This a follow up question/puzzle from an earlier question here How does rounding affect Fibonacci-ish sequences?
Explanation
For this new question, one needs only to understand a particular transformation
$$ f\big( \begin{bmatrix} a\\b\\c\\d \end{bmatrix}\big) = \begin{bmatrix} a+\text{floor}(\frac{d}{2})\\a\\b\\c \end{bmatrix} $$ So for example, this sequence is generated by repeatedly applying the transformation $$ \begin{bmatrix}5\\0\\0\\0\end{bmatrix}\rightarrow \begin{bmatrix}5\\5\\0\\0\end{bmatrix}\rightarrow \begin{bmatrix}5\\5\\5\\0\end{bmatrix}\rightarrow \begin{bmatrix}5\\5\\5\\5\end{bmatrix}\rightarrow \begin{bmatrix}7\\5\\5\\5\end{bmatrix}\rightarrow \begin{bmatrix}9\\7\\5\\5\end{bmatrix}\rightarrow \begin{bmatrix}11\\9\\7\\5\end{bmatrix}\rightarrow \begin{bmatrix}13\\11\\9\\7\end{bmatrix}\rightarrow \begin{bmatrix}16\\13\\11\\9\end{bmatrix}\rightarrow \ ... $$ Because the function rounds down $d$, it's very hard to predict the exact values after many iterations. Interestingly though -- certain initial values eventually converge to the same sequence
For example if we continue the sequence above $$ \cdots\ \rightarrow \begin{bmatrix}16\\13\\11\\9\end{bmatrix}\rightarrow \begin{bmatrix}20\\16\\13\\11\end{bmatrix}\rightarrow \begin{bmatrix}25\\20\\16\\13\end{bmatrix}\rightarrow \begin{bmatrix}31\\25\\20\\16\end{bmatrix}\rightarrow \begin{bmatrix}39\\31\\25\\20\end{bmatrix}\rightarrow \begin{bmatrix}49\\39\\31\\25\end{bmatrix}\rightarrow \ \cdots $$ And compare it to the sequence of another initial value $$ \begin{bmatrix}9\\0\\0\\0\end{bmatrix}\rightarrow \begin{bmatrix}9\\9\\0\\0\end{bmatrix}\rightarrow \ \cdots \ \rightarrow \begin{bmatrix}25\\21\\17\\13\end{bmatrix}\rightarrow \begin{bmatrix}31\\25\\21\\17\end{bmatrix}\rightarrow \begin{bmatrix}39\\31\\25\\21\end{bmatrix}\rightarrow \begin{bmatrix}49\\39\\31\\25\end{bmatrix}\rightarrow \ \cdots $$ The sequences hit the same point. More specifically, we will say that $$ f^{13}\big(\begin{bmatrix}5\\0\\0\\0\end{bmatrix}\big) = f^{10}\big(\begin{bmatrix}9\\0\\0\\0\end{bmatrix}\big) $$ This means the sequence for $5$ and the sequence for $9$ will be the same as each other if continued further.
I've been calling numbers like this "cow-equivalent" because this iteration models the population growth of animals in Minecraft – when bred every 5 minutes by the player. A related-but-simpler transformation is also discussed here Exponential growth of cow populations in Minecraft
Here we'll use the "$\equiv$" symbol to denote cow-equivalency. So $$a\equiv b \quad\text{if and only if there exists}\quad m, n \in \mathbb{N}$$ $$\text{ such that}$$ $$f^m(a\cdot\vec{e_1})=f^n(b\cdot\vec{e_1})$$
Questions
A quick computer search shows that there are triplets of cow-equivalent numbers like $$34\equiv 259\equiv 406$$ because $$f^{35}(\begin{bmatrix}34\\0\\0\\0\end{bmatrix})= f^{26}(\begin{bmatrix}259\\0\\0\\0\end{bmatrix})= f^{24}(\begin{bmatrix}406\\0\\0\\0\end{bmatrix})= \begin{bmatrix}113171\\90268\\72000\\57429\end{bmatrix}$$ Thus the primary question here is
- Does a quadruplet of cow-equivalent numbers exist?
Other interesting/related questions
- Are all numbers potentially cow-equivalent?
- Is the triplet above – or any other cow-equivalence set – complete?
- Is there a deterministic test one can apply for cow-equivalence?