Binomial transforms, floretions, and a divisors property

[Creighton Dement]

Dear Seqfans, 

The binomial transform of a sequence ((a(n)) is defined as  b(n)=SUM
C(n,k)a(k), k=0..n

Let x = x_0'i + x_1'j + x_2'k} + x_3i' + x_4j' + x_5k' + x_6'ii' +
x_7'jj' + x_8'kk' + x_9'ij' + x_{10}'ik' + x_{11}'ji' + x_{12}'jk' +
x_{13}'ki' + x_{14}'kj' + x_{15}e be a member of the floretion algebra
over the reals such that a(n) = tes[x^(n+1)] is an integer for all n >=
0. "tes" is defined as tes(x) = x_{15}. Further define tesseq[x] =
(tes[x^(1)], tes[x^(2)], tes[x^(3)], ) and x^0 = e. 

Conjecture: If a' = (1, a(0), a(1), ) = (1, tes[x], tes[x^2], ) is an
integer sequence then  b' = (1, b(0), b(1), b(2), ) = (1, tes[x+e],
tes[(x+e)^2, ) is the binomial transform of a'. 


Proof. tes is a linear mapping. Moreover, x and the unit e commute.
Thus,   
SUM C(n,k)a'(k), k=0..n = SUM C(n,k)tes[x^(k)], k=0..n 
= tes[SUM C(n,k)x^k], k=0..n 
= tes[(x+e)^k], k=0..n 
= b', q.e.d.


This is actually somthing which has confused me for a long time: Why was
tesseq[x+1] sometimes but not always the binomial transform of
tesseq[x]? I figured the reason must have lain really deep and just
tried my best to work around it. It now seems it was all just a matter
of how I'd defined the initial term
of the sequence tesseq[x] = (tes[x], tes[x^2], tes[x^3], tes[x^4],
tes[x^5], ) *not* 
tesseq[x] = (tes[x^0], tes[x^1], tes[x^2]. )


One more idea I had today which could be -if it should work-
interesting.

A conjecture I made in a previous post to the 
seqfan list (I believe under the heading "Mersenne numbers and the
divisors property" ) was that if (a(n)) satisfies a linear 2nd order
recurrence relation, then if s divides a(m) and s divides a(n) then s
also divides a(2n-m)  for n > m (I also posted this same question to
sci.math.)  My idea: try to reformulate the question so that it asks
about a property of floretions, instead.   

What would a "floretion proof" of this property look like? 

Here is a quick list of some of the conjectures made:
Define E =  + .25'i + .25i' + .25'ii' + .25'jj' + .25'kk' + .25'jk' +
.25'kj' + .25e
Use FAMP to show that E^2 = F = .25'ii' + .25'jj' + .25'kk' + .25e and
F^2 = F. 

Conjecture (MAIN): If tesseq[x] is an integer sequence then tesseq[x]
satisfies a linear 4th order recurrence relation or less. 
 
Conjecture II: If tesseq[x] is an integer sequence then tesseq[E*x]
satisfies a 
linear 2nd order recurrence relation. It appears there exists at least
one second element D (which is not related to E by some cyclic
transformation) with this same property.   

Proposition:
If x and y are floretions, then tesseq[x*y] = tesseq[y*x] (a proof of
this simple but important property, in my opinion, has been found). 

Question is: how "big" is the set of 2nd order sequences generated the
sequences tesseq[E*x] (or tesseq[D*x]? In any case, it will apparently
be possible to show that the set is not trivial.

  
A question related to (and/or equivilant to- depending on how big the
set of sequences is) the one above in terms of floretions is:

If s | tes[(E*x)^m] and s | tes[(E*x)^n] does it also follow that s |
tes[(E*x)^(2*m - n)]? This, in turn, could make it possible to use some
of the conjectures, above. To demonstrate the point with a wild example,
we might find that "all we really have left to show" is: If y is a
floretion such that s | tes[E*y] then s | tes[E*y^2] (or some similar
property) - showing these would presumably be much easier than the
original problem. Of course, we would still have all the conjectures
from above to prove.   


Sincerely, 
Creighton