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Problem: $(\text{a})$ Let $k, n \geq 1$. Find the number of sequences $\emptyset=S_0, S_1, \ldots, S_k$ of subsets of $[n]$ if for all $1 \leq i \leq k$ we have either $(\text{i}) \ S_{i-1} \subset S_i$ and $\left|S_i-S_{i-1}\right|=1$, or $(\text{ii}) \ S_i \subset S_{i-1}$ and $\left|S_{i-1}-S_i\right|=1$.

$(\text{b})$ Suppose that we add the additional condition that $S_k=\emptyset$. Show that now the number $f_k(n)$ of sequences is given by \begin{align*} f_k(n)=\frac{1}{2^n} \sum_{i=0}^n\binom{n}{i}(n-2 i)^k. \end{align*} Note that $f_k(n)=0$ if $k$ is odd.

For $(\text{a})$, let the number of elements in $S_i$ be $s$. There are two ways to obtain $S_{i+1}$. The first is to choose one element from the remaining $n-s$ elements and add it to $S_i$; the second is to remove one element from the $s$ elements. Therefore, there are $n$ choices at each step, so the number of sequences that satisfy the condition is $n^k$.

However, I don't have many ideas for $(\text{b})$. I believe it is related to Dyck paths or Catalan numbers. The difference from Dyck paths is that here, each step has a weight associated with it. When $k$ is odd, it is easy to understand why $f_k(n)=0$, because an odd number of steps will always change the parity of the number of elements. Perhaps this problem can be solved by using a recurrence relation similar to $D_n=\sum_{a=1}^n D_{a-1} D_{n-a}$ and generating function.

Scorpio
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1 Answers1

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Given that OP mentioned generating functions, consider $Q(x_1, x_2, \ldots x_n ) = ( x_1 + x_2 + \ldots + x_n)^k$.
We want to find the sum of coefficients where each power is even. (WHY?)

Recall that in 1-variable, if we wanted to find the sum of the even coefficients of $f(x) = \sum a_i x^i$, we simply take $ \frac{ f(1) + f(-1) } { 2 }$.
In $n-$variables, we have a similar formula, where we take the sum over all $Q(\pm 1, \pm 1, \ldots \pm 1)$ and weight it accordingly by $ \frac{1}{2^n}$.

Applying it to this scenario, we obtain

$$\frac{ \sum {n\choose i } (n-2i)^k } { 2^n}. $$

Furthermore, to show that $f_k(n) = 0$ if $k$ is odd, observe that $ { n \choose i } ( n-2i)^k = (-1)^k { n \choose n-i} (n-2(n-i) ) ^ k $. So if $k$ is odd, the terms pair up and cancel out.
(Note that when $n$ is even, we have to deal with the $i = n/2$ case that doesn't pair up, but then $(n-2i) = 0 $ so it still works out.)

Notes

  1. To explain the WHY?, create a bijection between a valid sequence of subsets, and a string of length $k$ of choices from $x_i$.
  2. For (a), the equivalent problem in this reskinning is to find the sum of coefficients of all terms of $Q(x_1, x_2, \ldots, x_n ) = ( x_1 + x_2 + \ldots + x_n)^k$, which is just $Q(1, 1, \ldots, 1) = n^k$, agreeing with OP's claim.
  3. The $n-$variables count is arguably where the magic happens, so make sure you understand it. I'd be surprised if it was covered in the first chapter, but you said "Volume 1 chapter" so it could be a much later chapter.
Calvin Lin
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