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This is related to the Collatz function which can be written $$T(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ 3n+1 & \text{if } n\equiv 1 \pmod{2} .\end{cases}$$

All I did was arrange $T(n)$ into a triangle like this:

$$\text{Collatz}_\triangle = \begin{matrix} &&&&&&&&&166\ldots\\ &&&&&&&&136&27\ldots\\ &&&&&&&18&22&160\ldots\\ &&&&&&14&106&130&26\ldots\\ &&&&&64&82&17&21&154\ldots \\ &&&&46&10&13&100&124&\color{red}{25}\ldots \\ &&&5&7&58&76&\color{red}{16}&20&148\ldots \\ &&3&28&40&\color{red}{9}&12&94&118&24\ldots \\ &10&16&\color{red}{4}&6&52&70&15&19&142\ldots \\ 4&\color{red}{1}&2&22&34&8&11&88&112&23\ldots \end{matrix}$$

Consider knight moves as in a game of chess on $\text{Collatz}_\triangle$ starting at $1$ - so this would be over two and then up 1. I get the sequence $1,4,9,16,25,36,\ldots$. Which are the squares. It also appears that the squares appear on odd numbered columns in the triangle. There are other similar sequences. For example starting at $22$ go up one and over two then you get the sequence $22,52,94,148,\ldots$ which appears to be A163433. You can just as easily start at the number $5$ and again go over two and up one then you $5,10,17,26,37,50,\ldots$ or A002522 (or numbers of the form $n^2+1$).

Question: Are there a combinatorial interpretation as to why knight moves on a triangular arrangement of the first iteration of the Collatz function give these sequences?

The motivation here is pure curiosity.

Note: I am not saying that the squares can only occur in this manner. Check out $16$ and $64$.

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  • $\begingroup$ How come some numbers are repeated? $\endgroup$
    – scott
    Dec 23, 2016 at 21:18

2 Answers 2

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These numbers are given by the sequence $a_n=T\left(2n^2\right)$, $n\geq 1$, which explains why they're the squares. This isn't really related to the Collatz sequence, but rather to the arrangement of the natural numbers in the triangle.

$$\text{Simple}_\triangle = \begin{matrix} &&&&&&&&&55\ldots\\ &&&&&&&&45&54\ldots\\ &&&&&&&36&44&53\ldots\\ &&&&&&28&35&43&52\ldots\\ &&&&&21&27&34&42&51\ldots \\ &&&&15&20&26&33&41&\color{red}{50}\ldots \\ &&&10&14&19&25&\color{red}{32}&40&49\ldots \\ &&6&9&13&\color{red}{18}&24&31&39&48\ldots \\ &3&5&\color{red}{8}&12&17&23&30&38&47\ldots \\ 1&\color{red}{2}&4&7&11&16&22&29&37&46\ldots \end{matrix}$$

The numbers in red are $2n^2$.

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Note that the entry in your triangle in column $i$ and row $j$ (from the bottom) is $$ T\left(i(i-1)/2+j\right). $$ Your knight move sequence starting at positions $(2,1)$ consists of positions $(2k,k)$ for $k\geq1$, so the entries are $$ T(2k(2k-1)/2+k)=T(2k^2)=k^2. $$ In general, note that your knight move doesn't change the parity of the argument to $T$, since $$ \left[(i+2)(i+1)/2+j+1\right]-\left[i(i-1)/2+j\right] =2i+2. $$ So along one of these sequences, we will only use one of the rules $T(n)=n/2$ or $T(n)=3n+1$. Thus the result will be a quadratic function of $k$.

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