Logarithms of a binomial series: A Stirling number approach

TL;DR

This paper introduces a novel approach to compute powers of the logarithm of the Catalan generating function using Stirling cycle numbers, offering an alternative expression in terms of harmonic numbers.

Contribution

It presents a new method for calculating logarithmic powers of the Catalan generating function via Stirling cycle numbers and harmonic numbers.

Findings

Derived explicit formulas using Stirling cycle numbers.

Expressed the generating function in terms of higher order harmonic numbers.

Provided a new combinatorial approach to analyze Catalan-related functions.

Abstract

The $p$-th power of the logarithm of the Catalan generating function is computed using the Stirling cycle numbers. Instead of Stirling numbers, one may write this generating function in terms of higher order harmonic numbers.

Equations

$$(\log C(z){)}^2=\sum _{n\ge 1}(\genfrac{}{}{0ex}{}{2n}{n})(H_{2n-1}-H_n)\frac{z^n}{n},$$

$$C(z)=\frac{1-\sqrt{1-4z}}{2z}=\sum _{n\ge 0}\frac{1}{n+1}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)z^n\text{and}{H}_n=\sum _{1\le k\le n}\frac{1}{k}$$

$$x^{\underline{n}}=\sum _{0\le k\le n}\left[\genfrac{}{}{0ex}{}{n}{k}\right](-1{)}^{n-k}{x}^k.$$

$$u^m=\sum _{n\ge m}\frac{m}{n}\left(\genfrac{}{}{0ex}{}{2n}{n-m}\right)z^n$$

$$F(z,\alpha )=\sum _{p\ge 0}\frac{{\alpha }^p}{p!}(\log C(z){)}^p=\exp (\alpha \log C(z))=C^{\alpha }(z)=(1+u{)}^{\alpha }=\sum _{m\ge 0}\left(\genfrac{}{}{0ex}{}{\alpha }{m}\right)u^m.$$

$$\left(\genfrac{}{}{0ex}{}{\alpha }{m}\right)=\frac{1}{m!}{\alpha }^{\underline{m}}=\frac{1}{m!}\sum _{0\le k\le m}(-1{)}^{m-k}[\genfrac{}{}{0ex}{}{m}{k}]{\alpha }^k.$$

$$F(z,\alpha )=\sum _{0\le k\le m}\frac{1}{m!}(-1{)}^{m-k}[\genfrac{}{}{0ex}{}{m}{k}\left]{\alpha }^k\sum _{n\ge m}\frac{m}{n}\right(\genfrac{}{}{0ex}{}{2n}{n-m})z^n.$$

$$(\log C(z){)}^p=p![{\alpha }^p]F(z,\alpha )=\sum _{p\le m\le n}\frac{p!}{m!}(-1{)}^{m-p}\left[\genfrac{}{}{0ex}{}{m}{p}\right]\frac{m}{n}\left(\genfrac{}{}{0ex}{}{2n}{n-m}\right)z^n.$$

$$\log C(z)=[{\alpha }^1]F(z,\alpha )=\sum _{1\le m\le n}\frac{1}{m!}(-1{)}^{m-1}[\genfrac{}{}{0ex}{}{m}{1}\left]\frac{m}{n}\right(\genfrac{}{}{0ex}{}{2n}{n-m})z^n.$$

$$\log C(z)=[{\alpha }^1]F(z,\alpha )=\frac{1}{2}\sum _{n\ge 1}\frac{1}{n}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)z^n.$$

$$2[{\alpha }^2]F(z,\alpha )$$

$$=2\sum _{2\le m\le n}{H}_{m-1}(-1{)}^m\frac{1}{n}(\genfrac{}{}{0ex}{}{2n}{n-m})z^n$$

$$=2\sum _{1\le j<m\le n}\frac{1}{j}(-1{)}^m\frac{1}{n}(\genfrac{}{}{0ex}{}{2n}{n-m})z^n$$

$$=2\sum _{1\le j<n}\frac{1}{j}(-1{)}^{j-1}\frac{1}{n}(\genfrac{}{}{0ex}{}{2n-1}{n-j-1})z^n.$$

$$\left(\genfrac{}{}{0ex}{}{2n}{n}\right)(H_{2n-1}-H_n)=2\sum _{1\le j<n}\frac{(-1{)}^{j-1}}{j}\left(\genfrac{}{}{0ex}{}{2n-1}{n-j-1}\right).$$

$$\frac{1}{n!}\genfrac{[}{]}{0.0pt}{}{n+1}{r+1}=(-1)^{r}\sum_{\{r\}}\prod\limits_{j=1}^{l}\frac{(-1)^{i_{j}}}{i_{j}!}\bigg{(}\frac{H_{n}^{(r_{j})}}{r_{j}}\bigg{)}^{i_{j}}.$$

$$H_n^{(i)}=\sum _{1\le k\le n}\frac{1}{k^i}.$$

$$\frac{1}{n!}\left[\genfrac{}{}{0ex}{}{n+1}{2}\right]$$

$$\frac{1}{n!}\left[\genfrac{}{}{0ex}{}{n+1}{3}\right]$$

$$\frac{1}{n!}\left[\genfrac{}{}{0ex}{}{n+1}{4}\right]$$

$$\frac{1}{n!}\left[\genfrac{}{}{0ex}{}{n+1}{5}\right]$$

$$(\log C(z){)}^p=\sum _{p\le m\le n}\frac{1}{(m-1)!}[\genfrac{}{}{0ex}{}{m}{p}](-1{)}^{m-p}\frac{p!}{n}\left(\genfrac{}{}{0ex}{}{2n}{n-m}\right)z^n$$

$$C_{\lambda }(z)=\sum _{n\ge 0}\left(\genfrac{}{}{0ex}{}{1+n\lambda }{n}\right)\frac{z^n}{1+n\lambda }.$$

$$u^m=\sum _{n\ge m}\left(\genfrac{}{}{0ex}{}{\lambda n+m}{n}\right)\frac{m}{\lambda n+m}{z}^n.$$

$$F(z,\alpha )$$

$$=\sum _{0\le k\le m\le n}\frac{1}{m!}(-1{)}^{m-k}[\genfrac{}{}{0ex}{}{m}{k}\left]{\alpha }^k\right(\genfrac{}{}{0ex}{}{\lambda n+m}{n})\frac{m}{\lambda n+m}{z}^n.$$

$$(\log C_{\lambda }(z){)}^p=p![{\alpha }^p]F(z,\alpha )=\sum _{p\le m\le n}\frac{p!}{m!}(-1{)}^{m-p}\left[\genfrac{}{}{0ex}{}{m}{p}\right]\left(\genfrac{}{}{0ex}{}{\lambda n+m}{n}\right)\frac{m}{\lambda n+m}{z}^n.$$

Full text

1. Introduction

Knuth [3] proposed the exciting formula

[TABLE]

where

[TABLE]

with the generating function of Catalan numbers and harmonic numbers.

This formula was recently extended by Chu [5] to general exponents $p$. Note that Knuth talked about the exponent $1$ in his christmas lecture from 2014 [4].

We present here a very simple approach to this question using Stirling cycle numbers; recall [2] that they transform falling powers into ordinary powers viz.

[TABLE]

2. The expansion of the $p$-th power

The substitution $z=\frac{u}{(1+u)^{2}}$ was presented in [1] and it is extremely useful when dealing with Catalan numbers and Catalan statistics. Then $C(z)=1+u$, and, by the Lagrange inversion formula [7],

[TABLE]

for $m\geq 1$. For $m=0$ the formula is still true when taking a limit. We now consider the bivariate generating function

[TABLE]

But

[TABLE]

Therefore

[TABLE]

The desired formula follows from reading off coefficients of $\alpha^{p}$:

[TABLE]

3. Special cases

For $p=1$, we get the instance of the Christmas lecture:

[TABLE]

Since $\genfrac{[}{]}{0.0pt}{}{m}{1}=(m-1)!$, this leads to

[TABLE]

Now we turn to the instance $p=2$ from [3]. (Note that $\genfrac{[}{]}{0.0pt}{}{m}{2}=(m-1)!H_{m-1}$.)

[TABLE]

To obtain the form proposed by Knuth, we still need to prove that

[TABLE]

Modern computer algebra systems readily simplify the difference of these two sides to 0, as expected.

4. Connection with harmonic numbers — the general case

In [6], there is the general formula

[TABLE]

Here, the sum is over all partitions of $r$: $r=i_{1}r_{1}+\cdots+i_{l}r_{l}$, with parts $r_{1}>\dots>r_{l}\geq 1$ and positive integers $i_{1},\dots,i_{l}$. As an example, the partitions of $r=4$ are $4$, $3+1$, $2+2$, $2+1+1$, $1+1+1+1$, written alternatively as $1\cdot 4$, $1\cdot 3+1\cdot 1$, $2\cdot 2$, $1\cdot 2+2\cdot 1$, $4\cdot 1$.

There appear higher order harmonic numbers as well:

[TABLE]

Here are the first few instances:

[TABLE]

This allows to replace $\frac{1}{(m-1)!}\genfrac{[}{]}{0.0pt}{}{m}{p}$ in

[TABLE]

by an expression involving $H_{m-1}^{(1)},\dots,H_{m-1}^{(p-1)}$.

5. Extension

If instead of $u=z(1+u)^{2}$ we work with $u=z(1+u)^{\lambda}$, then we deal with the generating function of extended (generalized) Catalan numbers

[TABLE]

From [2], we infer that

[TABLE]

So

[TABLE]

The desired formula follows from reading off coefficients of $\alpha^{p}$:

[TABLE]