Mathematrec

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

The sum of, say, the first four numbers in row \(n-1\) 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.