Growth of the Bell numbers

Question

Question How do the Bell numbers grow? Specifically if $b(n)$ denotes the $n$-th Bell number then which function $g(n)$ satisfies $b(n)=O(g(n))$ for $n\to \infty$?

My Attempt

At first i tried to come up with some usefull boundaries. A rather obvious one presented to be that for all $n\in \mathbb{N}$, $b(n)\leq n!$. I did prove this via induction over $n$. In the following ${n\brace k}$ denotes the stirling numbers of the second kind.

$ \underline{n=0} $ $$ b(0)=1\leq 0!=1 \implies \text{ True} $$ $ \underline{n\implies n+1} $ $$ \begin{equation*} \begin{split} b(n+1)&\leq (n+1)!\\ \iff \hspace{73pt} \sum\limits_{k=0}^{n+1}{n+1\brace k}&\leq(n+1)!\\ \iff \qquad \underbrace{\sum\limits_{k=0}^{n+1}{n\brace k-1}}_{b(n)}+\sum\limits_{k=0}^{n+1}k{n\brace k}&\leq (n+1)!\\ \overset{(n+1)!=nn!+n!}{\iff} \hspace{35pt} \underbrace{\sum\limits_{k=0}^{n+1}\frac{k}{n}{n\brace k}}_{{n\brace n+1}=0/\frac{k}{n}\leq 1\implies \leq b(n)}&\leq n!\implies\text{ True} \end{split} \end{equation*} $$ So the Bell numbers grow at most factorially

Now Iam going to use the following function for the Bell numbers $b(n)=\frac{1}{e}\sum_{r=0}^{\infty}\frac{r^n}{r!}$.

Let $q\in\mathbb{R}_+$ and $C>0$, then there exists a $N\in\mathbb{N}$ such that for all $n\geq N$

$$ \begin{equation*} \begin{split} \frac{b(n)}{q^n}=\frac{1}{e}\sum\limits_{r=0}^{\infty}\left(\frac{r}{q}\right)^n\frac{1}{r!}&=\frac{1}{e}\biggl[\sum\limits_{r=0}^{\lceil q+1 \rceil}\left(\frac{r}{q}\right)^n\frac{1}{r!}+\underbrace{\sum\limits_{r=\lceil q+2\rceil}^{\infty}\left(\frac{r}{q}\right)^n\frac{1}{r!}}_{\geq 0}\biggr]\\ &\geq \frac{1}{e}\sum\limits_{r=0}^{\lceil q+1\rceil}\left(\frac{r}{q}\right)^n\frac{1}{r!}\geq \frac{1}{e(\lceil q+1\rceil)!}\left(\frac{\lceil q+1\rceil}{q}\right)^n\geq C \end{split} \end{equation*} $$ This shows that the Bell numbers grow faster than any exponential function of the form $q^n$. In particular it follows that $e^n=o(b(n))$ for $n\to\infty$. So the Bell numbers grow faster than $e^n$ but not faster than $n!$, but how exactly they grow I wasnt able to determine. Your help is much appreciated.

Answer 1

If you look at sequence $A000110$ in $OEIS$, you will find a very asymptotics for Bell numbers (it was proposed in year 2002 by Benoit Cloitre and corrected by Vaclav Kotesovecin year 2013.

It write $$B_n=b^n e^{b-n-\frac{1}{2}}\sqrt{\frac{b}{b+n}}\qquad \text{with}\qquad b=e^{W\left(n-\frac{1}{2}\right)}$$ where $W(.)$ is the main branch of Lambert function.

I think that, as a closed form, this one is better than what is given in the linked Wikipedia. The absolute relative error is smaller than $0.05$% for $n=10$.

Using the series expansion of Lambert function, then the formula proposed by de Bruijn in year 1981 given in the Wikipedia page.

Answer 2

To get the precise asymptotic growth is not so easy and is usually done using the saddle point method. To get the asymptotic growth up to a factor of $n$, with upper and lower bounds, can be done in a relatively low-tech elementary way as follows. First, a lower bound:

Lower bound: We have $$B_n \ge \frac{k^n}{k!}$$ for every positive integer $k$.

Note that this is slightly better (by a factor of $e$) than what we get from picking a single term in Dobinski's formula.

Proof. By Burnside's lemma, $\frac{k^n}{k!}$ is a lower bound on the number of orbits of the action of $S_k$ on the set of functions $[n] \to [k]$, where $[n] = \{ 1, 2, \dots n \}$. These orbits correspond to partitions of $n$ into at most $k$ non-empty subsets, so there are at most $B_n$ of them. $\Box$

For fixed $k$ this already shows (without Dobinski's formula) that $B_n$ grows faster than exponential. But the real significance of this bound is that we can optimize $k$ as a function of $n$. We consider the function $f(k) = \frac{k^n}{k!}$ and compute the effect of increasing $k$ by $1$, namely

$$\frac{f(k+1)}{f(k)} = \frac{(k+1)^n}{k^n (k+1)} = \frac{\left( 1 + \frac{1}{k} \right)^n}{k+1}.$$

We want $k = k_n$ to be the smallest positive integer such that this ratio is $\le 1$, so that increasing $k$ no longer improves the bound. Equivalently, if $x_n$ denotes the exact real solution to the equation

$$x_n + 1 = \left( 1 + \frac{1}{x_n} \right)^n$$

then $k_n = \lceil x_n \rceil$. Applying the trusty exponential inequality $(1 + x)^n \le e^{nx}$ to the RHS gives

$$x_n \le x_n + 1 \le \exp \left( \frac{n}{x_n} \right) \Rightarrow$$ $$\frac{n}{x_n} \exp \left( \frac{n}{x_n} \right) \ge n.$$

The equation $we^w = n$ has solution $w = W(n)$ where $W$ is the Lambert $W$ function; this is, as far as I know, the simplest argument which explains why the Lambert $W$ function appears in the asymptotics of the Bell numbers. $\frac{n}{x_n}$ is slightly bigger than this, which gives

$$\begin{align*} \frac{n}{x_n} &\ge W(n) \Leftrightarrow \\ x_n &\le \frac{n}{W(n)} \Rightarrow \\ k_n &\le \boxed{ \left\lceil \frac{n}{W(n)} \right\rceil }. \end{align*}$$

We can prove a matching lower bound on $k_n$ but we don't really need it; using this choice for $k$ is enough, and gives:

Corollary: We have $$\boxed{ B_n \ge \frac{\left\lceil \frac{n}{W(n)} \right\rceil^n}{\left\lceil \frac{n}{W(n)} \right\rceil!} }.$$

This lower bound has the correct asymptotic growth up to a factor of $O(\sqrt{n})$ or so. You can get something that looks more like the other bounds on Wikipedia or in Claude's answer and comment using Stirling's formula. I like this bound because it has a nice conceptual interpretation: heuristically speaking, what it's saying is that a random set partition resembles the set partition obtained from the fibers of a random function $[n] \to [k]$ where $k \approx \frac{n}{W(n)}$. This suggests the following, all of which are true:

• the expected number of parts of a random set partition is roughly $\frac{n}{W(n)}$,
• the expected size of a random part of a random set partition is roughly $W(n)$,
• the expected number of parts of size $k$ has roughly a Poisson shape $\frac{W(n)^k}{k!}$.

In turn, one way to motivate the significance of the number $k \approx \frac{n}{W(n)}$ here is that it inverts the coupon collection time! It is in fact the solution to $n = k \log k$, so it tells us when a random function $[n] \to [k]$ is likely to be surjective, hence when the corresponding partition of $[n]$ is likely to not contain empty subsets.

There is a matching (up to a factor of $O(n)$) upper bound which can be derived by a similar argument, but we need to reason about the decomposition $B_n = \sum S(n, k)$ of the Bell numbers into the Stirling numbers of the second kind; I will edit it in later.

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Edit: Here is a sketch of the upper bound. We use:

Upper bound: We have $$S(n, k) \le \frac{k^n}{k!}$$ for every $1 \le k \le n$.

Proof. $S(n, k) k!$ is the number of surjective functions $[n] \to [k]$, which is at most the number of all functions. $\Box$

Corollary: We have $$B_n \le \sum_{k=1}^n \frac{k^n}{k!}.$$

Note the similarity to Dobinski's formula, but without the factor of $e^{-1}$, and we get to truncate to $n$ terms. Now we can apply the following simple idea: just bound this sum by $n$ times its largest term. We already defined this largest term to occur at $k = k_n$, so we get:

Corollary: We have $$\boxed{ B_n \le n \frac{k_n^n}{k_n!} }$$ where $k_n$ maximizes $\frac{k^n}{k!}$.

This is within exactly a factor of $n$ from our lower bound, which was actually $B_n \ge \frac{k_n^n}{k_n!}$. From here we finally need a lower bound on $k_n$ to locate it more precisely; I believe but have not carefully checked that it is always either equal to $\left\lceil \frac{n}{W(n)} \right\rceil$ or $\left\lfloor \frac{n}{W(n)} \right\rfloor$.

This means either our lower or our upper bound must be correct to within a factor of $\sqrt{n}$. A natural guess is that we should take the geometric mean of our bounds to get a better estimate, giving

$$B_n \approx \sqrt{n} \frac{k_n^n}{k_n!}$$

which we can test the accuracy of numerically. For $n = 10$ we have

$$B_{10} = \boxed{ 115975 }$$ $$x_{10} = \frac{10}{W(10)} = 5.73 \dots $$ $$k_{10} = 6$$ $$\frac{6^{10}}{6!} = 83980.8$$ $$10 \frac{6^{10}}{6!} = 839808$$ $$\sqrt{10} \frac{6^{10}}{6!} = 265570.6 \dots $$

which is not bad but suggests that a better estimate, taking the hint from Dobinski's formula, would be to divide by $e$; this gives

$$e^{-1} \sqrt{10} \frac{6^{10}}{6!} = \boxed{ 97698.0 \dots }$$

and suggests (without proof) the estimate

$$\boxed{ B_n \approx e^{-1} \sqrt{n} \frac{ \left( \frac{n}{W(n)} \right)^n}{ \left( \frac{n}{W(n)} \right)!} }$$

which I believe (but have not checked carefully) is correct asymptotically or close to it, possibly off by a small constant factor. Comparing to the other known asymptotics can be done, again, using Stirling's.

I think we should be able to recover the correct square root factor rigorously from Dobinski's formula by estimating the contributions of the terms other than the dominant term $k = k_n$ more closely; they should end up looking approximately like a Gaussian times the dominant term, so we'll get a factor that looks like a Gaussian integral.