Maximizing weighted sums of binomial coefficients using generalized continued fractions

Abstract

Let $m,\,r\in {\mathbb {Z}}$ and $\omega \in {\mathbb {R}}$ satisfy $0\leqslant r\leqslant m$ and $\omega \geqslant 1$. Our main result is a generalized continued fraction for an expression involving the partial binomial sum $s_m(r) = \sum _{i=0}^r\binom{m}{i}$. We apply this to create new upper and lower bounds for $s_m(r)$ and thus for $g_{\omega,m}(r)=\omega ^{-r}s_m(r)$. We also bound an integer $r_0 \in \{0,\,1,\,\ldots,\,m\}$ such that $g_{\omega,m}(0)<\cdots < g_{\omega,m}(r_0-1)\leqslant g_{\omega,m}(r_0)$ and $g_{\omega,m}(r_0)>\cdots >g_{\omega,m}(m)$. For real $\omega \geqslant \sqrt 3$ we prove that $r_0\in \{\lfloor \frac {m+2}{\omega +1}\rfloor,\,\lfloor \frac {m+2}{\omega +1}\rfloor +1\}$, and also $r_0 =\lfloor \frac {m+2}{\omega +1}\rfloor$ for $\omega \in \{3,\,4,\,\ldots \}$ or $\omega =2$ and $3\nmid m$.

1. Introduction

Given a real number $\omega \geqslant 1$ and integers $m,\,r$ satisfying $0\leqslant r\leqslant m$, set

(1.1)\begin{equation} s_m(r):=\sum_{i=0}^r\binom{m}{i}\quad\text{and}\quad g(r)= g_{\omega,m}(r):=\omega^{{-}r}s_m(r), \end{equation}

where the binomial coefficient $\binom{m}{i}$ equals $\prod _{k=1}^i\frac {m-k+1}{k}$ for $i>0$ and $\binom{m}{0}=1$. The weighted binomial sum $g_{\omega,m}(r)$ and the partial binomial sum $s_m(r)=g_{1,m}(r)$ appear in many formulas and inequalities, e.g. the cumulative distribution function $2^{-m}s_m(r)$ of a binomial random variable with $p=q=\tfrac {1}{2}$ as in remark 5.3, and the Gilbert–Varshamov bound [6, Theorem 5.2.6] for a code $C\subseteq \{0,\,1\}^n$. Partial sums of binomial coefficients are found in probability theory, coding theory, group theory, and elsewhere. As $s_m(r)$ cannot be computed exactly for most values of $r$, it is desirable for certain applications to find simple sharp upper and lower bounds for $s_m(r)$. Our interest in bounding $2^{-r}s_m(r)$ was piqued in [4] by an application to Reed–Muller codes $\textrm{RM}(m,\,r)$, which are linear codes of dimension $s_m(r)$.

Our main result is a generalized continued fraction $a_0 + \mathcal{K}_{i=1}^{r} \frac {b_i}{a_i}$ (using Gauss’ Kettenbruch notation) for $Q:=\frac {(r+1)}{s_m(r)}\binom{m}{r+1}$. From this we derive useful approximations to $Q,\, 2+\frac {Q}{r+1}$, and $s_m(r)$, and with these find a maximizing input $r_0$ for $g_{\omega,m}(r)$.

The $j$th tail of the generalized continued fraction $\mathcal{K}_{i=1}^{r} \frac {b_i}{a_i}$ is denoted by ${\mathcal {T}}_j$ where

(1.2)\begin{equation} {\mathcal{T}}_j:=\mathop{{\vcenter{\hbox{ $\mathcal{K}$}}}}_{i=j}^{r} \frac{b_i}{a_i}= \frac{b_j}{a_j +\dfrac{b_{j+1}}{a_{j+1} +\dfrac{b_{j+2}}{\llap{{\ddots}} {{a_{r-1} + \dfrac{b_r}{a_r}}}}}} =\frac{b_j}{a_j+{\mathcal{T}}_{j+1}} \quad\textrm{and}\ 1\leqslant j\leqslant r. \end{equation}

If ${\mathcal {T}}_j=\frac {B_j}{A_j}$, then ${\mathcal {T}}_j=\frac {b_j}{a_j+{\mathcal {T}}_{j+1}}$ shows $b_jA_j-a_jB_j={\mathcal {T}}_{j+1}B_j$. By convention we set ${\mathcal {T}}_{r+1}=0$.

It follows from $\binom{m}{r-i}=\binom{m}{r}\prod _{k=1}^i\frac {r-k+1}{m-r+k}$ that $x^i\binom{m}{r}\leqslant \binom{m}{r-i}\le y^i\binom{m}{r}$ for $0\leqslant i\leqslant r$ where $x:=\frac {1}{m}$ and $y:=\frac {r}{m-r+1}$. Hence $\frac {1-x^{r+1}}{1-x}\binom{m}{r}\leqslant s_m(r)\leqslant \frac {1-y^{r+1}}{1-y}\binom{m}{r}$. These bounds are close if $\frac {r}{m}$ is near 0. If $\frac {r}{m}$ is near $\tfrac {1}{2}$ then better approximations involve the Berry–Esseen inequality [7] to estimate the binomial cumulative distribution function $2^{-m}s_m(r)$. The cumulative distribution function $\Phi (x)=\frac {1}{\sqrt {2\pi }}\int _{-\infty }^xe^{-t^2/2}\,dt$ is used in remark 5.3 to show that $|2^{-m}s_m(r)-\Phi (\frac {2r-m}{\sqrt {m}})|\leqslant \frac {0.4215}{\sqrt {m}}$ for $0\leqslant r\leqslant m$ and $m\ne 0$. Each binomial $\binom{m}{i}$ can be estimated using Stirling's approximation as in [10, p. 2]: $\binom{m}{i}=\frac {C_i^m}{\sqrt {2\pi p(1-p)m}}\left (1+O\left (\frac {1}{m}\right )\right )$ where $C_i=\frac {1}{p^p(1-p)^{1-p}}$ and $p=p_i=i/m$. However, the sum $\sum _{i=0}^r\binom{m}{i}$ of binomials is harder to approximate. The preprint [11] discusses different approximations to $s_m(r)$.

Sums of binomial coefficients modulo prime powers, where $i$ lies in a congruence class, can be studied using number theory, see [5, p. 257]. Theorem 1.1 below shows how to find excellent rational approximations to $s_m(r)$ via generalized continued fractions.

Theorem 1.1 Fix $r,\,m\in {\mathbb {Z}}$ where $0\leqslant r\leqslant m$ and recall that $s_m(r)=\sum _{i=0}^r\binom{m}{i}$.

1. (a) If $b_i=2i(r+1-i)$, $a_i=m-2r+3i$ for $0\leqslant i\leqslant r$, then

   \[ Q:=\frac{(r+1)\binom{m}{r+1}}{s_m(r)}=a_0+{\vcenter{\hbox{ $\mathcal{K}$}}}_{i=1}^{r} \frac{b_i}{a_i}. \]

2. (b) If $1\leqslant j\leqslant r$, then ${\mathcal {T}}_j=R_j/R_{j-1}>0$ where the sum $R_j:= 2^j j!\sum _{k=0}^{r-j}\binom{r-k}{j}\binom{m}{k}$ satisfies $b_j R_{j-1} - a_j R_j = R_{j+1}$. Also, $(m-r)\binom{m}{r}-a_0R_0=R_1$.

Since $s_m(m)=2^m$, it follows that $s_m(m-r)=2^m-s_m(r-1)$ so we focus on values of $r$ satisfying $0\leqslant r\leqslant \lfloor \frac {m}{2}\rfloor$. Theorem 1.1 allows us to find a sequence of successively sharper upper and lower bounds for $Q$ (which can be made arbitrarily tight), the coarsest being $m-2r\leqslant Q\leqslant m-2r+\frac {2r}{m-2r+3}$ for $1\leqslant r<\frac {m+3}{2}$, see proposition 2.3 and corollary 2.4.

The fact that the tails ${\mathcal {T}}_1,\,\ldots,\,{\mathcal {T}}_r$ are all positive is unexpected as $b_i/a_i$ is negative if $\frac {m+3i}{2}< r$. This fact is crucial for approximating ${\mathcal {T}}_1=\mathcal{K}_{i=1}^{r} \frac {b_i}{a_i}$, see theorem 1.3. Theorem 1.1 implies that ${\mathcal {T}}_1{\mathcal {T}}_2\cdots {\mathcal {T}}_r=R_r/R_0$. Since $R_0=s_m(r)$, $R_r=2^rr!$, ${\mathcal {T}}_j=\frac {b_j}{a_j+{\mathcal {T}}_{j+1}}$ and $\prod _{j=1}^rb_j=2^r(r!)^2$, the surprising factorizations below follow c.f. remark 2.1.

Corollary 1.2 We have $s_m(r)\prod _{j=1}^r{\mathcal {T}}_j=2^rr!$ and $r!s_m(r)=\prod _{j=1}^r(a_j+{\mathcal {T}}_{j+1})$.

Suppose that $\omega >1$ and write $g(r)=g_{\omega,m}(r)$. We extend the domain of $g(r)$ by setting $g(-1)=0$ and $g(m+1)=\frac {g(m)}{\omega }$ in keeping with (1.1). It is easy to prove that $g(r)$ is a unimodal function c.f. [2, § 2]. Hence there exists some $r_0\in \{0,\,1,\,\ldots,\,m\}$ that satisfies

(1.3)\begin{align} & g_{\omega,m}({-}1)<\cdots< g_{\omega,m}(r_0-1)\leqslant g_{\omega,m}(r_0)\quad\text{and}\qquad\notag\\& g_{\omega,m}(r_0)>\cdots>g_{\omega,m}(m+1). \end{align}

As $g(-1)< g(0)=1$ and $\left (\frac {2}{\omega }\right )^m=g(m)>g(m+1)=\frac {2^m}{\omega ^{m+1}}$, both chains of inequalities are non-empty. The chains of inequalities (1.3) serve to define $r_0$.

We use theorem 1.1 to show that $r_0$ is commonly close to $r':=\lfloor \frac {m+2}{\omega +1}\rfloor$. We always have $r'\leqslant r_0$ (by lemma 3.3) and though $r_0-r'$ approaches $\frac {m}{2}$ as $\omega$ approaches 1 (see remark 4.4), if $\omega \geqslant \sqrt {3}$ then $0\leqslant r_0 - r'\leqslant 1$ by the next theorem.

Theorem 1.3 If $\omega \geqslant \sqrt {3}$, $m\in \{0,\,1,\,\ldots \}$ and $r':=\lfloor \frac {m+2}{\omega +1}\rfloor$, then $r_0\in \{r',\,r'+1\}$, that is

\[ g(0)<\cdots< g(r'-1)\leqslant g(r'),\quad\text{and}\quad g(r'+1)>g(r'+2)>\cdots>g(m). \]

Sharp bounds for $Q$ seem powerful: they enable short and elementary proofs of results that previously required substantial effort. For example, our proof in [4, Theorem 1.1] for $\omega =2$ of the formula $r_0=\lfloor \frac {m}{3}\rfloor +1$ involved a lengthy argument, and our first proof of theorem 1.4 below involved a delicate induction. By this theorem there is a unique maximum, namely $r_0=r'=\lfloor \frac {m+2}{\omega +1}\rfloor$ when $\omega \in \{3,\,4,\,5,\,\ldots \}$ and $\omega \ne m+1$, c.f. remark 4.2. In particular, strict inequality $g_{\omega,m}(r'-1)< g_{\omega,m}(r')$ holds.

Theorem 1.4 Suppose that $\omega \in \{3,\,4,\,5,\,\ldots \}$ and $r'=\lfloor \frac {m+2}{\omega +1}\rfloor$. Then

\[ g_{\omega,m}(0)<\cdots< g_{\omega,m}(r'-1)\leqslant g_{\omega,m}(r')>g_{\omega,m}(r'+1)>\cdots>g_{\omega,m}(m), \]

with equality if and only if $\omega =m+1$.

Our motivation was to analyse $g_{\omega,m}(r)$ by using estimates for $Q$ given by the generalized continued fraction in theorem 1.1. This gives tighter estimates than the method involving partial sums used in [4]. The plots of $y=g_{\omega,m}(r)$ for $0\leqslant r\leqslant m$ are highly asymmetrical if $\omega -1$ and $m$ are small. However, if $m$ is large the plots exhibit an ‘approximate symmetry’ about the vertical line $r=r_0$ (see figure 1). Our observation that $r_0$ is close to $r'$ for many choices of $\omega$ was the starting point of our research.

Figure 1. Plots of $y=g_{\omega,24}(r)$ for $0\leqslant r\leqslant 24$ with $\omega \in \{1,\,\frac 32,\,2,\,3\}$, and $y=g_{3,12}(r)$

Byun and Poznanović [2, Theorem 1.1] compute the maximizing input, call it $r^*$, for the function $f_{m,a}(r):=(1+a)^{-r}\sum _{i=0}^r\binom{m}{i}a^i$ where $a\in \{1,\,2,\,\ldots \}$. Their function equals $g_{\omega,m}(r)$ only when $\omega =1+a=2$. Some of their results and methods are similar to those in [4] which studied the case $\omega =2$. They prove that $r^*=\lfloor \frac {a(m+1)+2}{2a+1}\rfloor$ provided $m\not \in \{3,\,2a+4,\,4a+5\}$ or $(a,\,m)\ne (1,\,12)$ when $r^*=\lfloor \frac {a(m+1)+2}{2a+1}\rfloor -1$.

In Section 2 we prove theorem 1.1 and record approximations to our generalized continued fraction expansion. When $m$ is large, the plots of $y=g_{\omega,m}(r)$ are reminiscent of a normal distribution with mean $\mu \approx \frac {m}{\omega +1}$. Section 3 proves key lemmas for estimating $r_0$, and applies theorem 1.1 to prove theorem 1.4. Non-integral values of $\omega$ are considered in Section 4 where theorem 1.3 is proved. In Section 5 we estimate the maximum height $g(r_0)$ using elementary methods and estimations, see lemma 5.1. A ‘statistical’ approximation to $s_m(r)$ is given in remark 5.3, and it is compared in remark 5.4 to the ‘generalized continued fraction approximations’ of $s_m(r)$ in proposition 2.3.

2. Generalized continued fraction formulas

In this section we prove theorem 1.1, namely that $Q:=\frac {r+1}{s_m(r)}\binom{m}{r+1}=a_0+{\mathcal {T}}_1$ where ${\mathcal {T}}_1=\mathcal{K}_{i=1}^r\frac {b_i}{a_i}$. The equality $s_m(r)=\frac {r+1}{a_0+{\mathcal {T}}_1}\binom{m}{r+1}$ is noted in corollary 2.2.

A version of theorem 1.1(a) was announced in the SCS2022 Poster room, created to run concurrently with vICM 2022, see [9].

Proof Proof of theorem 1.1

Set $R_{-1}=Q\,s_m(r)=(r+1)\binom{m}{r+1}=(m-r)\binom{m}{r}$ and

\[ R_j=2^j j!\sum_{k=0}^{r-j}\binom{r-k}{j}\binom{m}{k} \quad\text{for } 0\leqslant j\leqslant r+1. \]

Clearly $R_0=s_m(r)$, $R_{r+1}=0$ and $R_j>0$ for $0\leqslant j\leqslant r$. We will prove in the following paragraph that the quantities $R_j$, $a_j=m-2r+3j$, and $b_j=2j(r+1-j)$ satisfy the following $r+1$ equations, where the first equation (2.1) is atypical:

(2.1)\begin{align} R_{{-}1}-a_0R_0& =R_1, \end{align}

(2.2)\begin{align} b_jR_{j-1}-a_jR_j& =R_{j+1}\quad\text{where } 1\leqslant j\leqslant r. \end{align}

Assuming (2.2) is true, we prove by induction that ${\mathcal {T}}_{j}=R_{j}/R_{j-1}$ holds for $r+1\geqslant j\geqslant 1$. This is clear for $j=r+1$ since ${\mathcal {T}}_{r+1}=R_{r+1}=0$. Suppose that $1\leqslant j\leqslant r$ and ${\mathcal {T}}_{j+1}=R_{j+1}/R_{j}$ holds. We show that ${\mathcal {T}}_{j}=R_{j}/R_{j-1}$ holds. Using (2.2) and $R_j>0$ we have $b_jR_{j-1}/R_j-a_j=R_{j+1}/R_j={\mathcal {T}}_{j+1}$. Hence $R_j/R_{j-1}=b_j/(a_j+{\mathcal {T}}_{j+1})={\mathcal {T}}_j$, completing the induction. Equation (2.1) gives $Q=R_{-1}/R_0=a_0+R_1/R_{0}=a_0+{\mathcal {T}}_1$ as claimed. Since $R_j>0$ for $0\leqslant j\leqslant r$, we have ${\mathcal {T}}_{j}=R_{j}/R_{j-1}>0$ for $1\leqslant j\leqslant r$. This proves the first half of theorem 1.1(b), and the recurrence ${\mathcal {T}}_j=b_j/(a_j+{\mathcal {T}}_{j+1})$ for $1\leqslant j\leqslant r$, proves part (a).

We now show that (2.1) holds. The identity $R_0=2^00!\sum _{k=0}^r\binom{m}{k}=s_m(r)$ gives

\begin{align*} R_{{-}1}-a_0R_0& =(r+1)\binom{m}{r+1}-(m-2r)\sum_{i=0}^r\binom{m}{i}\\ & =(r+1)\binom{m}{r+1}-\sum_{i=0}^r({-}i+m-i-2r+2i)\binom{m}{i}\\ & =\sum_{i=0}^{r} \left[(i+1)\binom{m}{i+1}-(m-i)\binom{m}{i}\right] +2\sum_{i=0}^{r-1}(r-i)\binom{m}{i}. \end{align*}

As $(i+1)\binom{m}{i+1}=(m-i)\binom{m}{i}$, we get $R_{-1}-a_0R_0=2\sum _{k=0}^{r-1}\binom{r-k}{1}\binom{m}{k}=R_1$.

We next show that (2.2) holds. In order to simplify our calculations, we divide by $C_{j}:= 2^j j!$. Using $(j+1)\binom{r-k}{j+1}=(r-k-j)\binom{r-k}{j}$ gives

\begin{align*} \frac{R_{j+1}}{C_j} & =\sum_{k=0}^{r-j-1}2(j+1)\binom{r-k}{j+1}\binom{m}{k}\\ & =\sum_{k=0}^{r-j}2(r-k-j)\binom{r-k}{j}\binom{m}{k}\\ & =\sum_{k=0}^{r-j+1}(j-k)\binom{r-k}{j}\binom{m}{k} - \sum_{k=0}^{r-j}(k-2r+3j)\binom{r-k}{j}\binom{m}{k} \end{align*}

noting that the term with $k=r-j+1$ in the first sum is zero as $\binom{j-1}{j}=0$. Using the abbreviation $L=\sum _{k=0}^{r-j}(k-2r+3j)\binom{r-k}{j}\binom{m}{k}$ and using the identity $j\binom{r-k}{j}=(r+1-j-k)\binom{r-k}{j-1}$ gives

\begin{align*} \frac{R_{j+1}}{C_j}& =\sum_{k=0}^{r-j+1} \left[(r+1-j-k)\binom{r-k}{j-1}-k\binom{r-k}{j}\right] \binom{m}{k} -L\\ & = \sum_{k=0}^{r-j+1} \left[(r+1-j)\binom{r-k}{j-1}-k\binom{r-k}{j-1} -k\binom{r-k}{j}\right]\binom{m}{k} -L\\ & = \sum_{k=0}^{r-j+1} \left[(r+1-j)\binom{r-k}{j-1}-k\binom{r-k+1}{j}\right]\binom{m}{k} -L. \end{align*}

However, $k\binom{m}{k}=(m-k+1)\binom{m}{k-1}$, and therefore,

\begin{align*} \sum_{k=0}^{r-j+1}k\binom{r-k+1}{j}\binom{m}{k} & =\sum_{k=1}^{r-j+1}(m-k+1)\binom{r-k+1}{j}\binom{m}{k-1}\\ & =\sum_{\ell=0}^{r-j}(m-\ell)\binom{r-\ell}{j}\binom{m}{\ell}. \end{align*}

Thus

\begin{align*} \frac{R_{j+1}}{C_{j}} & =\sum_{k=0}^{r-j+1}(r-j+1)\binom{r-k}{j-1}\binom{m}{k} -\sum_{k=0}^{r-j}(m-k)\binom{r-k}{j}\binom{m}{k} -L\\ & =\sum_{k=0}^{r-j+1}(r-j+1)\binom{r-k}{j-1}\binom{m}{k} -\sum_{k=0}^{r-j}(m-k+k-2r+3j)\binom{r-k}{j}\binom{m}{k}\\ & =\sum_{k=0}^{r-j+1}(r-j+1)\binom{r-k}{j-1}\binom{m}{k} - \sum_{k=0}^{r-j}\overbrace{(m-2r+3j)}^{a_j}\binom{r-k}{j}\binom{m}{k}\\ & =\frac{\overbrace{2j(r-j+1)}^{b_j}2^{j-1}(j-1)!}{C_{j}}\sum_{k=0}^{r-j+1} \binom{r-k}{j-1}\binom{m}{k} - \sum_{k=0}^{r-j}a_j\binom{r-k}{j}\binom{m}{k} \end{align*}

Hence $\frac {R_{j+1}}{C_{j}}=\frac {b_{j}R_{j-1}}{C_{j}}-\frac {a_{j}R_{j}}{C_{j}}$ for $1\leqslant j\leqslant r$. When $j=r$, our convention gives $R_{r+1}=0$. This proves part (b) and completes the proof of part (a).

Remark 2.1 View $m$ as an indeterminant, so that $r!s_m(r)$ is a polynomial in $m$ over ${\mathbb {Z}}$ of degree $r$. The factorization $r!s_m(r)=\prod _{j=1}^r(a_j+{\mathcal {T}}_{j+1})$ in corollary 1.2 involves the rational functions $a_j+{\mathcal {T}}_{j+1}$. However, theorem 1.1(b) gives ${\mathcal {T}}_{j+1}=\frac {R_{j+1}}{R_{j}}$, so that $a_j+{\mathcal {T}}_{j+1}=\frac {a_{j}R_{j}+R_{j+1}}{R_j}=\frac {b_{j}R_{j-1}}{R_j}$. This determines the numerator and denominator of the rational function $a_j+{\mathcal {T}}_{j+1}$, and explains why we have $\prod _{j=1}^r(a_j+{\mathcal {T}}_{j+1})= \frac {R_0}{R_r}\prod _{j=1}^rb_{j}=r!s_m(r)$. This is different from, but reminiscent of, the ratio $p_{j+1}/p_j$ described on p. 26 of [8]. $\diamond$

Corollary 2.2 If $r,\,m\in {\mathbb {Z}}$ and $0< r< m$, then

\[ s_m(r):=\sum_{i=0}^r\binom{m}{i} =\frac{(r+1)\binom{m}{r+1}}{m-2r+{\mathcal{T}}_1} \quad\text{where } {\mathcal{T}}_1={\vcenter{\hbox{ $\mathcal{K}$}}}_{i=1}^r\frac{2i(r+1-i)}{m-2r+3i}>0. \]

If $r=0$, then $s_m(r)=\frac {(r+1)\binom{m}{r+1}}{m-2r+{\mathcal {T}}_1}$ is true, but ${\mathcal {T}}_1=\mathcal{K}_{i=1}^r\frac {2i(r+1-i)}{m-2r+3i}=0$.

We will need some additional tools such as proposition 2.3 and corollary 2.4 below in order to prove theorem 1.3.

Since $s_m(m-r)=2^m-s_m(r-1)$ approximating $s_m(r)$ for $0\leqslant r\leqslant m$ reduces to approximating $s_m(r)$ for $0\leqslant r\leqslant \lfloor \frac {m}{2}\rfloor$. Hence the hypothesis $r<\frac {m+3}{2}$ in proposition 2.3 and corollary 2.4 is not too restrictive. Proposition 2.3 generalizes [8, Theorem 3.3].

Let ${\mathcal {H}}_j:=\mathcal{K}_{i=1}^j\frac {b_i}{a_i}$ denote the $j$th head of the fraction $\mathcal{K}_{i=1}^r\frac {b_i}{a_i}$, where ${\mathcal {H}}_0=0$.

Proposition 2.3 Let $b_i=2i(r+1-i)$ and $a_i=m-2r+3i$ for $0\leqslant i\leqslant r$. If $r<\frac {m+3}{2}$, then $a_0+{\mathcal {H}}_r=\frac {(r+1)\binom{m}{r+1}}{s_m(r)}$ can be approximated using the following chain of inequalities

\begin{align*} a_0+{\mathcal{H}}_0< a_0+{\mathcal{H}}_2<\cdots< a_0+{\mathcal{H}}_{2\lfloor r/2\rfloor} <& a_0+{\mathcal{H}}_{2\lfloor (r-1)/2\rfloor+1}<\\& \cdots< a_0+{\mathcal{H}}_3< a_0+{\mathcal{H}}_1. \end{align*}

Proof. Note that $r$ equals either $2\lfloor r/2\rfloor$ or $2\lfloor (r-1)/2\rfloor +1$, depending on its parity.

We showed in the proof of theorem 1.1 that $\frac {(r+1)\binom{m}{r+1}}{s_m(r)}=a_0+{\mathcal {H}}_r=a_0+\mathcal{K}_{i=1}^r\frac {b_i}{a_i}$. Since $r<\frac {m+3}{2}$, we have $a_i>0$ and $b_i>0$ for $1\leqslant i\leqslant r$ and hence $\frac {b_i}{a_i}>0$. A straightforward induction (which we omit) depending on the parity of $r$ proves that ${\mathcal {H}}_0<{\mathcal {H}}_2<\cdots <{\mathcal {H}}_{2\lfloor r/2\rfloor }<{\mathcal {H}}_{2\lfloor (r-1)/2\rfloor +1}<\cdots <{\mathcal {H}}_3<{\mathcal {H}}_1$. For example, if $r=3$, then

\[ {\mathcal{H}}_0=0<\frac{b_1}{a_1 +\dfrac{b_2}{a_2}} < \frac{b_1}{a_1 +\dfrac{b_2}{a_2 +\dfrac{b_3}{a_3}}} < \frac{b_1}{a_1}={\mathcal{H}}_1. \]

proves ${\mathcal {H}}_0<{\mathcal {H}}_2<{\mathcal {H}}_3<{\mathcal {H}}_1$ as the tails are positive. Adding $a_0$ proves the claim.

In asking whether $g_{\omega,m}(r)$ is a unimodal function, it is natural to consider the ratio $g_{\omega,m}(r+1)/g_{\omega,m}(r)$ of successive terms. This suggests defining

(2.3)\begin{equation} t(r)=t_{m}(r):= \frac{s_m(r+1)}{s_m(r)}=1+\frac{\binom{m}{r+1}}{s_m(r)}=1+\frac{Q}{r+1}. \end{equation}

We will prove in lemma 3.1 that $t(r)$ is a strictly decreasing function that determines when $g_{\omega,m}(r)$ is increasing or decreasing, and $t_{m}(r_0-1)\geqslant \omega >t_{m}(r_0)$ determines $r_0$.

Corollary 2.4 We have $m-2r\leqslant \frac {(r+1)\binom{m}{r+1}}{s_m(r)}$ for $r\geqslant 0$, and

\[ \frac {(r+1)\binom{m}{r+1}}{s_m(r)}\leqslant m-2r+\frac {2r}{m-2r+3}\quad{\rm for}\quad 0\leqslant r<\frac{m+3}{2}. \]

Hence $\frac {m+2}{r+1}\leqslant t_m(r)+1$ for $r\geqslant 0$, and

\[ \frac{m+2}{r+1}\leqslant t_m(r)+1\leqslant\frac{m+2}{r+1}+\frac{2r}{(r+1)(m-2r+3)} \quad\text{for $0\leqslant r<\frac{m+3}{2}$.} \]

Also $\frac {m+2}{r+1}< t_m(r)+1$ for $r>0$, and the above upper bound is strict for $1< r<\frac {m+3}{2}$.

Proof. We proved $Q=\frac {(r+1)\binom{m}{r}}{s_m(r)}=(m-2r)+\mathcal{K}_{i=1}^r\frac {2i(r+1-i)}{m-2r+3i}$ in theorem 1.1. Hence $m-2r=\frac {(r+1)\binom{m}{r+1}}{s_m(r)}$ if $r=0$ and $m-2r<\frac {(r+1)\binom{m}{r+1}}{s_m(r)}$ if $1\leqslant r<\frac {m+3}{2}$ by proposition 2.3. Clearly $m-2r<0\le \frac {(r+1)\binom{m}{r+1}}{s_m(r)}$ if $\frac {m+3}{2}\le r\le m$. Similarly $\frac {(r+1)\binom{m}{r+1}}{s_m(r)}= m-2r+\frac {2r}{m-2r+3}$ if $r=0,\,1$, and again proposition 2.3 shows that ${\frac {(r+1)\binom{m}{r+1}}{s_m(r)}< m-2r+\frac {2r}{m-2r+3}}$ if $1< r<\frac {m+3}{2}$. The remaining inequalities (and equalities) follow similarly since $t_m(r)+1=2+\frac {\binom{m}{r+1}}{s_m(r)}$ and $2+\frac {m-2r}{r+1}=\frac {m+2}{r+1}$.

3. Estimating the maximizing input $r_0$

Fix $\omega >1$. In this section we consider the function $g(r)=g_{\omega,m}(r)$ given by (1.1). As seen in table 1, it is easy to compute $g(r)$ if $r$ is near $0$ or $m$. For $m$ large and $r$ near $0$, we have ‘sub-exponential’ growth $g(r)\approx \frac {m^r}{r!\omega ^r}$. Similarly for $r$ near $m$, we have exponential decay $g(r)\approx \frac {2^m}{\omega ^r}$. The middle values require more thought.

Table 1. Values of $g_{w,m}(r)$

On the other hand, the plots $y=g(r)$, $0\leqslant r\leqslant m$, exhibit a remarkable visual symmetry when $m$ is large. The relation $s_m(m-r)=2^m-s_m(r-1)$ and the distorting scale factor of $\omega ^{-r}$ shape the plots. The examples in figure 1 show an approximate left–right symmetry about a maximizing input $r\approx \frac {m}{\omega +1}$. It surprised the authors that in many cases there exists a simple exact formula for the maximizing input (it is usually unique as corollary 3.2 suggests). In figure 1 we have used different scale factors for the $y$-axes. The maximum value of $g_{\omega,m}(r)$ varies considerably as $\omega$ varies (c.f. lemma 5.1), so we scaled the maxima (rounded to the nearest integer) to the same height.

Lemma 3.1 Recall that $g(r)=\omega ^{-r} s_m(r)$ by (1.1) and $t(r)=\frac {s_m(r+1)}{s_m(r)}$ by (2.3).

1. (a) $t(r-1)>t(r)>\frac {m-r}{r+1}$ for $0\leqslant r\leqslant m$ where $t(-1):=\infty$;

2. (b) $g(r)< g(r+1)$ if and only if $t(r)>\omega$;

3. (c) $g(r)\leqslant g(r+1)$ if and only if $t(r)\geqslant \omega$;

4. (d) $g(r)> g(r+1)$ if and only if $\omega > t(r)$;

5. (e) $g(r)\geqslant g(r+1)$ if and only if $\omega \geqslant t(r)$;

6. (f) if $\omega >1$ then some $r_0\in \{0,\,\ldots,\,m\}$ satisfies $t(r_0-1)\geqslant \omega > t(r_0)$, and this condition is equivalent to

   \[ g(0)<\cdots < g(r_0-1)\leqslant g(r_0) \quad {\rm and}\quad g(r_0)>\cdots >g(m). \]

Proof. (a) We prove, using induction on $r$, that $t(r-1)>t(r)>\binom{m}{r+1}/\binom{m}{r}$ holds for $0\leqslant r\leqslant m$. These inequalities are clear for $r=0$ as $\infty >m+1>m$. For real numbers $\alpha,\,\beta,\,\gamma,\,\delta >0$, we have $\alpha \delta -\beta \gamma >0$ if and only if $\frac {\alpha }{\beta }>\frac {\alpha +\gamma }{\beta +\delta }>\frac {\gamma }{\delta }$; that is, the mediant $\frac {\alpha +\gamma }{\beta +\delta }$ of $\frac {\alpha }{\beta }$ and $\frac {\gamma }{\delta }$ lies strictly between $\frac {\alpha }{\beta }$ and $\frac {\gamma }{\delta }$. If $0< r\leqslant m$, then by induction

\[ t(r-1)>\frac{\binom{m}{r}}{\binom{m}{r-1}} =\frac{m-r+1}{r}>\frac{m-r}{r+1}=\frac{\binom{m}{r+1}}{\binom{m}{r}}. \]

Applying the ‘mediant sum’ to $t(r-1)=\frac {s_m(r)}{s_m(r-1)}>\frac {\binom{m}{r+1}}{\binom{m}{r}}$ gives

\[ \frac{s_m(r)}{s_m(r-1)}>\frac{s_m(r)+\binom{m}{r+1}}{s_m(r-1)+\binom{m}{r}} =\frac{s_m(r+1)}{s_m(r)}=t(r) >\frac{\binom{m}{r+1}}{\binom{m}{r}}. \]

Therefore $t(r-1)>t(r)>\binom{m}{r+1}/\binom{m}{r}=\frac {m-r}{r+1}$ completing the induction, and proving (a).

(b,c,d,e) The following are equivalent: $g(r)< g(r+1)$; $\omega s_m(r)< s_m(r+1)$; and $\omega < t(r)$. The other claims are proved similarly by replacing < with $\leqslant$, >, $\geqslant$.

(f) Observe that $t(m)=\frac {s_m(m+1)}{s_m(m)}=\frac {2^m}{2^m}=1$. By part (a), the function $y=t(r)$ is decreasing for $-1\leqslant r\leqslant m$. Since $\omega >1$, there exists an integer $r_0\in \{0,\,\ldots,\,m\}$ such that $\infty =t(-1)>\cdots >t(r_0-1)\geqslant \omega > t(r_0)>\cdots > t(m)=1$. By parts (b,c,d,e) an equivalent condition is $g(0)<\cdots < g(r_0-1)\leqslant g(r_0)$ and $g(r_0)>\cdots >g(m)$.

The following is an immediate corollary of lemma 3.1(f).

Corollary 3.2 If $t(r_0-1)>\omega$, then the function $g(r)$ in (1.1) has a unique maximum at $r_0$. If $t(r_0-1)=\omega$, then $g(r)$ has two equal maxima, one at $r_0-1$ and one at $r_0$.

As an application of theorem 1.1 we show that the largest maximizing input $r_0$ for $g_{\omega,m}(r)$ satisfies $\lfloor \frac {m+2}{\omega +1}\rfloor \leqslant r_0$. There are at most two maximizing inputs by corollary 3.2.

Lemma 3.3 Suppose that $\omega >1$ and $m\in {\mathbb {Z}}$, $m\geqslant 0$. If $r':=\lfloor \frac {m+2}{\omega +1}\rfloor$, then

\[ g(-1)< g(0)<\dots < g(r'-1)\leqslant g(r'),\quad {\rm and}\quad g(-1)<\dots < g(r'-1)< g(r') \]

if $r'>1$ or $\omega \ne m+1$.

Proof. The result is clear when $r'=0$. If $r'=1$, then $r'\leqslant \frac {m+2}{\omega +1}$ gives $\omega \leqslant m+1$ or $g(0)\leqslant g(1)$. Hence $g(0)< g(1)$ if $\omega \ne m+1$. Suppose that $r'>1$. By lemma 3.1(c,f) the chain $g(0)<\dots < g(r')$ is equivalent to $g(r'-1)< g(r')$, that is $t(r'-1)> \omega$. However, $t(r'-1)+1>\frac {m+2}{r'}$ by corollary 2.4 and $r'\leqslant \frac {m+2}{\omega +1}$ implies $\frac {m+2}{r'}\geqslant \omega +1$. Hence $t(r'-1)+1>\omega +1$, so that $t(r'-1)> \omega$ as desired.

Proof Proof of theorem 1.4

Suppose that $\omega \in \{3,\,4,\,\ldots \}$. Then $g(0)<\dots < g(r'-1)\leqslant g(r')$ by lemma 3.3 with strictness when $\omega \ne m+1$. If $\omega =m+1$, then $r'=\lfloor \frac {m+2}{\omega +1}\rfloor =1$ and $g(0)=g(1)$ as claimed. It remains to show that $g(r')>g(r'+1)>\cdots >g(m)$. However, we need only prove that $g(r')> g(r'+1)$ by lemma 3.1(f), or equivalently $\omega > t(r')$ by lemma 3.1(d).

Clearly $\omega \geqslant 3$ implies $r'\leqslant \frac {m+2}{\omega +1}\leqslant \frac {m+2}{4}$. As $0\leqslant r'<\frac {m+3}{2}$, corollary 2.4 gives

\[ \frac{m+2}{r'+1}+\frac{2r'}{(r'+1)(m-2r'+3)}\geqslant t(r')+1. \]

Hence $\omega +1> t(r')+1$ holds if $\omega +1> \frac {m+2}{r'+1}+\frac {2r'}{(r'+1)(m-2r'+3)}$. Since $\omega +1$ is an integer, we have $m+2=r'(\omega +1)+c$ where $0\leqslant c\leqslant \omega$. It follows from $0\leqslant r'\leqslant \frac {m+2}{4}$ that $\frac {2r'}{m-2r'+3}<1$. This inequality and $m+2\leqslant r'(\omega +1)+\omega$ gives

\[ m+2+\frac{2r'}{m-2r'+3}< r'(\omega+1)+\omega+1=(r'+1)(\omega+1). \]

Thus $\omega +1>\frac {m+2}{r'+1}+\frac {2r'}{(r'+1)(m-2r'+3)}\geqslant t(r')+1$, so $\omega >t(r')$ as required.

Remark 3.4 The proof of theorem 1.4 can be adapted to the case $\omega =2$. If $m+2=3r'+c$ where $c\leqslant \omega -1=1$, then $\frac {2r'}{m-2r'+3}=\frac {2r'}{r'+c+1}<2$, and if $c=\omega =2$, then a sharper ${\mathcal {H}}_2$-bound must be used. This leads to a much shorter proof than [4, Theorem 1.1]. $\diamond$

4. Non-integral values of $\omega$

In this section, we prove that the maximum value of $g(r)$ is $g(r')$ or $g(r'+1)$ if $\omega \geqslant \sqrt {3}$. Before proving this result (theorem 1.3), we shall prove two preliminary lemmas.

Lemma 4.1 Suppose that $\omega >1$ and $r':=\lfloor \frac {m+2}{\omega +1}\rfloor$. If $\frac {m+2}{r'+1}\geqslant \sqrt {3}+1$, then

\[ g({-}1)< g(0)<\cdots< g(r'-1)\leqslant g(r'), \quad\text{and}\quad g(r'+1)>g(r'+2)>\cdots>g(m). \]

Proof. It suffices, by lemma 3.1(f) and lemma 3.3 to prove that $g(r'+1)>g(r'+2)$. The strategy is to show $\omega >t(r'+1)$, that is $\omega +1>t(r'+1)+1$. However, $\omega +1>\frac {m+2}{r'+1}$, so it suffices to prove that $\frac {m+2}{r'+1}\geqslant t(r'+1)+1$. Since $r'+1\leqslant \frac {m+2}{\sqrt {3}+1}<\frac {m+2}{2}$, we can use corollary 2.4 and just prove that $\frac {m+2}{r'+1}\geqslant \frac {m+2}{r'+2}+\frac {2r'+2}{(r'+2)(m-2r'+1)}$. This inequality is equivalent to $\frac {m+2}{r'+1}\geqslant \frac {2r'+2}{m-2r'+1}$. However, $\frac {m+2}{r'+1}\geqslant \sqrt {3}+1$, so we need only show that $\sqrt {3}+1\geqslant \frac {2(r'+1)}{m-2r'+1}$, or equivalently $m-2r'+1\geqslant (\sqrt {3}-1)(r'+1)$. This is true since $\frac {m+2}{r'+1}\geqslant \sqrt {3}+1$ implies $m-2r'+1\geqslant (\sqrt {3}-1)r'+\sqrt {3}>(\sqrt {3}-1)(r'+1)$.

Remark 4.2 The strict inequality $g(r'-1)< g(r')$ holds by lemma 3.3 if $r'>1$ or $\omega \ne m+1$. It holds vacuously for $r'=0$. Hence adding the additional hypothesis that $\omega \ne m+1$ if $r'=1$ to lemma 4.1 (and theorem 1.3), we may conclude that the inequality $g(r'-1)\leqslant g(r')$ is strict.

Remark 4.3 In lemma 4.1, the maximum can occur at $r'+1$. If $\omega =2.5$ and $m=8$, then $r'=\lfloor \frac {10}{3.5}\rfloor =2$ and $\frac {m+2}{r'+1}=\frac {10}{3}\geqslant \sqrt {3}+1$ however $g_{2.5,8}(2)=\frac {740}{125}< \frac {744}{125}=g_{2.5,8}(3)$. $\diamond$

Remark 4.4 The gap between $r'$ and the largest maximizing input $r_0$ can be arbitrarily large if $\omega$ is close to 1. For $\omega >1$, we have $r'=\lfloor \frac {m+2}{\omega +1}\rfloor <\frac {m+2}{2}$. If $1<\omega \leqslant \frac {1}{1-2^{-m}}$, then $g(m-1)\leqslant g(m)$, so $r_0=m$. Hence $r_0-r'>\frac {m-2}{2}$.

Remark 4.5 Since $r'\leqslant \lfloor \frac {m+2}{\omega +1}\rfloor < r'+1$, we see that $r'+1\approx \frac {m+2}{\omega +1}$, so that $\frac {m+2}{r'+1}\approx \omega +1$. Thus lemma 4.1 suggests that if $\omega \gtrsim \sqrt {3}$, then $g_{\omega,m}(r)$ may have a maximum at $r'$ or $r'+1$. This heuristic reasoning is made rigorous in theorem 1.3. $\diamond$

Remark 4.6 Theorem 1.1 can be rephrased as $t_m(r)=\frac {s_m(r+1)}{s_m(r)}=\frac {m-r+1}{r+1}+\frac {{\mathcal {K}}_m(r)}{r+1}$ where

(4.1)\begin{equation} {\mathcal{K}}_m(r)=\mathop{\vcenter{\hbox{ $\mathcal{K}$}}}_{i=1}^{r} \frac{2i(r+1-i)}{m-2r+3i}= \frac{\displaystyle 2r}{\displaystyle m-2r+3 +\frac{\displaystyle 4r-4}{\displaystyle m-2r+6 +\frac{\displaystyle 6r-12}{\llap{{\ddots}} {{\displaystyle m+r-3 + \frac{2r}{m+r}}}}}}. \end{equation}

The following lemma repeatedly uses the expression $\omega >t_m(r+1)$. This is equivalent to $\omega >\frac {m-r}{r+2}+\frac {{\mathcal {K}}_m(r+1)}{r+2}$, that is $(\omega +1)(r+2)>m+2+{\mathcal {K}}_m(r+1)$. $\diamond$

Lemma 4.7 Let $m\in \{0,\,1,\,\ldots \}$ and $r'=\lfloor \frac {m+2}{\omega +1}\rfloor$. If any of the following three conditions are met, then $g_{\omega,m}(r'+1)>\cdots >g_{\omega,m}(m)$ holds: (a) $\omega \geqslant 2$, or  (b) $\omega \geqslant \frac {1+\sqrt {97}}{6}$ and $r'\ne 2$, or  (c) $\omega \geqslant \sqrt {3}$ and $r'\not \in \{2,\,3\}$.

Proof. The conclusion $g_{\omega,m}(r'+1)>\cdots >g_{\omega,m}(m)$ holds trivially if $r'+1\geqslant m$. Suppose henceforth that $r'+1< m$. Except for the excluded values of $r',\,\omega$, we will prove that $g_{\omega,m}(r'+1)>g_{\omega,m}(r'+2)$ holds, as this implies $g_{\omega,m}(r'+1)>\cdots >g_{\omega,m}(m)$ by lemma 3.1(f). Hence we must prove that $\omega >t_m(r'+1)$ by lemma 3.1(d).

Recall that $r'\leqslant \frac {m+2}{\omega +1}< r'+1$. If $r'=0$, then $m+2<\omega +1$, that is $\omega >m+1>t(1)$ as desired. Suppose now that $r'=1$. There is nothing to prove if $m=r'+1=2$. Assume that $m>2$. Since $m+2<2(\omega +1)$, we have $2< m<2\omega$. The last line of remark 4.6 and (4.1) give the desired inequality:

\[ \omega>\frac{m}{2}\geqslant\frac{m-1}{3}+\frac{4}{3\left(m-1+\dfrac{4}{m+2}\right)}=t_m(2). \]

In summary, $g_{\omega,m}(r'+1)>\cdots >g_{\omega,m}(m)$ holds for all $\omega >1$ if $r'\in \{0,\,1\}$.

We next prove $g_{\omega,m}(r'+1)>g_{\omega,m}(r'+2)$, or equivalently $\omega >t_m(r'+1)$ for $r'$ large enough, depending on $\omega$. We must prove that $(\omega +1)(r'+2)>m+2+\mathcal{K}_m(r'+1)$ by remark 4.6. Writing $m+2=(\omega +1)(r'+\varepsilon )$ where $0\leqslant \varepsilon <1$, our goal, therefore, is to show $(\omega +1)(2-\varepsilon )>{\mathcal {K}}_m(r'+1)$. Using (4.1) gives

\[ {\mathcal{K}}_m(r'+1)=\frac{2(r'+1)}{m-2(r'+1)+3+{\mathcal{T}}} =\frac{2(r'+1)}{(\omega+1)(r'+\varepsilon)-2(r'+1)+1+{\mathcal{T}}} \]

where ${\mathcal {T}}>0$ by theorem 1.1 as $r'>0$. Rewriting the denominator using

\[ (\omega+1)(r'+\varepsilon)-2(r'+1)=(\omega-1)(r'+1)-(\omega+1)(1-\varepsilon), \]

our goal $(\omega +1)(2-\varepsilon )>{\mathcal {K}}_m(r'+1)$ becomes

\[ (\omega+1)(2-\varepsilon)\left[(\omega-1)(r'+1)-(\omega+1)(1-\varepsilon)+1+{\mathcal{T}}\right] >2(r'+1). \]

Dividing by $(2-\varepsilon )(r'+1)$ and rearranging gives

\[ (\omega^2-1)+\frac{(\omega+1)(1+{\mathcal{T}})}{r'+1}>\frac{2}{2-\varepsilon} +\frac{(\omega+1)^2(1-\varepsilon)}{r'+1}. \]

This inequality may be written $(\omega ^2-1)+\lambda >\frac {2}{2-\varepsilon }+\mu (1-\varepsilon )$ where $\lambda =\frac {(\omega +1)(1+{\mathcal {T}})}{r'+1}>0$ and $\mu =\frac {(\omega +1)^2}{r'+1}>0$. We view $f(\varepsilon ):=\frac {2}{2-\varepsilon }+\mu (1-\varepsilon )$ as a function of a real variable $\varepsilon$ where $0\leqslant \varepsilon <1$. However, $f(\varepsilon )$ is concave as the second derivative $f''(\varepsilon )=\frac {4}{(2-\varepsilon )^3}$ is positive for $0\leqslant \varepsilon <1$. Hence the maximum value occurs at an end point: either $f(0)=1+\mu$ or $f(1)=2$. Therefore, it suffices to prove that $(\omega ^2-1)+\lambda >\max \{2,\,1+\mu \}$.

If $2\geqslant 1+\mu$, then the desired bound $(\omega ^2-3)+\lambda >0$ holds as $\omega \geqslant \sqrt {3}$. Suppose now that $2<1+\mu$. We must show $(\omega ^2-1)+\lambda >1+\mu$, that is $\omega ^2-2>\mu -\lambda =\frac {(\omega +1)(\omega -{\mathcal {T}})}{r'+1}$. Since ${\mathcal {T}}>0$, a stronger inequality (that implies this) is $\omega ^2-2\geqslant \frac {(\omega +1)\omega }{r'+1}$. The (equivalent) quadratic inequality $r'\omega ^2-\omega -2(r'+1)\geqslant 0$ in $\omega$ is true provided $\omega \geqslant \frac {1+\sqrt {1+8r'(r'+1)}}{2r'}$. This says $\omega \geqslant 2$ if $r'=2$, and $\omega \geqslant \frac {1+\sqrt {97}}{6}$ if $r'=3$. If $r'\geqslant 4$, we have

\[ \frac{1+\sqrt{1+8r'(r'+1)}}{2r'}= \frac{1}{2r'}+\sqrt{\frac{1}{4(r')^2}+2\left(1+\frac{1}{r'}\right)} \leqslant\frac{1}{8}+\sqrt{\frac{1}{64}+\frac{5}{2}}<\sqrt{3}. \]

The conclusion now follows from the fact that $2>\frac {1+\sqrt {97}}{6} >\sqrt {3}$.

Proof Proof of theorem 1.3

By lemma 4.1 it suffices to show that $g(r'+1)>g(r'+2)$ holds when $r'+1< m$ and $\omega \geqslant \sqrt {3}$. By lemma 4.7(a), we can assume that $\sqrt {3}\leqslant \omega <2$ and $r'\in \{2,\,3\}$. For these choices of $\omega$ and $r'$, we must show that $\omega >t_m(r'+1)$ by lemma 3.1 for all permissible choices of $m$. Since $(\omega +1)r'\leqslant m+2<(\omega +1)(r'+1)$, when $r'=2$ we have $5<2(\sqrt {3}+1)\leqslant m+2<9$ so that $4\leqslant m\leqslant 6$. However, $t_m(3)$ equals $\tfrac {16}{15},\,\tfrac {31}{26},\,\tfrac {19}{14}$ for these values of $m$. Thus $\sqrt {3}>t_m(3)$ holds as desired. Similarly, if $r'=3$, then $8<3(\sqrt {3}+1)\leqslant m+2<12$ so that $7\leqslant m\leqslant 9$. In this case $t_m(4)$ equals $\tfrac {40}{33},\,\tfrac {219}{163},\,\tfrac {191}{128}$ for these values of $m$. In each case $\sqrt {3}>t_m(4)$, so the proof is complete.

Remark 4.8 We place remark 4.4 in context. The conclusion of theorem 1.3 remains true for values of $\omega$ smaller than $\sqrt {3}$ and not ‘too close to 1’ and $m$ is ‘sufficiently large’. Indeed, by adapting the proof of lemma 4.7 we can show there exists a sufficiently large integer $d$ such that $m > d^4$ and $\omega > 1 + \frac {1}{d}$ implies $g(r'+d) > g(r'+d+1)$. This shows that $r' \leqslant r_0 \leqslant r' +d$, so $r_0-r'\leqslant d$. We omit the technical proof of this fact. $\diamond$

Remark 4.9 The sequence, $a_0+{\mathcal {H}}_1,\,\ldots,\,a_0+{\mathcal {H}}_r$ terminates at $\frac {r+1}{s_m(r)}\binom{m}{r+1}$ by theorem 1.1. We will not comment here on how quickly the alternating sequence in proposition 2.3 converges when $r<\frac {m+3}{2}$. If $r=m$, then $a_0=-m$ and $\frac {m+1}{s_m(m+1)}\binom{m}{m+1}=0$, so theorem 1.1 gives the curious identity ${\mathcal {H}}_m=\mathcal{K}_{i=1}^m\frac {2i(m+1-i)}{3i-m}=m$. If $\omega$ is less than $\sqrt {3}$ and ‘not too close to 1’, then we believe that $r_0$ is approximately $\lfloor \frac {m+2}{\omega +1}+ \frac {2}{\omega ^2 -1}\rfloor$, c.f. remark 4.8.

5. Estimating the maximum value of $g_{\omega,m}(r)$

In this section we relate the size of the maximum value $g_{\omega,m}(r_0)$ to the size of the binomial coefficient $\binom{m}{r_0}$. In the case that we know a formula for a maximizing input $r_0$, we can readily estimate $g_{\omega,m}(r_0)$ using approximations, such as [10], for binomial coefficients.

Lemma 5.1 The maximum value $g_{\omega,m}(r_0)$ of $g_{\omega,m}(r)$, $0\leqslant r\leqslant m$, satisfies

\[ \frac{1}{(\omega-1)\omega^{r_0}}\binom{m}{r_0+1} < g_{\omega,m}(r_0) \leqslant \frac{1}{(\omega-1)\omega^{r_0-1}}\binom{m}{r_0}. \]

Proof. Since $g(r_0)$ is a maximum value, we have $g(r_0-1)\leqslant g(r_0)$. This is equivalent to $(\omega -1)s_m(r_0-1)\leqslant \binom{m}{r_0}$ as $s_m(r_0)=s_m(r_0-1)+\binom{m}{r_0}$. Adding $(\omega -1)\binom{m}{r_0}$ to both sides gives the equivalent inequality $(\omega -1) s_m(r_0)\leqslant \omega \binom{m}{r_0}$. This proves the upper bound.

Similar reasoning shows that the following are equivalent: (a) $g(r_0)> g(r_0+1)$; (b) $(\omega -1)s_m(r_0)> \binom{m}{r_0+1}$; and (c) $g_{\omega,m}(r_0)>\frac {1}{(\omega -1)\omega ^{r_0}}\binom{m}{r_0+1}$.

In theorem 1.3 the maximizing input $r_0$ satisfies $r_0=r'+d$ where $d\in \{0,\,1\}$. In such cases when $r_0$ and $d$ are known, we can bound the maximum $g_{\omega,m}(r_0)$ as follows.

Corollary 5.2 Set $r' := \lfloor \frac {m+2}{\omega +1}\rfloor$ and $k := m+2 - (\omega +1)r'$. Suppose that $r_0 = r'+d$ and $G=\frac {1}{(\omega -1)\omega ^{r_0-1}}\binom{m}{r_0}$. Then

\[ 0\leqslant k< \omega +1, d\geqslant 0\quad {\rm and}\quad 1-\frac {1+d-\frac {d+2-k}{\omega }}{r_0+1} <\frac {g_{\omega,m}(r_0)}{G} \leqslant 1. \]

Proof. By lemma 3.3, $r_0 = r'+d$ where $d\in \{0,\,1,\,\ldots \}$. Since $r' = \lfloor \frac {m+2}{\omega +1}\rfloor$, we have $m+2 = (\omega +1)r'+k$ where $0\leqslant k< \omega +1$. The result follows from lemma 5.1 and $m = (\omega +1)(r_0-d)+k-2$ as $\binom{m}{r_0+1}=\frac {m-r_0}{r_0+1}\binom{m}{r_0}$ and $\frac {m-r_0}{r_0+1}$ equals

\[ \frac{\omega(r_0-d)-d+k-2}{r_0+1} =\omega-\frac{\omega+\omega d+d+2-k}{r_0+1} =\omega\left(1-\frac{1+d+\frac{d+2-k}{\omega}}{r_0+1}\right). \]

The following remark is an application of the Chernoff bound, c.f. [11, Section 4]. Unlike theorem 1.1, it requires the cumulative distribution function $\Phi (x)$, which is a non-elementary integral, to approximate $s_m(r)$. It seems to give better approximations only for values of $r$ near $\frac {m}{2}$, see remark 5.4.

Remark 5.3 We show how the Berry–Esseen inequality for a sum of binomial random variables can be used to approximate $s_m(r)$. Let $B_1,\,\ldots,\,B_m$ be independent identically distributed binomial variables with a parameter $p$ where $0< p<1$, so that $P(B_i=1)=p$ and $P(B_i=0)=q:= 1-p$. Let $X_i:= B_i-p$ and $X:=\frac {1}{\sqrt {mpq}}(\sum _{i=1}^m X_i)$. Then

\[ E(X_i)=E(B_i)-p=0,\quad E(X_i^2)=pq,\quad {\rm and}\quad E(|X_i|^3)=pq(p^2+q^2). \]

Hence $E(X)=\frac {1}{\sqrt {mpq}}(\sum _{i=1}^m E(X_i))=0$ and $E(X^2)=\frac {1}{mpq}(\sum _{i=1}^m E(X_i^2))=1$. By [7, Theorem 2] the Berry–Esseen inequality applied to $X$ states that

\begin{align*} |P(X\leqslant x)-\Phi(x)|& \leqslant\frac{Cpq(p^2+q^2)}{(pq)^{3/2}\sqrt{m}} \\ & =\frac{C(p^2+q^2)}{\sqrt{mpq}} \quad\text{for all } m\in\{1,2,\ldots\} \text{ and } x\in{\mathbb{R}}, \end{align*}

where the constant $C:= 0.4215$ is close to the lower bound $C_0=\frac {10+\sqrt {3}}{6\sqrt {2\pi }}=0.4097\cdots$ and $\Phi (x)=\frac {1}{\sqrt {2\pi }}\int _{-\infty }^x e^{-t^2/2}\,dt=\frac {1}{2}\left (1+{\rm erf}\left (\frac {x}{\sqrt {2}}\right )\right )$ is the cumulative distribution function for standard normal distribution.

Writing $B=\sum _{i=1}^mB_i$ we have $P(B\leqslant b)=\sum _{i=0}^{\lfloor b\rfloor }\binom{m}{i}p^iq^{m-i}$ for $b\in {\mathbb {R}}$. Thus $X=\frac {B-mp}{\sqrt {mpq}}$ and $x=\frac {b-mp}{\sqrt {mpq}}$ satisfy

\[ \left|P(B\leqslant b)-\Phi\left(\frac{b-mp}{\sqrt{mpq}}\right)\right| \leqslant\frac{C(p^2+q^2)}{\sqrt{mpq}} \quad\text{for all $m\in\{1,\,2,\,\ldots\}$ and $b\in{\mathbb{R}}$} .\]

Setting $p=q=\tfrac {1}{2}$, and taking $b=r\in \{0,\,1,\,\ldots,\,m\}$ shows

\[ \left|2^{{-}m}s_m(r)-\Phi\left(\frac{2r-m}{\sqrt{m}}\right)\right| \leqslant\frac{0.4215}{\sqrt{m}} \quad\text{for } m\in\{1,2,\ldots\}. \]

Remark 5.4 Let $a_0+{\mathcal {H}}_k$ be the generalized continued fraction approximation to $\frac {(r+1)\binom{m}{r+1}}{s_m(r)}$ suggested by theorem 1.1, where ${\mathcal {H}}_k:=\mathcal{K}_{i=1}^k\frac {b_i}{a_i}$, and $k$ is the depth of the generalized continued fraction. We compare the following two quantities:

\[ e_{m,r,k}:= 1-\frac{(r+1)\binom{m}{r+1}}{(a_0+{\mathcal{H}}_k)s_m(r)} \quad\text{and}\quad E_{m,r}:=\left|1-\frac{2^m\Phi(\frac{2r-m}{\sqrt{m}})}{s_m(r)} \right| \leqslant\frac{0.4215\cdot 2^m}{\sqrt{m}\,s_m(r)}. \]

The sign of $e_{m,r,k}$ is governed by the parity of $k$ by proposition 2.3. We shall assume that $r\leqslant \frac {m}{2}$. As $\frac {2^m}{s_m(r)}$ ranges from $2^m$ to about 2 as $r$ ranges from 0 to $\lfloor \frac {m}{2}\rfloor$, it is clear that the upper bound for $E_{m,r}$ will be huge unless $r$ satisfies $\frac {m-\varepsilon }{2}\le r\le \frac {m}{2}$ where $\varepsilon$ is ‘small’ compared to $m$. By contrast, the computer code [3] verifies that the same is true for $E_{m,r}$, and shows that $|e_{m,r,k}|$ is small, even when $k$ is tiny, when $0\le r<\frac {m-\varepsilon }{2}$, see table 2. Hence the ‘generalized continued fraction’ approximation to $s_m(r)$ is complementary to the ‘statistical’ approximation, as shown in table 2. The reader can extend table 2 by running the code [3] written in the Magma [1] language, using the online calculator http://magma.maths.usyd.edu.au/calc/, for example.

Table 2. Upper bounds for $|e_{m,r,k}|$ and $E_{m,r}$ for $m=10^4$