In a partition of the subsets of {1,2,…,n} into symmetric chains, how many chains have only k subset in them?

In a partition of the subsets of {1,2,…,n} into symmetric chains, how many chains have only 1 subset in them? How many chains have only 2 subsets in them? How about k?

I tried to solve it using recursive rule like (n,k)=(n-1,k-1)+(n-1,k+1), and got (n,k)=sum_{i=0 to n-1}C(n-1,i)(1,k-(n-1)+2i). However, my solution gets wrong result in some case since it doesn't consider about the (1,2) under (x,0) in the binary tree.
I've got a solution from classmates(as shown in the picture) but we don't understand where does 2|A1|+l-1=n come from.

Any idea about this problem? Thanks!

#Combinatorics

Easy Math Editor

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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$