Inspired by Pascal, I put on some shackles and a thorny belt. Inspiration came pouring in, and I thought of the following triangle:
$ \begin{array}{rcccccccccc} & & & & & 1\\\ & & & & 1 & & 1\\\ & & & 1 & & \frac{1}{2} & & 1\\\ & & 1 & & \frac{2}{3} & & \frac{2}{3} & & 1\\\ & 1 & & \frac{3}{5} & & \frac{3}{4} & & \frac{3}{5} & & 1\\\ 1 & & \frac{5}{8} & & \frac{20}{27} & & \frac{20}{27} & & \frac{5}{8} & & 1\\\ & ... & & & &... & & & & ... & \end{array} $
Let's call the corresponding entry ${n \choose k}$ because there is clearly no danger of confusion. The construction rule is very simple. Instead of having ${n+1 \choose k} = {n \choose k-1} + {n \choose k},$ as usual, we have ${n+1 \choose k} = \frac{1}{{n \choose k-1} + {n \choose k}}.$
It's easy to see by induction that all entries of the triangle lie between $1/2$ and $1$. Also, for fixed $k$, ${n \choose k}$ converges to a limit $C_k$. For example, $C_1=1/\phi$, where $\phi$ is the golden ratio (this should be obvious! think of Fibonacci numbers...). We can determine easily $C_{k+1}$ in terms of $C_k$ by taking the limit in the construction rule, which yields $C_k=(C_{k-1}+C_k)^{-1}$. In particular, all of the $C_k$'s are algebraic numbers.
I would like to know if anybody here can prove interesting properties of this triangle, or of the numbers $C_k$.
Enjoy! I'm going to take off the shackles and belt now.