Analytically, why does Sierpinski's triangle appear in Pascal's triangle?

Question

It is well-known that if you isolate the even from the odd terms in Pascal's triangle, you obtain an 'approximation' of Sierpinski’s triangle. Better stated, from user Glorfindel in an answer to this post (edits for correctness):

Pascal's triangle

If one takes Pascal's triangle with $2^n$ rows and colors the even numbers white, and the odd numbers black, the result is an approximation to the Sierpinski triangle. More precisely, the limit as n approaches infinity of this parity-colored $2^n$-row Pascal triangle is the Sierpinski triangle.

My question: what is the analytic/formal proof that this is the case? I can find many resources saying that this happens, but none explain why.

I presume the explanation is because every certain number of rows, the binomial coefficient run through every value is odd, and It follows from Pascal’s identity that even numbers would then appear in a descending pattern resulting in triangles; however, I don’t think this fully explains it.

Answer 1

The following argument is not a proof, but it may be converted to a proof. We will write $0,1$ for the binomial coefficients taken modulo $2$, and use tautologically the two colors $0$ and $1$ to color the coefficients.

The coefficients are considered as elements of the field $\Bbb F_2$. Most of the time we use only its additive structure, i.e. the underlying abelian group structure $\Bbb Z/2$, but at some point I want to use the Frobenius morphism in characteristic two.

The line number zero in the Pascal triangle is simply $$1$$ but it may be useful to think it as $$ \dots\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ \dots $$ or as $$ \dots\ 0\ 0\ 0\ 0\ 0\ 0\ \color{red}{\blacktriangle}\ 0\ 0\ 0\ 0\ 0\ 0\ \dots $$ It corresponds to $(1+x)^0$. Recall that the $n$-th row corresponds to the coefficients of $(1+x)^n$.

For $n=2$, let $N=2^2$ and build the Pascal triangle over the field $\Bbb F_2$ up to row number $2^n-1 = N-1 = 3$:

0:            1
1:          1   1
2:        1   0   1
3:      1   1   1   1

The last line is a line of ones since

• it corresponds to the coefficients of $(1 + x)^3$ modulo 2, and
• the next line (line number $N = 2^n = 4$ in this case) corresponds to the coefficients of $(1+x)^{2^n} = 1^{2^n} + x^{2^n}$ modulo 2, so there survive only two extremal 1 values (leftmost and rightmost) and all other coefficients are zero modulo 2.

This works for general $n$. Notice that if we know line $N$ then there is only one possible configuration of this line for line $N-1$ to be a line of only 1s.

Now, put a white $\nabla$ over the only 0 entry and three $\blacktriangle$ over the 1 entries over the Pascal triangle drawn above, so that it looks like $$\blacktriangle\\ \blacktriangle\nabla\blacktriangle $$ and compare it with the second triangle in Sierpinski triangle evolution.

As mentioned before, the next line (line number $2^2=4$) is a line with just two 1 entries and the rest are $0$. Here is a picture:

0:              1
1:            1   1
2:          1   0   1
3:        1   1   1   1
4:      1   0   0   0   1 

And it is useful to think about it like: $$ \color{red}{\blacktriangle}\\ 1\ 1\\ 1\ 0\ 1\\ 1\ 1\ 1\ 1\\ \color{red}{\blacktriangle}\ 0\ 0\ 0\ \color{red}{\blacktriangle} $$

Now we play again the "game of life", where a 1 bit gives life to the diagonally-placed entries in the next row. If two 1 entries "interfere" in the same diagonally-placed entry in the next row, the result is a 0 (the sum modulo 2).

Then the game works for the two $\color{red}{\blacktriangle}$ entries in the last row, as it worked for the upper $\color{red}{\blacktriangle}$, of course until they interfere. We know exactly where the interference is: one step before the usage of the next Frobenius power, $(1+x)^8 \equiv 1+x^8 \mod 2$; the step previous to this row is at $(1+x)^7 \equiv 1 + x + x^2 + \dots + x^7 \mod 2$.

We have the following situation, where the last row corresponds to the coefficients of $(1+x)^7$ modulo 2:

0:                  1
1:                1   1
2:              1   0   1
3:            1   1   1   1
4:          1   0   0   0   1 
5:        1   1   0   0   1   1
6:      1   0   1   0   1   0   1
7:    1   1   1   1   1   1   1   1

Now place a white $\nabla$ on the zero entries in the middle, so that the upper bar of $\nabla$ corresponds to the zero entries in the row $1\ 0\ 0\ 0\ 1$, and compare with the next picture in loc. cit. - we have reached the next stage. The next line is again a $1\ 0\ 0\ \dots\ 0\ 0\ 1$ line, and its picture is: $$ \color{red}{\blacktriangle}\\ 1\ 1\\ 1\ 0\ 1\\ 1\ 1\ 1\ 1\\ 1\ 0\ 0\ 0\ 1\\ 1\ 1\ 0\ 0\ 1\ 1\\ 1\ 0\ 1\ 0\ 1\ 0\ 1\\ 1\ 1\ 1\ 1\ 1\ 1\ 1\ 1\\ \color{red}{\blacktriangle}\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ \color{red}{\blacktriangle} $$ The extremal $1$ values in the last row will suffer now the same copy+paste operation from the upper vertex, and we get a bigger $\nabla$ in the middle, we see the upper line of the white $\nabla$ in the last row, and the copy+paste game goes on...

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The inductive construction above shows a self-similarity of the "picture" from one step ($N=2^n$) to the next one ($2N=2^{n+1}$), and it is natural to expect a self-similarity in the limit (after a rigorous definition of the limit.) Note that the construction above is parallel to the construction of the Sierpinsky triangle.)

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If it is really needed for a special purpose, I may try to become analytic, and establish an analytic formula for points, using their $2$-adic coordinates in the barycentric coordinates of the points inside the Sierpinski triangle - in the level $N$ and in the limi (It is hard to write it down, and the proof of the convergence will not be intuitive.)