The binomial numbers, the Rule 60 celluar automata, and the Sierpinski triangle

Consider the pattern of the parity of the binomial numbers. Color the even numbers white and odd numbers black. You get a Sierpinski Triangle.

We will prove this using spacetime-geometric reasoning. It would help a lot if you have done some geometry of special relativity.

Define the parity function $f(t, n) = C(t, n) \mod 2$, defined on $\mathbb{N}\times\mathbb{Z}$.

We can consider $f(t, n)$ as encoding a 1-d celluar automata, with $t \ge 0, n \in \mathbb{Z}$. Then we have the evolution rule:

$$f(t+1, n) = f(t, n) + f(t, n-1) \mod 2$$

This is just the Rule 60 celluar automata. 

Then it's easy to prove the following:

Light speed = 1

The state of any cell $n$ at time $t$ is determined by any constant-time-surface across its past lightcone. This is essentially a discrete version of physicist's idea of "past Cauchy surface": The state of any point in spacetime is determined by any surface that cuts through its past lightcone. 

(Figure probably from Roger Penrose)

Lemma: for any cell, if all cells in any of its past Cauchy surface are dead, then that cell is also dead. 
Proof: induct on $\Delta t$, how far in the past that Cauchy surface is.

Now begin the world with a single living cell at $n=0, t=0$. For cell at $n < 0$ or $n > t$, their past-Cauchy-surfaces at $t=0$ are all dead, so they are also dead.

Now induct to show Sierpinski. 

At time $2^k - 1$, by induction, there is a solid row from $n = 0$ to $n = 2^k - 1$. So at time $2^k$, there is a big die-off, leaving only living cells at $n = 0$ and $2^k$.

At time $t + 2^k$, consider a cell at $0 \le n \le 2^k - 1$. Its relation to its past Cauchy surface at time $2^k$ are exactly the same as the relation of cell $n$ at time $t$ with its past Cauchy surface at time $0$. Same for a cell at $n + 2^k$.

QED