# On the Bell Numbers

Edit (first version was incorrectly stated. Thank you Douglas and others for your corrections) Let $B_n$ be the $n$th Bell number (the number of partitions of a set with $n$ members). For each $n > 3$, I have a set $A_n$ of size $|A_n|=B_n$. I then have a subset $A'_n \subset A_n$ where $|A'_n|=B_n-B_{n-1}$. I would like to say something about the size of $A'_n$ relative to the size of $A_n$. For instance, it seems that $lim_{n \to \infty} \frac{B_{n-1}}{B_n}=0$. Can I make a stronger statement about the ratio of successive Bell numbers? How can I formalize the statement "for sufficiently large $n$, most of $A_n$ is in $A'_n$."

• That conjecture is very far off. Do you mean ordinary partitions instead of set partitions? The asymptotics are known for both. Mar 4, 2010 at 15:05
• The conjecture is true for ordinary partitions of integers. Mar 4, 2010 at 15:44

It's easy to see that $$B_n \ge 2 B_{n-1}$$ since we always have a choice of whether to add $$n$$ to the same part as $$n-1$$ or not. Since the number of parts in a typical set partition of size $$n-1$$ grows, the choices for adding $$n$$ to a new or existing part grow, so

$$\lim_{n\to\infty} B_{n-1}/B_n = 0.$$ There are asymptotics in the Wikipedia article on the Bell numbers, but it may not be obvious how to work with the Lambert $$W$$-function in that expression, or how to bound $$B_{n-1}/B_n$$. A faster proof that the limit is $$0$$ can be obtained from Dobiński's formula, that $$B_n$$ is the $$n$$th moment of a Poisson distribution with mean $$1$$:

For any $$c \in \mathbb R$$, the Poisson distribution has positive probability of being greater than $$c$$. So, for large enough $$n$$, the $$n$$th moment $$B_n$$ is at least $$c^n$$. By Jensen's inequality, the moments satisfy

$$B_n^{\frac{n+1}{n}} \le B_{n+1}$$

$$c \le \sqrt[n]{B_n} \le \frac {B_{n+1}}{B_n}$$

A precise estimate of the relative ratios of consecutive Bell numbers is proved by Asai, Nobuhiro; Kubo, Izumi; Kuo, Hui- Hsiung in Acta Appl. Math. 63 (2000), no. 1-3, 7987: $$1\leq \frac{B(n)B(n+2)}{B^2(n+1)}\leq \frac{n+2}{n+1}$$

• The sequence $C(n)=2^n$ satisfies $C(n)C(n+2)/C^2(n+1)$=1, but this tells us little about $C(n-1)/C(n)$. Dec 10, 2019 at 11:30