 
> 
> My goal over the past few weeks has been to show that if E = .25('i +
> 'i + 'ii' + 'jj' + 'kk' + 'jk' + 'kj' +  e) and x is any floretion,
> then tesseq[Ex] satisfies a second order linear recurrence relation
> with constant coefficients. Note that E = .25('i + i' + 'jk' + 'kj) +
> F  - hopefully it will be possible to use the above results in one way
> or another.  If you have any ideas for me (including those on how to
> simplify the notation given as much as possible), I would enjoy
> hearing them.
> 

When x, above, is a "pure quaternion" of the form x = x_0 'i + x_1 'j +
x_2 'k, 
the conjecture can be formulated more explicitely:

Claim:
 If x = x_0 'i + x_1 'j + x_2 'k , and E = .25('i + 'i + 'ii' + 'jj' +
'kk' + 'jk' + 'kj' +  e) 

then 

tes[(Ex)^n] = (x_0)*tes[(Ex)^{n-1}] + (x_1)*(x_2)*tes[(Ex)^{n-2}], n > 2
with tes[(Ex)] = -x_0.

I was delighted to have stumbled upon the above as part of searching for
a proof for more general x. When x is more general, the formula becomes
more complicated  but remains a second order recurrence relation. For
ex.  if x = x_0 'i + x_1 'j + x_2 'k + x_3 i' + x_4 j' + x_5 k' then
data collected shows the claim would become: 

tes[(Ex)^n] = (x_0 + x_3)*tes[(Ex)^{n-1}] + ((x_1)*(x_2) - (x_1)*(x_5) +
(x_4)*(x_5) - (x_2)*(x_4)*tes[(Ex)^{n-2}], n > 2  with tes[(Ex)] = -x_0
- x_3

(proof pending but no longer *appears* to be very difficult )
 
The above more or less immediately leads one to the following table:

Note that adding 3 sequences from the same row gives the zero sequence
and that the factor "2" keeps flip-flopping from left to right only to
disappear and then reappear.  Any clues as to why and / or what it
means? For mor e details (and a daring analogy), see 
Note 2.0.3 and the previous proposition at
http://www.crowdog.de/Flointseq.pdf 
 
 $ 0 & (0, 1, 0, 1, ) & (1, 0, 1, 0, 1) & (-1, -1, -1, -1, -1)$ \\

 $ 1 & (1, 3, 4, 7, 11, 18) & ( 1, -1, 0, -1, -1, -2, -3, -5) & 2*(-1,
-1, -2, -3, -5, -8, -13)$ \\

 $ 2 & (1, 3, 7, 17, 41, 99) & 2*(0, -1, -2, -5, -12, -29) &  (-1, -1,
-3, -7, -17, -41)$ \\

 $ 3 & (3, 11, 36, 119, 393) & (-1, -9, -28, -93, -307) &  2*(-1, -1,
-4, -13, -43, -142)$\\

 $ 4 & ( 2, 9, 38, 161, 682, 2889) & (-1, -8, -33, -140, -593) &   (-1,
-1, -5, -21, -89, -377)$ \\

$ 5 & ( 5, 27, 140, 727, 3775) & (-3, -25, -128, -665, -3453) &  2*(-1,
-1, -6, -31, -161, -836)$ \\

$ 6 & (3, 19, 117, 721, 4443, 27379) & 2*( -1, -9, -55, -339, -2089,
-12873) &   (-1, -1, -7, -43, -265, -1633)$ \\

$ 7 & (7, 51, 364, 2599, 18557) & (  -5, -49, -348, -2485, -17743) &  
2*(-1, -1, -8, -57, -407, -2906)$ \\