Pascal's triangle contains huge numbers. I tried to "tame" it (make the numbers smaller) by "killing" the largest numbers, i.e. the central binomial coefficients: $2,6,20,70,252,924,\dots$.
That is, as I built the rows (each number being the sum of the two numbers above it), whenever a row had a centre number (except the $1$ at the very top), I made the centre number $0$ instead of adding the two numbers above it.
So instead of:
$$
1\\
1\quad 1\\
1\quad \color{red}{2}\quad 1\\
1\quad 3\quad 3\quad 1\\
1\quad 4\quad \color{red}{6}\quad 4\quad 1\\
1\quad 5\quad 10\quad 10\quad 5\quad 1\\
1\quad 6\quad 15\quad \color{red}{20} \quad 15\quad 6\quad 1\\
1\quad 7\quad 21\quad 35\quad 35\quad 21\quad 7\quad 1\\
\cdots
$$
I got:
$$
1\\
1\quad 1\\
1\quad \color{red}{0}\quad 1\\
1\quad 1\quad 1\quad 1\\
1\quad 2\quad \color{red}{0}\quad 2\quad 1\\
1\quad 3\quad 2\quad 2\quad 3\quad 1\\
1\quad 4\quad 5\quad \color{red}{0} \quad 5\quad 4\quad 1\\
1\quad 5\quad 9\quad 5\quad 5\quad 9\quad 5\quad 1\\
\cdots
$$
I thought I had gotten rid of those nasty central binomial coefficients. But then, curious about the properties of my "tamed" version of Pascal's number, I decided to calculate the sum of each row, and I got:
$$1,2,\color{red}{2},4,\color{red}{6},12,\color{red}{20},40,\color{red}{70},140,\color{red}{252},504,\color{red}{924},\dots$$
Those central binomial coefficents found their way back!
Is there an explanation for this?
Context: Recently I've been interested in Pascal's triangle, asking questions here, here and here.
Best Answer
Let $g_n(x)$ be the generating function of row $n$ of your triangle:
$g_0(x) = 1$, $g_1(x) = 1+x$, $g_2(x) = 1 + x^2$, etc.
It seems [empirically: I don't have a proof] these satisfy the recurrence
$$4 x (1+x) n g_n(x) - 4 x n g_{n + 1}(x) - (n+3)(1+x) g_{n+2}(x) + (n + 3) g_{n+3}(x) = 0 $$
The sum of row $n$ is then $g_n(1)$, which satisfies
$$ 8 n g_n(1) - 4 n g_{n+1}(1) - 2 (n+3) g_{n+2}(1) + (n+3) g_{n+3}(1) = 0 $$
and the solution of this with initial conditions $g_0(1) = 1$, $g_1(1) = 2$, $g_2(1) = 2$ is
$$ g_{2n}(1) = {2n \choose n},\ g_{2n+1}(1) = 2 {2n \choose n} $$
EDIT: OK, here's a proof. Let $h_k(x)$ be $g_k(x)$ with the signs of the coefficients of $x^i$ for $i > k/2$ changed. Thus $$ \eqalign{h_1(x) &= 1-x \cr h_2(x) &= 1 - x^2\cr h_3(x) &= 1 + x - x^2 - x^3\cr h_4(x) &= 1 + 2 x - 2 x^3 - x^4\cr} $$ The coefficient of $x^{k}$ in $h_n(x)$ is $-$ the coefficient of $x^{n-k}$ there. It can then be seen that the coefficients of $h_n$ form the entries of a "Pascal's triangle" starting with the row $(1,-1)$; there's no need to kill the central coefficients because the symmetry makes them automatically $0$. Thus $h_n(x) = (1-x) (1+x)^{n-1}$. So the coefficient of $x^k$ in $g_n(x)$ is ${n-1 \choose k} - {n-1 \choose k-1}$ for $k < n/2$ (with ${n-1 \choose -1} = 0$) and ${n-1 \choose k-1} - {n-1 \choose k}$ for $k > n/2$. Adding these up, taking advantage of telescoping, we have $$ \eqalign{g_{2n}(1) &= 2 \sum_{k=0}^{n-1} \left({2n-1 \choose k} - {2n-1 \choose k-1}\right) = 2{2n-1 \choose n-1} = {2n \choose n}\cr g_{2n+1}(1) &= 2 \sum_{k=0}^{n} \left({2n \choose k} - {2n \choose k-1}\right) = 2{2n \choose n}\cr}$$