Counting monotonically increasing sequence that $a(i) < i$

Question

How many sequences $(a(1),\dots,a(n))$ of integers are there such that $0 \leq a(1) \leq a(2)\le \dots \leq a(n)$ and $a(i) < i$ for each $i\in \{1,\dots,n\}$?

My guess is it's related to Catalan numbers.

Answer 1

Hint

The Catalan numbers enumerate the set of lattice paths from $(0,0)$ to $(n,n)$, where each step is one unit up or right, and which stay below the line $y=x$. Try to find a bijection between such paths and the sequences you are counting. The conditions "$a(i)<i$" and "the path must be below the line $y=x$" are strikingly similar$\dots$

Further hint

When $n=2$, there are two possible lattice paths:

   |      _|
_ _|    _|  

and there are two possible sequences:

$$ (0,0) \qquad and\qquad (0,1) $$

Notice that the path on the left stays pretty low to the ground (until the very end), while the path on the right is gradually increasing. Similarly, the sequence $(0,0)$ stays low, while $(0,1)$ gradually increases.

When $n=3$, there are $5$ paths:

     |       |     _|       |       _|
     |      _|    |      _ _|     _|
_ _ _|  _ _|   _ _|    _|       _|

sorted in terms of how quickly they increase. Similarly, there are $5$ sequences:

$$ (0,0,0)\quad (0,0,1)\quad(0,0,2)\quad(0,1,1)\quad(0,1,2) $$ Try to see show the general trend of the paths corresponds to the general trend of the sequences, and use that to construct a bijection that works for any $n$.

Solution

We prove this using a bijection. Given a monotonic sequence $(a(1),\dots,a(n))$ with $a(i)<i$, we will product a unique lattice path from $(0,0)$ to $(n,n)$ which stays at or below the line $y=x$, such that all lattice paths are generated. Since it is well-known that the number of such lattice paths is $C_n=\frac1{n+1}\binom{2n}n$, this proves that there are $C_n$ monotonic sequences with $a(i)<i$.

Specifically, given $(a(1),\dots,a(n))$, let $P$ be the lattice path such that, for each fixed $x\in \{0,1,\dots,n-1\}$, the maximum value of $y$ such that $(x,y)$ is a point on $P$ is $a(x)+1$. This uniquely defines $P$.