I’ve now watched that 3Blue1Brown video and it explains how the Moser circle problem’s solution for $l n$ points is just the sum of the first 5 numbers in the $l n^{th}$ row of Pascal’s triangle, or $l R(n,5) = \sum_{i=0}^{4}{{n}\choose{i}}$, and how this explains why the first five numbers as well as the tenth one are powers of 2, since the $l n^{th}$ row of Pascal’s triangle sums to $l 2^n$. It doesn’t talk about more general p2 polynomials, but it’s immediately apparent the sequence generated by $l p2(m)$ is the sums of the first $l m$ numbers in each row, or $l R(n,m) = \sum_{i=0}^{m-1}{{n}\choose{i}}$. Hence entry $l 2m-1$ (counting from 0) is also a power of 2.

This isn’t too surprising; the finite differences approach to generating the sequences seems a lot like Pascal’s triangle anyway. Essentially we’d gotten the sequences using:

$$R(n,m) = 1\quad m=1$$

$$R(n,m) = 1\quad n = 0$$

$$R(n,m) = R(n-1,m-1)+R(n-1,m)\quad m >1, n > 0$$

But $l R(n,m) = \sum_{i=0}^{m-1}{{n}\choose{i}}$ gives the first two of these. As for the third, it’s just saying the sum of the first $l m-1$ entries and the first $l m$ entries in row $l n-1$ of Pascal’s triangle is equal to the sum of the first $l m$ entries in row $l n$, and you can see that visually:

The sum of, say, the first four numbers in row $l n-1$ is

$$\left({n-1\choose{0}} + {n-1\choose{1}}\right), \left({n-1\choose{2}} + {n-1\choose{3}}\right) = {n\choose{1}} + {n\choose{3}}$$

while the sum of the first five numbers in the same row is

$$\left(0 + {n-1\choose{0}}\right) + \left({n-1\choose{1}} + {n-1\choose{2}}\right) + \left({n-1\choose{3}} + {n-1\choose{4}}\right) = {n\choose{0}} + {n\choose{2}} + {n\choose{4}}$$

And then it’s no surprise the Sierpinski triangle rears its head, because it’s already embedded in Pascal’s triangle .