Representing Collatz sequences with Pascal's simplices modulo 2? In this binomial coefficient pdf (from a Russian summer school?), I found a curious identity (page 2, solution on page 10) pointed out by the authors, which ties the modulo 2 Pascal's triangle (first image below) to numbers of the form: $3^{a_1} + 2\cdot 3^{a_2} + \cdots + 2^n \cdot 3^{a_n}$ where $a_1 > a_2 > \cdots > a_n$. Specifically these $a_j$ come from the binary representation of a row index of this triangle. This is very similar to the series one obtains from iterates of the Collatz function ($2^{a_1} + 3\cdot 2^{a_2} + \cdots + 3^n \cdot 2^{a_n}$). Except that we obtain every power of two and only some powers of three. Whereas for the Collatz sums the reverse is true. The identity arises due to each triangle in Pascal's triangle mod 2 being followed by two 'child' triangles of equal size. Hence the sum of the first 2^n rows has 3^n '1's.
So I thought of whether this correspondence could be made closer, and wagered that Pascal's pyramid (next picture) modulo 2 (final picture) is the logical choice. Here each pyramid is followed by three child pyramids. Hence one obtains a correspondence of the form: $4^{a_1} + 3\cdot 4^{a_2} + \cdots + 3^n \cdot 4^{a_n}$ by a similar argument as in the linked pdf.
My question is whether there is an approach to turn these 4s into 2s (to therefore create a representation of the Collatz sequences vs simplical side length) either thru algebra or by using a related approach? The problem is that its hard to conceive of a simplex or fractal that has the required properties: branching factor three, with a doubling factor between some object and its child objects. I guess there might be a proof that no such construction could exist at least in the stated way. Thanks for reading!



 A: You can represent $n$ steps of the conjecture as an $n-$simplex by:

*

*Starting with an arrow moving in any direction of your choice, and

*extending your vertex in the current direction one step for each division by $2$, and

*turning at right angles to every existing vertex whenever you encounter a $3x+1$.

This is hard to visualise for n greater than 3, but if you study "Hilbert's hotel", you will find ways to contruct bijections from the integers into ever-greater dimensional shapes. An appropriate bijection from $\omega^{<\omega}\to\Bbb N$ can be thought of as giving you a way of representing the up to infinitely-dimensional simplex in a way you can visualise.
One way I have found useful visualising this is via functions from $\omega^{<\omega}$ which is the set of finitely long integer strings into $\Bbb N$.
You get one such example if you count the number of successive ones and zeroes in any given integer's binary representation.  So e.g. $111011_2=59$ has three ones, one zero and then two ones, so its sequence in $\omega^{<\omega}$ is $3,1,2$ so in this function $3,1,2\mapsto59$
It's not as visual as your $2$- and $3$-dimensional simplices, but it's the same principle and the difficulty visualising infinite-dimensionality is one of the things that makes the problem hard.
