Quaternions and 2nd order recurrence relations

[Creighton Dement]

Hello / Bonjour, 

I should have also noted that in the claim, as it stands

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

tes[(Ex)^n] = -(x_0)tes[(Ex)^(n-1)] - (x_1)(x_2)tes[(Ex)^(n-2)] with
4tes[Ex] = -x_0

************

not all initial values are possible since 4tes[(Ex)^2] = -2(x_1)(x_2) +
(x_0)^2. 

However, this may not be much of a problem. As stated before, if x has
the more general form
x = = (x_0)'i + (x_1)'j + (x_2)'k + (x_3)'i + (x_4)'j + (x_5)'k 

then it appears that the claim generalizes to 

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


I haven't yet checked, but surely there will be more freedom to choose
the 2nd term of the sequence in the latter case. In my opinion, the best
news appears to be that only minor changes to the proof will be
necessary, since - where before (in the "pure quaternion" case)  use was
made of the properties:
 
Define F = .25('ii' + 'jj' + 'kk' + e) and G = .25('i + i' + 'jk' +
kj'), 
 Then E = F + G and the following relations can be easily derived 
(Fx)^2 = 0; GxG = (-x_0)G ; tes[(Gx)^n] = (-x_0)^n  ;  tes[GxFx] =
-(x_1)(x_2) ;
tes[Fx] = 0.

in the general case we still have (Fx)^2 = 0. The rest becomes GxG =
(-x_0 - x_3)G,  tes[(Gx)^2] = (-x_0 - x_3)^n, etc. It follows that we
can reuse all of the above reasoning for (a + b)^n. 

     
Returning to the formula for tes[(a + b)^n], I  previously listed

tes[(a + b)^7] = tes[b^7] + 7tes[b^6a] + 7tes{b^4aba] + 7tes[b^3ab^2a] +
7tes[b^2ababa] 
Check: Lucas(7) = 29 = 1 + 7 + 7 + 7 + 7, ; 

tes[(a + b)^6] = tes[b^6] + 6tes[b^5a] + 6tes[b^3aba] + 3tes[b^2ab^2a] +
2tes[ababab]
Check: Lucas(6) = 18 = 1 + 6 + 6 + 3 + 2

tes[(a + b)^5] = tes[b^5] + 5tes[b^4a] + 5tes[b^2aba]   

tes[(a + b)^4] = tes[b^4] + 4tes[b^3a] + 2tes[baba]   
 
tes[(a + b)^3)] = tes[b^3] + 3tes[b^2a] 


The coefficients of the terms tes[(a + b)^2n] appear more complicated
than those of tes[(a + b)^(2n+1)]. 

For the odd case, I conjecture that

tes[(a + b)^(2n+1)] = tes[b^(2n+1) + (2n+1)tes[b^(2n+1)a] +
(2n+1)(tes...) + (2n+1)(tes...) + 

Since the total number of terms in tes[(a + b)^n] was seen to be
Lucas(n), if the conjecture is correct a direct corollary would be that 

2n+1 divides Lucas(2n+1) - 1 = A000204(2n+1) - 1. This appears to check
(I'm sure this is already known, though I didn't seem to find it
directly under A000204 at first glance).  

A final remark: it is perhaps interesting to see how the formula tes[(a
+ b)^7] (as an example) breaks down in the case of x = (x_0)'i + (x_1)'j
+ (x_2)'k

tes[(a + b)^7] = tes[b^7] + 7tes[b^6a] + 7tes{b^4aba] + 7tes[b^3ab^2a] +
7tes[b^2ababa] 
= -(x_0)^7 + 7*(x_0)^5*(x_1)*(x_2) - 14*(x_0)^3*(x_1)^2*(x_2)^2 +
7*(x_0)*(x_1)^3*(x_2)^3 

For x = -'i + -'j + 'k (x_0 = -1, x_1 = -1, x_2 = 1), the formula
returns 29 = Lucas(7) . 

For x = -2'i + -'j + 'k (x_0 = -1, x_1 = -1, x_2 = 1), the formula
returns 478 =  2*A001333(8)  (not A001333(7), since OEIS gives an
additional initial term ). 


The rest of the problem, at least for odd n, seems to boil down to
figuring out how many b's are in the various terms 
of tes[(a + b)^(2n+1)], any ideas?

Many thanks, 
Creighton