Find the number of possible sequences $(a_1,a_2,...,a_n)$ in terms of $n$, with the following conditions:

(i)$a_1=a_2=0$

(ii)$a_1\leq a_2\leq ...\leq a_n$

(iii)$a_i\leq (i-2)$

## Sequence count

### Sequence count

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- Phlembac Adib Hasan
**Posts:**1016**Joined:**Tue Nov 22, 2011 7:49 pm**Location:**127.0.0.1-
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### Re: Sequence count

Start the sequence from $a_0$ and toss out the first term. So the third condition turns into $a_i\leq i-1$. Now consider the following bijection: take any $A$ dominated sequence of length $n$ and set $a_i$ as the number of $B$'s preceding $i$-th $A$. Since the sequence is $A$ dominating, there will be at most $i-1$ $B$'s before $i$-th $A$. And obviously the derived sequence will be non-decreasing.

Also, from a given sequence of $a_i$s, it is possible to reverse this move and make an $A$ dominated sequence. Thus the number of sequences is $C_n=\frac{1}{n+1}\binom {2n}{n}$

And here are two examples of the bijection, for clarification:

$AABABB\to 0,0,1$

$AABBAB\to 0,0,2$

Also, from a given sequence of $a_i$s, it is possible to reverse this move and make an $A$ dominated sequence. Thus the number of sequences is $C_n=\frac{1}{n+1}\binom {2n}{n}$

And here are two examples of the bijection, for clarification:

$AABABB\to 0,0,1$

$AABBAB\to 0,0,2$

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- asif e elahi
**Posts:**183**Joined:**Mon Aug 05, 2013 12:36 pm**Location:**Sylhet,Bangladesh

### Re: Sequence count

We consider a $n\times n$ board and color the upper $a_{i}$ cells black of column $i-1$ for $i=2,3......n$. Now the board is divided into $2$ parts. This makes a path from the upper left vertex to the lower right vertex of length $2n-2$.The $3rd$ condition implies that the black squares don't intersect the main diagonal containing the upper left vertex. It's easy to see that this is a catalan path.(Just rotate the board $45^{\circ}$ in counterclockwise direction). Again any such catalan path indicates a sequence of length $n-1$. So we have got a bijection between the sequences and the catalan paths of length $2n-2$.

The number of sequences= number of catalan paths of length $2n-2$ = $\dfrac{1}{n}\binom{2n-2}{n-1}$.

The number of sequences= number of catalan paths of length $2n-2$ = $\dfrac{1}{n}\binom{2n-2}{n-1}$.