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Let $F(n)$ be the number of all partitions of $[n]$ with no singleton blocks.
Prove that $$\lim_{n\to\infty}\frac{F(n)}{B(n)}=0$$

According to this question, we can know $$F(n+1)=\sum_{i=0}^{n-1}(-1)^iB(n-i)$$ Bell numbers have the property $$B(n+1)=\sum_{i=0}^n\begin{pmatrix} n\\i \end{pmatrix}B(i)$$ so $$\begin{aligned} \lim_{n\to\infty}\frac{F(n+1)}{B(n+1)}&=\lim_{n\to\infty}\frac{\sum_{i=0}^{n-1}(-1)^iB(n-i)}{\sum_{i=0}^n\begin{pmatrix} n\\i \end{pmatrix}B(i)}\\ &=\frac{B(n)-B(n-1)\cdots}{B(n)+\begin{pmatrix} n\\n-1 \end{pmatrix}B(n-1)\cdots} \end{aligned} $$ I think $B(n)\gg B(n-1)$ so it is equal to $1$ not $0$.

How to prove it? Thanks!

RobPratt
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Leo SHEN
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1 Answers1

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Assume $n\gt2$.

Let $G(n)$ be the number of partitions of $[n]$ into at least $\sqrt n$ non-singleton blocks, and let $H(n)$ be the number of partitions of $[n]$ into at most $\sqrt n$ non-singleton blocks.

$F(n)\le G(n)+H(n)$, so $\max\{G(n),H(n)\}\ge\frac12F(n)$.

Let $B'(n)=B(n)-F(n)$, the number of partitions of $[n]$ with at least one singleton block.

$B'(n)\ge\sqrt nG(n)$ because we can turn any $G(n)$-partition into at least $\sqrt n$ different $B'(n)$-partitions by breaking one of its blocks into singletons, and each $B'(n)$-partition obtained in this way has a unique source.

$B'(n)\ge\sqrt nH(n)$, because we can turn any $H(n)$-partition into a $B'(n)$-partition by choosing an element of $[n]$ and making it a singleton block, and no $B'(n)$-partition is obtained in more than $\sqrt n$ ways.

It follows that $$B(n)-F(n)=B'(n)\ge\sqrt n\max\{G(n),H(n)\}\ge\frac{\sqrt n}2F(n),$$ so $$B(n)\ge\frac{\sqrt n+2}2F(n)$$ and $$\frac{F(n)}{B(n)}\le\frac2{\sqrt n+2}\to0\text{ as }n\to\infty.$$

bof
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