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 (in the case of quaternions) or this (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:

- floretion order 0: real numbers
- floretion order 1: quaterions $X = A i + B j + C k + D e$
- floretion order 2: the "original" floretions
- floretion order 3: not covered here (but embedded in 4th order case)
- floretion order 4 Note on downloading software in R: ver 1.01 has several improvements and will be uploaded in the next few days.

Relevant posts:

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.

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:

- 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 $Y$, $\text{tesseq}(X \cdot Y) = \text{tesseq}(Y \cdot X)$.
- 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)
- 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 $\text{tesseq} = 0$).
- provides a basis to start with a 4th order linear recurrence relation and pick a corresponding floretion which satisfies this relation.
- 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.