[seqfan] Floret's Equation
Creighton Kenneth Dement
creighton.k.dement at mail.uni-oldenburg.de
Tue Jan 12 18:02:43 CET 2010
"Floret's Equation" (named in honor of my former teacher) is a property
I've spent countless hours trying to prove in recent weeks, having
wondered about it for several years before that. I think I finally have
Let X be any floretion. Define z = tes(X).
The equation is:
3*X^4 = 12*z*X^3 +
(6*tes(X^2) - 24*z^2)*X^2 +
(32*z^3 - 24*z*tes(X^2) + 4*tes(X^3))*X +
(3*tes(X^4) - 16*z*tes(X^3) + 48*z^2*tes(X^2) - 6*(tes(X^2))^2 – 2*z^4)*ee
where ee is the unit (see http://www.mrob.com/pub/math/seq-floretion.html
for a review of terms).
If tes(X) = 0, the equation simplifies to
3*X^4 = 6*tes(X^2)*X^2 + 4*tes(X^3)*X + 3(tes(X^4)-2*(tes(X^2))^2)ee
The above equation in its general form should be applicable to
all 4th order linear recurrence sequences in the OEIS
and should combine with the following previously discussed properties:
1. tesseq(X + ee) is the binomial transform of the sequence tesseq(X)
2. For p prime, p divides tes(X^p) - (tes(X))^p
3. tesseq(X*Y) = tesseq(Y*X) for all X,Y
(and many more) to produce some, hopefully, elegant proofs. I understand
that many properties may be just as easy to prove with other methods.
However, the idea of being able to prove something using floretions alone
seems very appealing to me. For example, previously, matrices were used to
prove that if X is a floretion, then the sequence tesseq(X) = (tes(X),
tes(X^2), tes(X^3), ...) satisfies a 4th order linear recurrence. That I
always found too hard to prove direclty- but it's now a trivial
consequence of Floret's equation.
To demonstrate one of the simplest possible cases,
assume X = A'i + B'j + C'k is a "pure quaternion". Then tes(X) = 0 and the
formula 3*X^4 = 6tes(X^2)*X^2 + 4tes(X^3)*X + 3[ tes(X^4)-2(tes(X^2))^2
For Y = D'i + E'j + F'k and Z = G'i + H'j + I'k, tes(X*Y*Z) is easily
shown to be the determinant of the matrix [A,B,C;D,E,F;G,H,I]. Since the
determinat of [A,B,C;A,B,C;A,B,C] is zero, it follows that tes(X^3) = 0.
3*X^4 = 6tes(X^2)*X^2 + 3[ tes(X^4)-2(tes(X^2))^2 ]ee
= -6*(A^2 + B^2 + C^2)*X^2 - 3(2A^2*B^2 + 2*B^2*C^2 + 2*C^2*A^2 +
A^4 + B^4 + C^4)ee
gives the general linear recurrence relation for any "pure" quaternion
(note the factor of 3 divides off).
If we didn't know the formula for the inverse of a quaternion, we need
only multiply both sides of the above equation by X^(-1). If we wanted to
know what recurrence relation is satisfied by tesseq(X^(-1)), we need only
multiply both sides by (X^(-1))^4.
In the past, floretions were used to find sequences "related" to a given
sequence (satisfying a linear 4th order recurrence) and to set up
relationships between these without using generating functions/partial
fraction decomposition, etc. Moreover, it was easy to assume that if a
certain floretion generated a batch of, say, eight linear recurrence
sequences, half which were already listed in OEIS with the "nice" label,
then it would be safe to assume that the other 4 sequences should also (at
the very least) be included in the OEIS.
The difference now is that the equation allows one to go backwards, i.e.
starting with a a sequence of numbers satisfying a given 4th order linear
recurrence relation, find the set of floretions which generate that
sequence. Therefore, it should be quite possible to extend superseeker's
guess program to conjecture a floretion in the same way that it now
suggests a generating function.
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