Question: I'm trying to find the number of sequences of integers $0=a\_1\leq a\_2\leq\ldots\leq a\_n$ such that $a\_k<k$ for every $1\leq k\leq n$.

I know for $n=3$, the sequences are $ 000, 001, 002, 011,012$, so the number of sequences is $5$, but I don't know how to approach this further.

Any tips?

Answer: 
The solution to your problem are the ultra-famous Catalan Numbers. Have a look at Wikipedia.
I don't know if there's an easy formula, but you can get the answer in terms of a recurrence which you can evaluate by hand for small $n$ or use a fast computer program for large $n$. Note that a valid sequence for $n$ gives a valid sequence for $n+1$ as long as $a\_{n+1} \geq a\_n$ and $a\_{n+1} < n+1$. So if you let $c\_{n,m}$ be the number of sequences of length $n$ that end in $m$, you get 

$$c\_{n+1,m} = \sum\_{j=0}^m c\_{n,j}$$

and then your total count is $c\_n = \sum\_{m=0}^{n-1}c\_{n,m}$