An asymptotic Formula for the iterated exponential Bell Numbers

TL;DR

This paper derives an asymptotic formula for the leading coefficient of the iterated exponential Bell numbers, advancing understanding of their growth and structure.

Contribution

It provides the first explicit asymptotic formula for the leading coefficients of Bell's iterated exponential numbers.

Findings

Derived an asymptotic expression for the leading coefficient

Enhanced understanding of the growth rate of iterated exponential Bell numbers

Established a foundation for further coefficient analysis

Abstract

In 1938 E. T. Bell introduced "The Iterated Exponential Integers". He proved that these numbers may be expressed by polynomials with rational coefficients. However, Bell gave no formulas for any of the coefficients except the trivial one, which is always 1. Our task has been to find the coefficient of the leading term, giving asymptotic information about these numbers.

Tables

Table 1. Table 1: Higher order Bell numbers B n ( m ) superscript subscript 𝐵 𝑛 𝑚 B_{n}^{(m)} when 1 ≤ n ≤ 8 1 𝑛 8 1\leq n\leq 8 and 1 ≤ m ≤ 5 1 𝑚 5 1\leq m\leq 5

$m\backslash n$	1	2	3	4	5	6	7	8
1	1	2	5	15	52	203	877	4140
2	1	3	12	60	358	2471	19302	167894
3	1	4	22	154	1304	12915	146115	1855570
4	1	5	35	315	3455	44590	660665	11035095
5	1	6	51	561	7556	120196	2201856	45592666

Table 2. Table 2: B 3 ( m ) superscript subscript 𝐵 3 𝑚 B_{3}^{(m)} and ( 3 ! / 2 3 − 1 ) ​ m 3 − 1 3 superscript 2 3 1 superscript 𝑚 3 1 (3!/2^{3-1})m^{3-1} are compared. The computations are done with the aid of Maple 2016.

$m$	${10}^2$	${10}^5$	${10}^8$
$B_3^{\left(m\right)}$	15251	15000250001	15000000250000001
$\left(3!/2^{3-1}\right)m^{3-1}$	15000	15000000000	15000000000000000

Equations

$$E_m(x)=\sum _{n=0}^{\infty }{B}_n^{(m)}\frac{x^n}{n!}(m\ge 0),$$

$$B_n=\sum _{k=1}^{n}S(n,k).$$

$$B_n^{(m)}=\sum _{k=1}^n{B}_k^{(m-1)}S(n,k).$$

$$B_n^{(m)}=c_{n-1}{m}^{n-1}+c_{n-2}{m}^{n-2}+\cdots +c_0,$$

$$\begin{array}[]{lll}B_{n}^{(m)}-B_{n}^{(m-1)}&=&\displaystyle\sum\limits_{k=1}^{n-1}B_{k}^{(m-1)}S(n,k)\\[11.38092pt] {}&=&d_{n-2}m^{n-2}+d_{n-3}m^{n-3}+\cdots+d_{1}m+d_{0},\end{array}$$

$$\sum _{k=1}^m(B_n^{(k)}-B_n^{(k-1)})=B_n^{(m)}-1=\sum _{r=0}^{n-2}{d}_r\left(\sum _{k=1}^m{k}^r\right).$$

$$\sum _{k=1}^m{k}^r=\frac{1}{r+1}{m}^{r+1}+m\sum _{k=2}^{2\lfloor r/2\rfloor }\frac{b_k}{k}\left(\genfrac{}{}{0ex}{}{r}{k-1}\right)m^{r-k},$$

$$B_n^{(m)}\sim \frac{n!}{2^{n-1}}{m}^{n-1}(m\to \infty ).$$

$$c_{n-1}^{(n)}=\frac{n}{2}{c}_{n-2}^{(n-1)}.$$

Full text

1 Higher order Bell numbers

The iterated exponential numbers, also called higher order Bell numbers, were introduced by E. T. Bell in [1]:

Definition**.**

The $m$-th order Bell numbers, $B_{n}^{(m)}\quad(m,n=0,1,\ldots)$, are given by the exponential generating functions

[TABLE]

where $E_{0}(x)=\exp(x)$ and $E_{m+1}(x)=\exp(E_{m}(x)-1)\quad(m\geq 0)$.

Obviously we have $B_{n}^{(0)}=1$ and $B_{0}^{(m)}=E_{m}(0)=1$. In Table 1 $B_{n}^{(m)}$ is computed for a few values of $m$ and $n$.

We note that $E_{1}(x)=\exp(\exp(x)-1)$ is the exponential generating function of the first order Bell numbers $B_{n}^{(1)}$, representing the total number $B_{n}$ of partitions of an $n$-set (see for example [2, p. 24]). We also point out that the Stirling numbers of the second kind, $S(n,k)$, represent the number of $k$-partitions of an $n$-set, so that

[TABLE]

In [1, p. 544] E. T. Bell proved a generalization of (1) that connects higher order Bell numbers to Stirling numbers of the second kind by the following recursion relation:

[TABLE]

We will find (2) useful in the following section.

2 Polynomial expansions

The introduction of higher order Bell numbers does not immediately suggest the existence of a polynomial representation. Still, that was exactly what Bell was able to prove in [1, p.545]. Based on (2), we will carry out a somewhat simpler proof, containing a few useful details:

Lemma 1**.**

$B_{n}^{(m)}$* may be expressed in the form*

[TABLE]

where $c_{n-1},\ldots,c_{0}$ are rational numbers, independent of $m$.

Proof.

Our proof is by induction. First we observe that $B_{1}^{(m)}=E_{m}^{\prime}(0)=\prod_{k=0}^{m}E_{k}(0)=1$. Together with $B_{n}^{(0)}=1$, this means that we must have $c_{0}=1$ if (3) is to be correct. Now, suppose that (3) is true for $k<n$, so that $B_{k}^{(m)}$ is a polynomial of degree $k-1$ with rational coefficients and constant term equal to 1. By rewriting (2), we have

[TABLE]

where the $d_{i}$’s are rational coefficients. Summing over $m$, the telescoping property gives

[TABLE]

It is well known (see [3, p. 525]) that

[TABLE]

where the $b_{k}$’s are the (rational) Bernoulli numbers. This completes the induction step and proves the lemma. ∎

3 An asymptotic formula of the higher order Bell numbers

When $n$ is fixed, it is straightforward to see that $\lim_{m\rightarrow\infty}B_{n}^{(m)}/B_{n}^{(m-1)}=1$. However, looking more carefully into the details of the proof of the lemma, we can sharpen the asymptotics:

Theorem 1**.**

Assume $n$ fixed. Then we have

[TABLE]

Proof.

Since we already have the polynomial expansion from the lemma, it remains to prove that $c_{n-1}=n!/2^{n-1}$. To avoid ambiguity we will use the notation $c_{n-1}=c_{n-1}^{(n)}$ for the leading coefficient of $B_{n}^{(m)}$. By (5) and (6) $c_{n-1}^{(n)}$ can be expressed as $d_{n-2}/(n-1)$. At the same time, because of (4), $d_{n-2}$ must be the $m^{n-2}$-coefficient of $B_{n-1}^{(m-1)}S(n,n-1)$, namely $c_{n-2}^{(n-1)}S(n,n-1)$. Since $S(n,n-1)=\binom{n}{2}=n/2$, this means that

[TABLE]

The initial value of this recurrence relation is $c_{0}^{(1)}=B_{1}^{(m)}=1$, which enables us to conclude that $c_{n-1}=c_{n-1}^{(n)}=n!/2^{n-1}$, proving (7). ∎

In Table 2 we see how the $m$-growth of $B_{n}^{(m)}$ is ”explained” by the leading term $(n!/2^{n-1})m^{n-1}$ when $n=3$.