
> 
> Update: I've marked this question as answered. If you are thinking "What the heck are floretions?", go right to the answer provided by the Grinch. I definitely should have added clearer information on the algebra as opposed to "hiding" it in the links. Yes, of course I will continue to study the space. The whole idea is that using a "very, very concrete basis" makes defining algorithms and operators (a main one being the basic cyclic transform $i \rightarrow j$, $j \rightarrow k$, $k \rightarrow i$ as it's well-defined for a floretion of any order) that lead to [this](https://mathoverflow.net/questions/224376/quaternions-ellipse-effect) (in the case of quaternions) or [this](https://www.youtube.com/watch?v=GKL4Lkw-rU0) (2nd order floretions) that much easier to setup and examine. Finding ways to make the elements of the algebra "come alive" and do strange things is and was my main motivation. There is no way I know to do that just by looking at sequences of integers generated by $(X, X^2, ...)$ for some floretion $X$. In that case we just generate "boring" linear recurrence relations of some order. However, precicely because a sequence of powers of $X$ is the simplest case of an "algorithm" I can imagine, I interpreted the question "What is Floret's Equation for a 4th Order floretion?" as being extremely important and fundamental!
> 
> 
> Note on the above mentioned cyclic transform and 4th order floretions. It is still an open question to me whether $Y$, $Z$ exist such that for all $X:$ $cyc(X) = Y \cdot X \cdot Z$.
> 
> 
> 


I'll spend a bit of time setting up the details and motivation of the actual question: "What is Floret's equation for a 4th order floretion?" by looking in particular at Floret's equation for 2nd order floretions. The reason: 


* it's without a doubt one of the most important questions I can currently think of on the topic
* most, I assume, are not very familiar with the topic at all


Regarding the first point: I'm a hobby mathematician so any knowledge outside my "comfort zone" of floretions is very limited. On the second point: 
As there is (as of now) no current tag for this topic, here are relevant posts and a very brief overview:


1. floretion order 0: real numbers
2. floretion order 1: quaterions $X = A
 i + B j + C k + D e$
3. floretion order 2: [the "original"
floretions](http://doku.floretions.fumba.eu/doku.php?id=secondeorder)
4. floretion order 3: not covered here (but embedded in 4th order case)
5. [floretion order 4](http://doku.floretions.fumba.eu/doku.php?id=fourthorder) Note on downloading software in R: ver 1.01 has several improvements and will be uploaded in the next few days.


Relevant posts:


* [What is the motivation and purpose of the Floretion Group?](https://mathoverflow.net/questions/210514/what-is-the-motivation-and-purpose-of-the-floretion-group)
* [Quaternions: Ellipse Effect](https://mathoverflow.net/questions/224376/quaternions-ellipse-effect)


A 2nd order floretion can be written in the form 


$X = Aie + Bje + Cke + Dei + Eej + Fek + Gii + Hjj + Ikk + Jij + Kik + Lji + Mjk + Nki + Okj + Pee$ 


where the coefficients can also be given via projection operators 


$A = \eta\_{ie}(X)$, $B = \eta\_{je}(X)$, $C = \eta\_{ke}(X)$, $D = \eta\_{ei}(X)$, $E = \eta\_{ej}(X)$, $F = \eta\_{ek}(X)$, $G = \eta\_{ii}(X)$ $(\cdots)$, $P = \eta\_{ee}(X)$ 


Note $\eta\_{ee}$ also often written as $tes$.


Start with a member of the quaternion algebra over the reals 
$X = Ai + Bj + Ck + Pe$. Then 


$X^2 = (Ai + Bj + Ck + Pe)(Ai + Bj + Ck + Pe) = (-A^2 - B^2 - C^2 + P^2)e + 2P(Ai + Bj + Ck)$


This is Floret's equation for **1st order floretions**:



> 
> $X^2 = \eta\_{e}(X^2)e + 2\eta\_{e}(X) (Ai + Bj + Ck)$
> 
> 
> 


Moving to 2nd order floretions, a previous goal was to prove that a sequence of (half/quarter) integers of the form 


$( \eta\_{ie}(X), \eta\_{ie}(X^2), \cdots, \eta\_{ie}(X^n)) = \text{iebaseseq}(X)$ 


$( \eta\_{ee}(X), \eta\_{ee}(X^2), \cdots, \eta\_{ee}(X^n)) = \text{tesseq}(X)$ 


for an appropriately chosen floretion $X$ was indeed given by a 4th order linear recurrence relation and that this did not need to be additionally *conjectured*. To give just one example [A104934](https://oeis.org/A104934). 


Now, we can come up with many basic "sequence identities" just by defining $\text{vesseq}(X) = \text{iebaseseq}(X) + 
 \text{jebaseseq}(X) + \cdots +
 \text{tesseq}(X)$
and looking for floretions X which generate Fibonacci / Pell / etc. numbers. Brief examples of identities obtained via 2nd order floretions:


* |2\*Fib(n) - 9\*Fib(n+1)| = 4\*A000032(n) + A000032(n+1)
* A000032(n) = Fib(n+3) - 2\*Fib(n)
* (Pell numbers) Define c(2n) = -A001108(n), c(2n+1) = -A001108(n+1) and d(2n) = d(2n+1) = A001652(n), then ((-1)^n)\*(c(n) + d(n)) = a(n). [Proof given by Max Alekseyev.]


For me this was a subtle point in the beginning: why is it still a "conjecture" when observing relationships among integer sequences which result from the above "vesseq" identity- it's a definition so there should be nothing to prove, right? The point is that nowhere in the identity does it state that we are dealing with 4th order linear recurrence relations. 
Knowing this means these and other relationships which come about by simple manipulation of terms in various identities do not need to be stated as conjectures.


So, six years ago I wrote a symbolic multiplier for floretions in Python. I conjectured that if sequences generated by second order floretions were 4th order linear recurrences, then comparing all coeffcients for $X$, $X^2$, $X^3$, $X^4$ should lead to something.


Floret's equation for **2nd order floretions**. Any $X$ satisfies, with $q = \eta\_{ee}(X)$, $r = \eta\_{ee}(X^2)$, $s = \eta\_{ee}(X^3), t = \eta\_{ee}(X^4)$:



> 
> $3X^4 = 12qX^{3} + (6r - 24q^2)X^{2} + (32q^3 - 24qr + 4s)X + (3t - 16qs + 48q^2r - 6r^2 – 32q^4)ee$
> 
> 
> 


The formula contributes to the above discussion in several ways:


1. proves $\text{tesseq}(X)$, $\text{iebaseseq}(X)$ etc. are 4th order recurrence relations. To get $\text{tes}(X^n)$ for higher n, just multiply both sides by a power of $X$. Note it as been shown that $\text{tesseq}(X+ee)$ is the binomial transform of $\text{tesseq}(X)$ where $\text{tesseq}(X-ee)$ is the corresponding inverse transform. Also, for any $X$ and $Y$, $\text{tesseq}(X \cdot Y) = \text{tesseq}(Y \cdot X)$.
2. gives a formula for the inverse of $X$, if it exists. Multiply by $X^{-1}$ and rearrange, dividing by the term $3t - 16qs + 48q^2r - 6r^2 – 32q^4$ (thus a necessary condition for invertibility is that this term is not zero)
3. proves that if $\text{tes}(X) = \text{tes}(X^2) = \text{tes}(X^3) = \text{tes}(X^4) = 0$, then $X^4 = 0$ (thus all higher terms of $\text{tesseq} = 0$).
4. provides a basis to start with a 4th order linear recurrence relation and pick a corresponding floretion which satisfies this relation.
5. leads to new symmetries relating Fibonacci / Pell numbers.


Example for point 5:


The first thing to notice is that projecting onto the unit axis by taking $\eta\_{ee}$ of both sides does not produce any new information (other than as a check that the formula is correct) as all terms cancel out. However, we can, of course, project onto any other axis. 


To get concrete, define 


$Y = \frac{1}{4}(ie + ei + ii+ jj + kk + kj + jk + ee)$, $Z = \alpha ie -je +ke$ and 
$X = Y \cdot Z$. Then choosing $\alpha = -1$ for the coefficient of $ie$ in $Z$ leads to 


$\text{tesseq}(X) = \frac{1}{4}(1,3,4,7,11,18,29,47,76,123,199,322,521,843,1364,2207, \cdots)$
and 
$\text{iebaseseq}(X) = \frac{-1}{4}(1,0,1,1,2,3,5,8,13,21,34,55,89,144,233,377, \cdots)$


By Floret's equation, we know the above sequences really do correspond to Lucas / Fibonacci numbers respectively and don't just appear to resemble these sequences. If we now rename $\text{lucseq}(X) = \text{tesseq}(X)$ and $\text{fibseq}(X) = \text{iebaseseq}(X)$, applying the projection operator $\eta\_{ie}$ to both sides of the equation results in



```
3*fibseq[n] = 

12*lucseq[1]*fibseq[n-1] +

(6*lucseq[2] - 24*(lucseq[1])^2)*fibseq[n-2] +

(32*(lucseq[1])^3 - 24*lucseq[1]*lucseq[2] + 4*lucseq[3])*fibseq[n-3] +

(3*lucseq[4] - 16*lucseq[1]*lucseq[3] + 48*(lucseq[1])^2*lucseq[2] - 6*(lucseq[2])^2 - 32*(lucseq[1])^4)*fibseq[n-4]

```

The neat thing is that changing to $\alpha = -2$ in $Z$ leads to Pell numbers (see below), but does absolutely nothing to change the structure of the above equation, which for $\alpha = -1$ returns the fibonacci sequence as a 4th order recurrence with coefficients given by the first four numbers of the lucas sequence. 


$\text{tesseq}(X) = \frac{1}{2}(1,3,7,17,41,99,239,577,1393,3363,8119,19601,47321,114243,275807,665857 \cdots)$
and 
$\text{iebaseseq}(X) = \frac{-1}{2}(1,0,1,2,5,12,29,70,169,408,985,2378,5741,13860,33461,80782 \cdots)$


So, with that long introduction, I formulate the actual question. Given a 4th order floretion
$X = \alpha\_1 ieEE + \alpha\_2 jeEE + \alpha\_3 keEE + \alpha\_4 eiEE + (\cdots) + \alpha\_{256}eeEE$, what is Floret's equation in this case? 


My conjecture is that the order of linear occurences generated by $X, X^2, X^3, ...$ corresponds to the square root of the number of positive base vectors involved ($eeEE$, $eiEE, \cdots $). As there are 256 such elements, the highest order linear recurrence would be 16. In the case of 3rd order floretions, there are 64 positive base elements so the highest order linear recurrence is presumably 8. In the future, I *might* be able calculate Floret's equation for the 3rd order case on my own. However, I feel the 4th order case is too hard for me alone without outside help. Since I'm now actively investigating 4th order floretions in R, I would love to have such an equation at my disposal. 



4th Order Floretions: Floret's Equation
$\def\F{\mathbb{F}}$
$\def\C{\mathbf{C}}$
$\def\H{\mathbb{H}}$


**Note: this answer is not really intended for the OP, but rather for regular mathoverflow readers who must be thinking "what the heck are the floretions?" and who doesn't want to bother reading up about them. Maybe this question will be closed, but I'm kinda really bored right now so I thought I would write this.**


**What are the floretions (of order $2$)?**
The (order two) floretions $\F^{(2)}$ are a rank $16$ associative algebra. The coefficient ring is (in the linked presentation of their definition) somewhat ambiguous.
However, for the purposes of algebraic relations between floretions and their powers, it suffices, using the Lefschetz principle, to work over the complex numbers. Then, by the Artin-Wedderburn theorem,
it follows that $\F^{(2)}\_{\C}$ is a direct sum of matrix algebras. In this case, it turns out that


$$\F^{(2)}\_{\C} \simeq M\_4(\C).$$


**What is Floret's equation for the $2$nd order floretions?** 


By the Cayley-Hamilton Theorem, any element of $M\_4(\C)$ and thus of $\F^{(2)}\_{\C}$ satisfies its degree four characteristic polynomial. Again, by the Lefschetz principle, the same holds for $\F^{(2)}$ over any ring.
The coefficient $\eta\_{ee}(X)$ for $X \in \F^{(2)}$ is $1/4$ times the trace of $X$. Hence Floret's equation is a reflection of the fact, first observed by Newton, that the coefficients of a degree $n$ (in this case $n = 4$) polynomial can be written as polynomials in the power basis of the roots.


**What is Floret's equation for the $4$th order floretions?** 
From what I gather, the floretions $\F^{(n)}$ are an associative algebra of dimension $4^n$. Moreover, the construction (as far as I can tell) is inductive starting with the quaternions $\H$. Indeed, I suspect (although I didn't even find a formal definition) that


$$\F^{(n)} \simeq \H \otimes \H \otimes \ldots \H.$$


In this case, it certainly follows that $\F^{(n)}\_{\C} \simeq M\_{2^n}(\C)$.
In which case, Floret's equation for the $n$th order floretions will again be given by
Cayley hamilton. I do not wish to look at <http://fumba.eu/sitelayout/multMelt.php>
too closely, but it is quite possible that the trace map 


$$T: \F^{(n)} \rightarrow R$$


for degree $n$ floretions over $R$ will be identified with 


$$2^{-n} \eta:=2^{-n} \eta\_{e \ldots (\text{$n$-times}) \ldots e}.$$


In this case, the corresponding equation will be, over rings $R$ in which $2^n!$ is invertible, the polynomial (but not obviously polynomial) relation:


$$X^{2^n} \exp \left(- \sum\_{m=1}^{\infty} \frac{2^{n} \eta(X^m)}{m X^m} \right) = 0,$$
or
$$X^{2^n} = 2^n \eta(X) X^{2^{n} - 1} + 2^{n-1} (\eta(X^2) - 2^n \eta(X)^2) X^{2^{n} - 2} $$
$$ + \frac{2^{n-1}}{3} (2^{2n} \eta(X)^3 - 3 \cdot 2^n \eta(X) \eta(X^2)
 + 2 \eta(X^3))X^{2^{n} - 3} + \ldots $$


**Are the floretions a new object in mathematics that I should be interested in?**
Not really, it's just a very very concrete example of an associate algebra given in terms of a 
very explicit basis. It's hard to imagine that one could find anything by explicit computation
that isn't already well understood in the more general classical theory of the Brauer group, or (more simply) the theory of matrices.


