3
$\begingroup$

Let $p_1^{a_1}p_2^{a_2}\cdots$ denotes the integer partition of $n$, i.e. $a_1p_1+a_2p_2+\cdots=n$. Or equivalently $m_1+m_2+\cdots=n$. It is known that the number of partitions of set $\{x_1,x_2,\cdots,x_n\}$ corresponding to $p_1^{a_1}p_2^{a_2}\cdots$ (i.e. there are $a_i$ length-$p_i$ parts in the partition) is $$\frac{n!}{(p_1!)^{a_1}\cdots(p_l!)^{a_l}a_1!\cdots a_l!}$$

Now, after obtain partitions the set $\{x_1,x_2,\cdots,x_n\}$, we consider assign values from $\{1,2,\cdots,L\}$ to the elements, such that if elements are in the same part, then the value should be the same; if the element are in different part, then the value should be different. Thus there are in total $\binom{L}{\sum_{i=1}^l a_i}(\sum_{i=1}^l a_i)!\frac{n!}{(p_1!)^{a_1}\cdots(p_l!)^{a_l}a_1!\cdots a_l!}$ valued-assigned set partitions.

For example consider $n=3$ and partition $\{x_1,x_2\}\{x_3\}$, we want to assign values from $\{1,\cdots,L\}$ to them such that $x_1=x_2$ and $x_1\neq x_3$, $x_2\neq x_3$. Thus there are $\binom{L}{2}2!$ ways of assignments.

Question 1 considers adding additional constraint on set partitions and compute how many valued-assigned set partitions are there. Specifically, constraint that $\sum a_i\binom{p_i}{2}=N$ where $N\in\{0,1,2,\cdots,\lfloor \frac{n^2-2\sqrt{2} n^{3/2}-2cn^{1+\epsilon}}{2} \rfloor\}$ where $C,0<\epsilon<\frac{1}{2}$ are constants. Equivalently, $\sum a_ip_i^2=2N+n$ or $\sum m_i^2=2N+n$.

The number of valued-assigned set partitions satisfying this constraint is

$$\sum_{\substack{n^1\neq p_1^{a_1} \cdots p_l^{a_l}\vdash n, \\ \sum_{i=1}^la_i\leq L},\\ \sum_{i=1}^l \binom{p_i}{2}a_i=N(n)}\binom{L}{\sum_{i=1}^l a_i}(\sum_{i=1}^l a_i)!\frac{n!}{(p_1!)^{a_1}\cdots(p_l!)^{a_l}a_1!\cdots a_l!}$$

How to find the asymptotic/non-asymptotic upper bound on this quantity?

Note: $L$ as increasing function of $n$ but it is ok if we consider $L$ to be a constant that does not depend on $n$ if it make problem simpler.

If this problem is hard to solve, I would be grateful to hear the reason, the thoughts.. Thanks in advance for any help!

$\endgroup$
9
  • $\begingroup$ could the down-voted be explained so that I can improve my question? $\endgroup$
    – tony
    Commented Sep 14, 2023 at 13:02
  • $\begingroup$ perhaps they thought your question was too broad in scope. fwiw this looks like you might want to search the term 'multinomial' $\endgroup$ Commented Sep 14, 2023 at 13:12
  • $\begingroup$ @SeanSanford Thanks. but the expression in my post is exact, and I have no idea how to make the post less broad.. $\endgroup$
    – tony
    Commented Sep 14, 2023 at 13:42
  • $\begingroup$ The downvote not being mine, I'm just guessing, but it seems to me that the problem with this post is the equation barf without motivational text. Why are you interested, where did this come from, and what would it do for you? $\endgroup$ Commented Sep 14, 2023 at 19:05
  • $\begingroup$ @DaveBenson Indeed, thank you! Now I added the motivation part. $\endgroup$
    – tony
    Commented Sep 15, 2023 at 8:07

0

You must log in to answer this question.