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AN EFFECTIVE BENFORD’S LAW FOR STRICTLY INCREASING AND LOG-CONCAVE SEQUENCES

Part of: Sequences and sets Multiplicative number theory Probabilistic theory: distribution modulo $1$; metric theory of algorithms

Published online by Cambridge University Press:  21 October 2026

TUYUN LAN Open the ORCID record for TUYUN LAN [Opens in a new window]  and
NIAN HONG ZHOU Open the ORCID record for NIAN HONG ZHOU [Opens in a new window]
TUYUN LAN
Affiliation:
School of Mathematics and Statistics, Guangxi Normal University , Guilin 541004, Guangxi, PR China e-mail: TuyunLan@outlook.com
NIAN HONG ZHOU*
Affiliation:
School of Mathematics and Statistics, Guangxi Normal University , Guilin 541004, Guangxi, PR China
*
e-mail: nianhongzhou@outlook.com
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Abstract

We establish an effective form of Benford’s law for a class of strictly increasing and log-concave sequences $(f_n)_{n\geq 0}$. We apply our result to the partition and plane partition functions, obtaining improvements on bounds due to Iyer for the first occurrence of prescribed leading digit strings.

Keywords

Benford’s lawleading digitsdiscrepancylog-concavity

MSC classification

Primary: 11K38: Irregularities of distribution, discrepancy
Secondary: 11B83: Special sequences and polynomials 11N64: Other results on the distribution of values or the characterization of arithmetic functions

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Type
Research Article
Information
Bulletin of the Australian Mathematical Society , First View , pp. 1 - 11
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Copyright
© The Author(s), 2026. Published by Cambridge University Press on behalf of Australian Mathematical Publishing Association Inc

1 Introduction

1.1 Main results

Let $b\geq 2$ b greater than or equals 2 be an integer and let $w\geq 1$ w greater than or equals 1 be an integer with base-b expansion

$$ \begin{align*} w=\sum_{0\leq i<t} w_i b^i, \end{align*} $$ w equals sigma summation Underscript 0 less than or equals i less than t Endscripts w Subscript i Baseline b Superscript i Baseline comma

where $t\geq 1$ t greater than or equals 1 , $w_i\in \{0,\ldots ,b-1\}$ w Subscript i Baseline element of StartSet 0 comma ellipsis comma b minus 1 EndSet and $w_{t-1}\neq 0$ w Subscript t minus 1 Baseline not equals 0 . Thus, w represents a base-b string of length t and

$$ \begin{align*} b^{t-1}\leq w<b^t. \end{align*} $$ b Superscript t minus 1 Baseline less than or equals w less than b Superscript t Baseline period

Throughout this paper, let $f=(f_k)_{k\geq 0}$ f equals left parenthesis f Subscript k Baseline right parenthesis Subscript k greater than or equals 0 be a sequence of positive real numbers. We say that $f_k$ f Subscript k begins with the base-b string w if there exists an integer h such that

$$ \begin{align*} wb^h\leq f_k<(w+1)b^h. \end{align*} $$ w b Superscript h Baseline less than or equals f Subscript k Baseline less than left parenthesis w plus 1 right parenthesis b Superscript h Baseline period

For $x\geq 0$ x greater than or equals 0 , define

$$ \begin{align*} S_{w,b,f}(x) := \#\{k:0\leq k<x, f_k \text{ begins with the base-}b \text{ string } w\}. \end{align*} $$ upper S Subscript w comma b comma f Baseline left parenthesis x right parenthesis colon equals number sign left brace k colon 0 less than or equals k less than x comma f Subscript k Baseline begins with the base minus b string w right brace period

We say that f satisfies Benford’s law in base b if, for every base-b string w,

$$ \begin{align*} \lim_{x\to\infty}\frac{S_{w,\,b,f}(x)}{x} = \rho_{w,\,b}, \end{align*} $$ limit Underscript x right arrow infinity Endscripts StartFraction upper S Subscript w comma b comma f Baseline left parenthesis x right parenthesis Over x EndFraction equals rho Subscript w comma b Baseline comma

where $\log _b$ log Subscript b denotes the logarithm to base b and

$$ \begin{align*} \rho_{w,\,b} := \log_b(w+1)-\log_b w = \log_b\bigg(1+\frac{1}{w}\bigg). \end{align*} $$ rho Subscript w comma b Baseline colon equals log Subscript b Baseline left parenthesis w plus 1 right parenthesis minus log Subscript b Baseline w equals log Subscript b Baseline left parenthesis 1 plus StartFraction 1 Over w EndFraction right parenthesis period

Let $n_0$ n 0 and n be nonnegative integers with $n\geq n_0+2$ n greater than or equals n 0 plus 2 . We say that a sequence $(f_k)_{n_0\leq k\leq n}$ left parenthesis f Subscript k Baseline right parenthesis Subscript n 0 less than or equals k less than or equals n of positive real numbers is strictly increasing and log-concave if

$$ \begin{align*} f_k>f_{k-1}\quad (n_0<k\le n), \end{align*} $$ f Subscript k Baseline greater than f Subscript k minus 1 Baseline left parenthesis n 0 less than k less than or equals n right parenthesis comma

and

$$ \begin{align*} f_k^2\geq f_{k-1}f_{k+1}\quad (n_0<k<n). \end{align*} $$ f Subscript k Superscript 2 Baseline greater than or equals f Subscript k minus 1 Baseline f Subscript k plus 1 Baseline left parenthesis n 0 less than k less than n right parenthesis period

We use the same terminology for an infinite sequence $(f_k)_{k\geq n_0}$ left parenthesis f Subscript k Baseline right parenthesis Subscript k greater than or equals n 0 when every finite subsequence $(f_k)_{n_0\leq k\leq n}$ left parenthesis f Subscript k Baseline right parenthesis Subscript n 0 less than or equals k less than or equals n , with $n\geq n_0+2$ n greater than or equals n 0 plus 2 , has these properties.

The main result of this paper is the following quantitative discrepancy estimate for the counting function $S_{w,b,f}(x)$ upper S Subscript w comma b comma f Baseline left parenthesis x right parenthesis .

Theorem 1.1. Let $n_0\geq 0$ n 0 greater than or equals 0 be an integer. Suppose that $(f_n)_{n\geq n_0}$ left parenthesis f Subscript n Baseline right parenthesis Subscript n greater than or equals n 0 is strictly increasing and log-concave. Then, for every integer $n>n_0$ n greater than n 0 ,

$$ \begin{align*} |S_{w,b,f}(n)-\rho_{w,\,b}n| \leq 2\log_b f_n+ \frac{\rho_{w,b}(1-\rho_{w,b})} {\log_b f_n-\log_b f_{n-1}} +\lambda_{w,b,f}(n_0), \end{align*} $$ StartAbsoluteValue upper S Subscript w comma b comma f Baseline left parenthesis n right parenthesis minus rho Subscript w comma b Baseline n EndAbsoluteValue less than or equals 2 log Subscript b Baseline f Subscript n Baseline plus StartFraction rho Subscript w comma b Baseline left parenthesis 1 minus rho Subscript w comma b Baseline right parenthesis Over log Subscript b Baseline f Subscript n Baseline minus log Subscript b Baseline f Subscript n minus 1 Baseline EndFraction plus lamda Subscript w comma b comma f Baseline left parenthesis n 0 right parenthesis comma

where

$$ \begin{align*} \lambda_{w,b,f}(n_0) :=2+n_0+2\rho_{w,b}-2\log_b f_{n_0}. \end{align*} $$ lamda Subscript w comma b comma f Baseline left parenthesis n 0 right parenthesis colon equals 2 plus n 0 plus 2 rho Subscript w comma b Baseline minus 2 log Subscript b Baseline f Subscript n 0 Baseline period

Theorem 1.1 yields the following Benford criterion, since

$$ \begin{align*} S_{w,b,f}(x)=S_{w,b,f}(\lceil x\rceil), \end{align*} $$ StartLayout 1st Row upper S Subscript w comma b comma f Baseline left parenthesis x right parenthesis equals upper S Subscript w comma b comma f Baseline left parenthesis left ceiling x right ceiling right parenthesis comma EndLayout

where $\lceil x\rceil $ left ceiling x right ceiling denotes the least integer $\ge x$ greater than or equals x .

Corollary 1.2. Under the hypotheses of Theorem 1.1, suppose further that

$$ \begin{align*} \frac{\log f_n}{n}\to 0 \quad\text{and}\quad n(\log f_n-\log f_{n-1})\to\infty \end{align*} $$ StartFraction log f Subscript n Baseline Over n EndFraction right arrow 0 and n left parenthesis log f Subscript n Baseline minus log f Subscript n minus 1 Baseline right parenthesis right arrow infinity

as $n\to \infty $ n right arrow infinity . Then, f satisfies Benford’s law in every base $b\geq 2$ b greater than or equals 2 . More precisely,

$$ \begin{align*} S_{w,b,f}(x) = \rho_{w,b}x+o_{w,b}(x) \quad (x\to\infty). \end{align*} $$ upper S Subscript w comma b comma f Baseline left parenthesis x right parenthesis equals rho Subscript w comma b Baseline x plus o Subscript w comma b Baseline left parenthesis x right parenthesis left parenthesis x right arrow infinity right parenthesis period

General sufficient criteria for Benford’s law have previously been established (see, for example, Anderson et al. [Reference Anderson, Rolen and Stoehr1]). The criterion in Corollary 1.2 is instead derived from the structural assumptions of strict increase and log-concavity. Moreover, Theorem 1.1 gives an explicit discrepancy estimate for the occurrence of each prescribed initial string.

For a sequence $f=(f_n)_{n\geq 0}$ f equals left parenthesis f Subscript n Baseline right parenthesis Subscript n greater than or equals 0 of positive real numbers, let $N_f(w,b)$ upper N Subscript f Baseline left parenthesis w comma b right parenthesis denote the least integer $n\geq 0$ n greater than or equals 0 such that $f_n$ f Subscript n begins with the string w in base b, whenever such an integer exists. Equivalently,

(1.1) $$ \begin{align} N_f(w,b)=\min\{n\geq1:S_{w,b,f}(n)>0\}-1. \end{align} $$ upper N Subscript f Baseline left parenthesis w comma b right parenthesis equals min left brace n greater than or equals 1 colon upper S Subscript w comma b comma f Baseline left parenthesis n right parenthesis greater than 0 right brace minus 1 period

Theorem 1.1 immediately yields the following upper bound.

Corollary 1.3. Under the hypotheses of Theorem 1.1, suppose that the set of integers $n>n_0$ n greater than n 0 satisfying

$$ \begin{align*} n> \frac{2}{\rho_{w,\,b}\log b}\log f_n + \frac{(1-\rho_{w,\,b})\log b} {\log f_n-\log f_{n-1}} + \frac{\lambda_{w,b,f}(n_0)}{\rho_{w,b}} \end{align*} $$ n greater than StartFraction 2 Over rho Subscript w comma b Baseline log b EndFraction log f Subscript n Baseline plus StartFraction left parenthesis 1 minus rho Subscript w comma b Baseline right parenthesis log b Over log f Subscript n Baseline minus log f Subscript n minus 1 Baseline EndFraction plus StartFraction lamda Subscript w comma b comma f Baseline left parenthesis n 0 right parenthesis Over rho Subscript w comma b Baseline EndFraction

is nonempty. Let $N_{w,b,f}$ upper N Subscript w comma b comma f denote its least element. Then,

$$ \begin{align*} N_f(w,b)\leq N_{w,b,f}. \end{align*} $$ StartLayout 1st Row upper N Subscript f Baseline left parenthesis w comma b right parenthesis less than or equals upper N Subscript w comma b comma f Baseline period EndLayout

1.2 Applications to partitions and plane partitions

For $n\geq 0$ n greater than or equals 0 , let $p(n)$ p left parenthesis n right parenthesis denote the number of partitions of n and let $\mathrm {PL}(n)$ upper P upper L left parenthesis n right parenthesis denote the number of plane partitions of n. Recall that a partition of n is a nonincreasing sequence of positive integers with sum n. A plane partition of n is an array of nonnegative integers $ (\pi _{i,j})_{i,j\geq 1} $ left parenthesis pi Subscript i comma j Baseline right parenthesis Subscript i comma j greater than or equals 1 whose rows and columns are weakly decreasing, and such that $ \sum _{i,j\geq 1}\pi _{i,j}=n $ sigma summation Underscript i comma j greater than or equals 1 Endscripts pi Subscript i comma j Baseline equals n . We refer the reader to [Reference Andrews2] for background on partitions and plane partitions.

The leading digits of both $p(n)$ p left parenthesis n right parenthesis and $\mathrm {PL}(n)$ upper P upper L left parenthesis n right parenthesis obey Benford’s law (see [Reference Anderson, Rolen and Stoehr1, Reference Douglass and Ono4]). Motivated by this phenomenon, Douglass and Ono [Reference Douglass and Ono4] asked for nontrivial upper bounds for $N_p(w,b)$ upper N Subscript p Baseline left parenthesis w comma b right parenthesis and $N_{\mathrm {PL}}(w,b)$ upper N Subscript upper P upper L Baseline left parenthesis w comma b right parenthesis . Here, w is a base-b digit string, and $N_p(w,b)$ upper N Subscript p Baseline left parenthesis w comma b right parenthesis , respectively $N_{\mathrm {PL}}(w,b)$ upper N Subscript upper P upper L Baseline left parenthesis w comma b right parenthesis , is the least index n for which $p(n)$ p left parenthesis n right parenthesis , respectively $\mathrm {PL}(n)$ upper P upper L left parenthesis n right parenthesis , begins with the digit string w in base b.

Let t be the length of w. Using exponential sums and transcendence theory, Luca [Reference Luca6, Reference Luca7] proved that

$$ \begin{align*} N_p(w,b) \leq \exp(2\cdot 10^{25}(t+12)(\log b)^2), \end{align*} $$ upper N Subscript p Baseline left parenthesis w comma b right parenthesis less than or equals exp left parenthesis 2 dot 10 Superscript 25 Baseline left parenthesis t plus 12 right parenthesis left parenthesis log b right parenthesis squared right parenthesis comma

and, using results concerning prime numbers, that

$$ \begin{align*} N_{\mathrm{PL}}(w,b)\leq b^{51t+688}. \end{align*} $$ upper N Subscript upper P upper L Baseline left parenthesis w comma b right parenthesis less than or equals b Superscript 51 t plus 688 Baseline period

More recently, Iyer [Reference Iyer5] developed an elementary discrepancy method and obtained the substantially sharper bounds

$$ \begin{align*} N_p(w,b)\leq 288b^{2t}+2 \end{align*} $$ upper N Subscript p Baseline left parenthesis w comma b right parenthesis less than or equals 288 b Superscript 2 t Baseline plus 2

and

$$ \begin{align*} N_{\mathrm{PL}}(w,b) \leq 130b^{3t}+29400b^{3t/2}. \end{align*} $$ upper N Subscript upper P upper L Baseline left parenthesis w comma b right parenthesis less than or equals 130 b Superscript 3 t Baseline plus 29400 b Superscript 3 t divided by 2 Baseline period

We now apply Theorem 1.1 to obtain explicit discrepancy estimates for $p(n)$ p left parenthesis n right parenthesis and $\mathrm {PL}(n)$ upper P upper L left parenthesis n right parenthesis . These estimates yield improved bounds for the first occurrence of a prescribed leading digit string.

The sequences $(p(n))_{n\geq 2}$ left parenthesis p left parenthesis n right parenthesis right parenthesis Subscript n greater than or equals 2 and $(\mathrm {PL}(n))_{n\geq 2}$ left parenthesis upper P upper L left parenthesis n right parenthesis right parenthesis Subscript n greater than or equals 2 are strictly increasing. Moreover, Nicolas [Reference Nicolas8] and DeSalvo and Pak [Reference DeSalvo and Pak3, Theorem 1.1] proved that $(p(n))_{n\geq 25}$ left parenthesis p left parenthesis n right parenthesis right parenthesis Subscript n greater than or equals 25 is log-concave, whereas Ono et al. [Reference Ono, Pujahari and Rolen9] proved that $(\mathrm {PL}(n))_{n\geq 12}$ left parenthesis upper P upper L left parenthesis n right parenthesis right parenthesis Subscript n greater than or equals 12 is log-concave. Consequently, Theorem 1.1 applies to $p(n)$ p left parenthesis n right parenthesis with $n_0=25$ n 0 equals 25 and to $\mathrm {PL}(n)$ upper P upper L left parenthesis n right parenthesis with $n_0=12$ n 0 equals 12 .

Corollary 1.4. For every integer $n\geq 100$ n greater than or equals 100 ,

$$ \begin{align*} |S_{w,b,p}(n)-\rho_{w,\,b}n| < \frac{4\pi}{\sqrt{6}\log b}\sqrt{n} +1.1\rho_{w,b}\sqrt{n}\log b+27. \end{align*} $$ StartAbsoluteValue upper S Subscript w comma b comma p Baseline left parenthesis n right parenthesis minus rho Subscript w comma b Baseline n EndAbsoluteValue less than StartFraction 4 pi Over StartRoot 6 EndRoot log b EndFraction StartRoot n EndRoot plus 1.1 rho Subscript w comma b Baseline StartRoot n EndRoot log b plus 27 period

For every integer $n\geq 2829$ n greater than or equals 2829 ,

$$ \begin{align*} |S_{w,\,b,\mathrm{PL}}(n)-\rho_{w,b}n| < \frac{6\zeta(3)^{1/3}}{2^{2/3}\log b}n^{2/3} +0.9\rho_{w,b}n^{1/3}\log b+14. \end{align*} $$ StartAbsoluteValue upper S Subscript w comma b comma upper P upper L Baseline left parenthesis n right parenthesis minus rho Subscript w comma b Baseline n EndAbsoluteValue less than StartFraction 6 zeta left parenthesis 3 right parenthesis Superscript 1 divided by 3 Baseline Over 2 Superscript 2 divided by 3 Baseline log b EndFraction n Superscript 2 divided by 3 Baseline plus 0.9 rho Subscript w comma b Baseline n Superscript 1 divided by 3 Baseline log b plus 14 period

Here, $\zeta (s)$ zeta left parenthesis s right parenthesis denotes the Riemann zeta function.

We derive the following upper bounds from Corollary 1.4.

Corollary 1.5. We have

$$ \begin{align*} N_p(w,b)\leq \frac{8\pi^2/3}{\rho_{w,b}^2(\log b)^2} +\frac{66}{\rho_{w,b}} +29(\log b)^2 \end{align*} $$ upper N Subscript p Baseline left parenthesis w comma b right parenthesis less than or equals StartFraction 8 pi squared divided by 3 Over rho Subscript w comma b Superscript 2 Baseline left parenthesis log b right parenthesis squared EndFraction plus StartFraction 66 Over rho Subscript w comma b Baseline EndFraction plus 29 left parenthesis log b right parenthesis squared

and

$$ \begin{align*} N_{\mathrm{PL}}(w,b)\leq \frac{54\zeta(3)}{\rho_{w,\,b}^3(\log b)^3} +\frac{65}{\rho_{w,\,b}} +15\rho_{w,\,b}(\log b)^3 +2\rho_{w,b}^3(\log b)^6+2697. \end{align*} $$ upper N Subscript upper P upper L Baseline left parenthesis w comma b right parenthesis less than or equals StartFraction 54 zeta left parenthesis 3 right parenthesis Over rho Subscript w comma b Superscript 3 Baseline left parenthesis log b right parenthesis cubed EndFraction plus StartFraction 65 Over rho Subscript w comma b Baseline EndFraction plus 15 rho Subscript w comma b Baseline left parenthesis log b right parenthesis cubed plus 2 rho Subscript w comma b Superscript 3 Baseline left parenthesis log b right parenthesis Superscript 6 Baseline plus 2697 period

Remark 1.6. The bounds in Corollary 1.5 depend on w and b through $\rho _{w,b}$ rho Subscript w comma b , and therefore retain more information than bounds depending only on the length t of w. Consequently, for a fixed digit string w, they can be substantially sharper than the corresponding bounds obtained by replacing $\rho _{w,b}$ rho Subscript w comma b with a length-based estimate. For comparison, since $b^{t-1}\leq w<b^t$ b Superscript t minus 1 Baseline less than or equals w less than b Superscript t , the bounds in Corollary 1.5 yield

$$ \begin{align*} N_p(w,b) < 27b^{2t} +66b^t\log b +29(\log b)^2 \end{align*} $$ upper N Subscript p Baseline left parenthesis w comma b right parenthesis less than 27 b Superscript 2 t Baseline plus 66 b Superscript t Baseline log b plus 29 left parenthesis log b right parenthesis squared

and

$$ \begin{align*} N_{\mathrm{PL}}(w,b) < 65b^{3t}+2697. \end{align*} $$ upper N Subscript upper P upper L Baseline left parenthesis w comma b right parenthesis less than 65 b Superscript 3 t Baseline plus 2697 period

These estimates improve the corresponding bounds of Iyer [Reference Iyer5].

2 Proof of Theorem 1.1

We begin by translating the leading-string condition into a condition on logarithms.

Lemma 2.1. Let $y>0$ y greater than 0 . Then, y begins with the base-b string w if and only if

$$ \begin{align*} \log_b y\in\mathcal E_{w,b}, \end{align*} $$ log Subscript b Baseline y element of script upper E Subscript w comma b Baseline comma

where

$$ \begin{align*} \mathcal E_{w,b} := \bigsqcup_{h\in\mathbb Z} [ h+\log_b w,\, h+\log_b w+\rho_{w,b} ). \end{align*} $$ script upper E Subscript w comma b Baseline colon equals square union Underscript h element of double struck upper Z Endscripts left bracket h plus log Subscript b Baseline w comma h plus log Subscript b Baseline w plus rho Subscript w comma b Baseline right parenthesis period

Proof. By definition, y begins with the string w if and only if there exists $h\in \mathbb Z$ h element of double struck upper Z such that $ wb^h\leq y<(w+1)b^h. $ w b Superscript h Baseline less than or equals y less than left parenthesis w plus 1 right parenthesis b Superscript h Baseline period Taking logarithms to base b, this is equivalent to

$$ \begin{align*} h+\log_b w \leq \log_b y < h+\log_b(w+1). \end{align*} $$ h plus log Subscript b Baseline w less than or equals log Subscript b Baseline y less than h plus log Subscript b Baseline left parenthesis w plus 1 right parenthesis period

Since $ \log _b(w+1)=\log _b w+\rho _{w,\,b}, $ log Subscript b Baseline left parenthesis w plus 1 right parenthesis equals log Subscript b Baseline w plus rho Subscript w comma b Baseline comma the result follows.

We now consider a general $1$ 1 -periodic set of the form

$$ \begin{align*} \mathcal E = \bigsqcup_{h\in\mathbb Z} [h+\alpha,h+\alpha+\rho), \quad \alpha\in\mathbb R,\ 0<\rho\leq 1. \end{align*} $$ script upper E equals square union Underscript h element of double struck upper Z Endscripts left bracket h plus alpha comma h plus alpha plus rho right parenthesis comma alpha element of double struck upper R comma 0 less than rho less than or equals 1 period

Lemma 2.2. Let $a<b$ a less than b and let $g:[a,b]\to [0,\infty )$ g colon left bracket a comma b right bracket right arrow left bracket 0 comma infinity right parenthesis be nondecreasing. Then,

$$ \begin{align*} \bigg| \int_a^b(\mathbf{1}_{\mathcal E}(u)-\rho)g(u)\,d u \bigg| \leq \rho(1-\rho)g(b^-). \end{align*} $$ StartLayout 1st Row StartAbsoluteValue integral Subscript a Superscript b Baseline left parenthesis bold 1 Subscript script upper E Baseline left parenthesis u right parenthesis minus rho right parenthesis g left parenthesis u right parenthesis d u EndAbsoluteValue less than or equals rho left parenthesis 1 minus rho right parenthesis g left parenthesis b Superscript minus Baseline right parenthesis period EndLayout

Here, $\mathbf {1}_{\mathcal E}(u)$ bold 1 Subscript script upper E Baseline left parenthesis u right parenthesis equals $1$ 1 if $u\in \mathcal E$ u element of script upper E and $0$ 0 otherwise.

Proof. Put $ q(u):=\mathbf {1}_{\mathcal E}(u)-\rho. $ q left parenthesis u right parenthesis colon equals bold 1 Subscript script upper E Baseline left parenthesis u right parenthesis minus rho period Then, q is $1$ 1 -periodic and has mean zero on every period. Let $ t:=\{u-\alpha \} $ t colon equals left brace u minus alpha right brace and define

$$ \begin{align*} Q(u):= \begin{cases} (1-\rho)t-\frac{1}{2}\rho(1-\rho), & 0\leq t<\rho, \\[10pt] \rho(1-t)-\frac{1}{2}\rho(1-\rho), & \rho\leq t<1. \end{cases} \end{align*} $$ upper Q left parenthesis u right parenthesis colon equals StartLayout Enlarged left brace 1st Row 1st Column left parenthesis 1 minus rho right parenthesis t minus one half rho left parenthesis 1 minus rho right parenthesis comma 2nd Column 0 less than or equals t less than rho comma 2nd Row 1st Column rho left parenthesis 1 minus t right parenthesis minus one half rho left parenthesis 1 minus rho right parenthesis comma 2nd Column rho less than or equals t less than 1 period EndLayout

Then, Q is continuous and $1$ 1 -periodic, with

$$ \begin{align*} Q'(u)=q(u) \ \text{for almost every }u, \quad |Q(u)|\leq\tfrac{1}{2}\rho(1-\rho). \end{align*} $$ StartLayout 1st Row upper Q prime left parenthesis u right parenthesis equals q left parenthesis u right parenthesis for almost every u comma StartAbsoluteValue upper Q left parenthesis u right parenthesis EndAbsoluteValue less than or equals one half rho left parenthesis 1 minus rho right parenthesis period EndLayout

By Lebesgue–Stieltjes integration by parts,

$$ \begin{align*} \int_a^b q(u)g(u)\,d u = Q(b)g(b^-)-Q(a)g(a^+) -\int_{(a,b)}Q(u)\,d g(u). \end{align*} $$ StartLayout 1st Row integral Subscript a Superscript b Baseline q left parenthesis u right parenthesis g left parenthesis u right parenthesis d u equals upper Q left parenthesis b right parenthesis g left parenthesis b Superscript minus Baseline right parenthesis minus upper Q left parenthesis a right parenthesis g left parenthesis a Superscript plus Baseline right parenthesis minus integral Underscript left parenthesis a comma b right parenthesis Endscripts upper Q left parenthesis u right parenthesis d g left parenthesis u right parenthesis period EndLayout

Since g is nondecreasing, $d g$ d g is a positive measure. Hence,

$$ \begin{align*} \bigg|\int_a^b q(u)g(u)\,d u\bigg| \leq \frac12\rho(1-\rho) \bigg( g(b^-)+g(a^+)+\int_{(a,b)}\,d g \bigg) = \rho(1-\rho)g(b^-). \end{align*} $$ StartLayout 1st Row StartAbsoluteValue integral Subscript a Superscript b Baseline q left parenthesis u right parenthesis g left parenthesis u right parenthesis d u EndAbsoluteValue less than or equals one half rho left parenthesis 1 minus rho right parenthesis left parenthesis g left parenthesis b Superscript minus Baseline right parenthesis plus g left parenthesis a Superscript plus Baseline right parenthesis plus integral Underscript left parenthesis a comma b right parenthesis Endscripts d g right parenthesis equals rho left parenthesis 1 minus rho right parenthesis g left parenthesis b Superscript minus Baseline right parenthesis period EndLayout

This completes the proof.

Proposition 2.3. Let $(c_k)_{0\leq k\leq n}$ left parenthesis c Subscript k Baseline right parenthesis Subscript 0 less than or equals k less than or equals n be a strictly increasing concave sequence of real numbers. Then,

$$ \begin{align*} |\#\{k:0\leq k<n, c_k\in\mathcal E\}-\rho n| \leq 2(c_n-c_0+1+\rho)+\frac{\rho(1-\rho)}{c_n-c_{n-1}}. \end{align*} $$ StartAbsoluteValue number sign left brace k colon 0 less than or equals k less than n comma c Subscript k Baseline element of script upper E right brace minus rho n EndAbsoluteValue less than or equals 2 left parenthesis c Subscript n Baseline minus c 0 plus 1 plus rho right parenthesis plus StartFraction rho left parenthesis 1 minus rho right parenthesis Over c Subscript n Baseline minus c Subscript n minus 1 Baseline EndFraction period

Proof. Define $F:[0,n]\to \mathbb R$ upper F colon left bracket 0 comma n right bracket right arrow double struck upper R by $F(k)=c_k$ upper F left parenthesis k right parenthesis equals c Subscript k for $0\leq k\leq n$ 0 less than or equals k less than or equals n and by requiring F to be affine on each interval, $[k,k+1]$ left bracket k comma k plus 1 right bracket . Thus,

$$ \begin{align*} F(x)=c_k+(x-k)(c_{k+1}-c_k) \quad(k\leq x\leq k+1). \end{align*} $$ upper F left parenthesis x right parenthesis equals c Subscript k Baseline plus left parenthesis x minus k right parenthesis left parenthesis c Subscript k plus 1 Baseline minus c Subscript k Baseline right parenthesis left parenthesis k less than or equals x less than or equals k plus 1 right parenthesis period

The slopes of F are positive and nonincreasing, so F is strictly increasing and concave. Hence, $G:=F^{-1}:[c_0,c_n]\to [0,n]$ upper G colon equals upper F Superscript negative 1 Baseline colon left bracket c 0 comma c Subscript n Baseline right bracket right arrow left bracket 0 comma n right bracket is convex; let $G'$ upper G prime be its nondecreasing piecewise-constant derivative. Write

$$ \begin{align*} \mathcal E\cap[c_0,c_n)=I_1\sqcup\cdots\sqcup I_m \end{align*} $$ script upper E intersection left bracket c 0 comma c Subscript n Baseline right parenthesis equals upper I 1 square cup midline horizontal ellipsis square cup upper I Subscript m

as a disjoint union of intervals. Each $I_j$ upper I Subscript j lies in some $[\ell +\alpha ,\ell +\alpha +\rho )$ left bracket script l plus alpha comma script l plus alpha plus rho right parenthesis meeting $[c_0,c_n)$ left bracket c 0 comma c Subscript n Baseline right parenthesis , whence $ m<c_n-c_0+1+\rho. $ m less than c Subscript n Baseline minus c 0 plus 1 plus rho period Since $F([0,n))=[c_0,c_n)$ upper F left parenthesis left bracket 0 comma n right parenthesis right parenthesis equals left bracket c 0 comma c Subscript n Baseline right parenthesis ,

$$ \begin{align*} \#\{k:0\leq k<n, c_k\in\mathcal E\} =\sum_{j=1}^m\#(\mathbb Z\cap G(I_j)). \end{align*} $$ number sign left brace k colon 0 less than or equals k less than n comma c Subscript k Baseline element of script upper E right brace equals sigma summation Underscript j equals 1 Overscript m Endscripts number sign left parenthesis double struck upper Z intersection upper G left parenthesis upper I Subscript j Baseline right parenthesis right parenthesis period

Moreover,

$$ \begin{align*} |G(I_j)|=\int_{I_j}G'(u)\,d u, \quad \rho n=\int_{c_0}^{c_n}\rho G'(u)\,d u. \end{align*} $$ StartLayout 1st Row StartAbsoluteValue upper G left parenthesis upper I Subscript j Baseline right parenthesis EndAbsoluteValue equals integral Underscript upper I Subscript j Baseline Endscripts upper G prime left parenthesis u right parenthesis d u comma rho n equals integral Subscript c 0 Superscript c Subscript n Baseline Baseline rho upper G prime left parenthesis u right parenthesis d u period EndLayout

Therefore,

$$ \begin{align*} \#\{k:0\leq k<n, c_k\in\mathcal E\}-\rho n = \int_{c_0}^{c_n}(\mathbf{1}_{\mathcal E}(u)-\rho)G'(u)\,d u+\sum_{j=1}^m(\#(\mathbb Z\cap G(I_j))-|G(I_j)|). \end{align*} $$ StartLayout 1st Row number sign left brace k colon 0 less than or equals k less than n comma c Subscript k Baseline element of script upper E right brace minus rho n equals integral Subscript c 0 Superscript c Subscript n Baseline Baseline left parenthesis bold 1 Subscript script upper E Baseline left parenthesis u right parenthesis minus rho right parenthesis upper G prime left parenthesis u right parenthesis d u plus sigma summation Underscript j equals 1 Overscript m Endscripts left parenthesis number sign left parenthesis double struck upper Z intersection upper G left parenthesis upper I Subscript j Baseline right parenthesis right parenthesis minus StartAbsoluteValue upper G left parenthesis upper I Subscript j Baseline right parenthesis EndAbsoluteValue right parenthesis period EndLayout

For every interval J, we have $ |\#(\mathbb Z\cap J)-|J||\leq 2. $ StartAbsoluteValue number sign left parenthesis double struck upper Z intersection upper J right parenthesis minus EndAbsoluteValue upper J StartAbsoluteValue EndAbsoluteValue less than or equals 2 period Since $G'$ upper G prime is nondecreasing, Lemma 2.2 gives

$$ \begin{align*} \bigg|\int_{c_0}^{c_n}(\mathbf{1}_{\mathcal E}(u)-\rho)G'(u)\,d u\bigg| \leq\rho(1-\rho)G'(c_n^-). \end{align*} $$ StartLayout 1st Row StartAbsoluteValue integral Subscript c 0 Superscript c Subscript n Baseline Baseline left parenthesis bold 1 Subscript script upper E Baseline left parenthesis u right parenthesis minus rho right parenthesis upper G prime left parenthesis u right parenthesis d u EndAbsoluteValue less than or equals rho left parenthesis 1 minus rho right parenthesis upper G prime left parenthesis c Subscript n Superscript minus Baseline right parenthesis period EndLayout

Finally, since F has slope $c_n-c_{n-1}$ c Subscript n Baseline minus c Subscript n minus 1 on $[n-1,n)$ left bracket n minus 1 comma n right parenthesis ,

$$ \begin{align*} G'(c_n^-)=\frac1{c_n-c_{n-1}}. \end{align*} $$ upper G prime left parenthesis c Subscript n Superscript minus Baseline right parenthesis equals StartFraction 1 Over c Subscript n Baseline minus c Subscript n minus 1 Baseline EndFraction period

The result follows.

Proof of Theorem 1.1.

Put $c_k:=\log _b f_{k+n_0}$ c Subscript k Baseline colon equals log Subscript b Baseline f Subscript k plus n 0 for all $k\geq 0$ k greater than or equals 0 . Then, $(c_k)_{k\geq 0}$ left parenthesis c Subscript k Baseline right parenthesis Subscript k greater than or equals 0 is strictly increasing and concave. By Lemma 2.1, $f_{k+n_0}$ f Subscript k plus n 0 begins with the base-b string w if and only if $c_k\in \mathcal E_{w,b}$ c Subscript k Baseline element of script upper E Subscript w comma b , where $\mathcal E_{w,b}$ script upper E Subscript w comma b is of the form considered in Proposition 2.3, with $ {\alpha =\log _b w} $ alpha equals log Subscript b Baseline w and $ \rho =\rho _{w,b}. $ rho equals rho Subscript w comma b Baseline period Applying Proposition 2.3 to $ (c_k)_{0\leq k\leq n-n_0} $ left parenthesis c Subscript k Baseline right parenthesis Subscript 0 less than or equals k less than or equals n minus n 0 and noting that $c_k=\log _b f_{k+n_0}$ c Subscript k Baseline equals log Subscript b Baseline f Subscript k plus n 0 for $k\geq 0$ k greater than or equals 0 , we obtain

$$ \begin{align*} &| \#\{0\leq k<n-n_0:c_k\in\mathcal E_{w,b}\} -\rho_{w,b}(n-n_0) |\\&\qquad \leq {} 2( \log_b f_n-\log_b f_{n_0}+1+\rho_{w,b} )+ \frac{\rho_{w,b}(1-\rho_{w,b})} {\log_b f_n-\log_b f_{n-1}}. \end{align*} $$ StartLayout 1st Row 1st Column Blank 2nd Column StartAbsoluteValue number sign StartSet 0 less than or equals k less than n minus n 0 colon c Subscript k Baseline element of script upper E Subscript w comma b Baseline EndSet minus rho Subscript w comma b Baseline left parenthesis n minus n 0 right parenthesis EndAbsoluteValue 2nd Row 1st Column Blank 2nd Column less than or equals 2 left parenthesis log Subscript b Baseline f Subscript n Baseline minus log Subscript b Baseline f Subscript n 0 Baseline plus 1 plus rho Subscript w comma b Baseline right parenthesis plus StartFraction rho Subscript w comma b Baseline left parenthesis 1 minus rho Subscript w comma b Baseline right parenthesis Over log Subscript b Baseline f Subscript n Baseline minus log Subscript b Baseline f Subscript n minus 1 Baseline EndFraction period EndLayout

Moreover,

$$ \begin{align*} \#\{k:0\leq k<n-n_0, c_k\in\mathcal E_{w,b}\} =S_{w,b,f}(n)-S_{w,b,f}(n_0) \end{align*} $$ StartLayout 1st Row number sign left brace k colon 0 less than or equals k less than n minus n 0 comma c Subscript k Baseline element of script upper E Subscript w comma b Baseline right brace equals upper S Subscript w comma b comma f Baseline left parenthesis n right parenthesis minus upper S Subscript w comma b comma f Baseline left parenthesis n 0 right parenthesis EndLayout

and

$$ \begin{align*} |S_{w,b,f}(n_0)-\rho_{w,b}n_0| \leq n_0. \end{align*} $$ StartAbsoluteValue upper S Subscript w comma b comma f Baseline left parenthesis n 0 right parenthesis minus rho Subscript w comma b Baseline n 0 EndAbsoluteValue less than or equals n 0 period

The triangle inequality therefore gives

$$ \begin{align*} |S_{w,b,f}(n)-\rho_{w,b}n| \leq{} 2(\log_b f_n-\log_b f_{n_0}) +2(1+\rho_{w,b}) + \frac{\rho_{w,b}(1-\rho_{w,b})} {\log_b f_n-\log_b f_{n-1}}+n_0, \end{align*} $$ StartAbsoluteValue upper S Subscript w comma b comma f Baseline left parenthesis n right parenthesis minus rho Subscript w comma b Baseline n EndAbsoluteValue less than or equals 2 left parenthesis log Subscript b Baseline f Subscript n Baseline minus log Subscript b Baseline f Subscript n 0 Baseline right parenthesis plus 2 left parenthesis 1 plus rho Subscript w comma b Baseline right parenthesis plus StartFraction rho Subscript w comma b Baseline left parenthesis 1 minus rho Subscript w comma b Baseline right parenthesis Over log Subscript b Baseline f Subscript n Baseline minus log Subscript b Baseline f Subscript n minus 1 Baseline EndFraction plus n 0 comma

which simplifies to the asserted bound.

3 Proofs of the applications

We use the following explicit estimate of Iyer [Reference Iyer5, Lemma 2.2].

Lemma 3.1. For every integer $n\geq 4$ n greater than or equals 4 ,

$$ \begin{align*} \bigg| \log p(n)-\frac{2\pi}{\sqrt6}\sqrt n+\log n -\log\bigg(\frac{\sqrt3}{12}\bigg) \bigg|\leq \frac4{\sqrt n}. \end{align*} $$ StartAbsoluteValue log p left parenthesis n right parenthesis minus StartFraction 2 pi Over StartRoot 6 EndRoot EndFraction StartRoot n EndRoot plus log n minus log left parenthesis StartFraction StartRoot 3 EndRoot Over 12 EndFraction right parenthesis EndAbsoluteValue less than or equals StartFraction 4 Over StartRoot n EndRoot EndFraction period

Set

$$ \begin{align*} C:=\pi\sqrt{\frac23},\quad \beta:=C(\sqrt2-1)-\frac{\log2}{\sqrt{100}} -\frac{4+2\sqrt2}{100}> 0.9249. \end{align*} $$ upper C colon equals pi StartRoot two thirds EndRoot comma beta colon equals upper C left parenthesis StartRoot 2 EndRoot minus 1 right parenthesis minus StartFraction log 2 Over StartRoot 100 EndRoot EndFraction minus StartFraction 4 plus 2 StartRoot 2 EndRoot Over 100 EndFraction greater than 0.9249 period

Lemma 3.2. For every integer $n\geq 100$ n greater than or equals 100 ,

$$ \begin{align*} \log p(n)\leq C\sqrt n,\quad \log p(n)-\log p(n-1)\geq\beta n^{-1/2}. \end{align*} $$ StartLayout 1st Row log p left parenthesis n right parenthesis less than or equals upper C StartRoot n EndRoot comma log p left parenthesis n right parenthesis minus log p left parenthesis n minus 1 right parenthesis greater than or equals beta n Superscript negative 1 divided by 2 Baseline period EndLayout

Proof. The first assertion follows from Lemma 3.1, since

$$ \begin{align*} -\log n+\log\bigg(\frac{\sqrt3}{12}\bigg)+\frac4{\sqrt n}<0 \quad(n\geq100). \end{align*} $$ minus log n plus log left parenthesis StartFraction StartRoot 3 EndRoot Over 12 EndFraction right parenthesis plus StartFraction 4 Over StartRoot n EndRoot EndFraction less than 0 left parenthesis n greater than or equals 100 right parenthesis period

For the second, put $d_n:=\log p(n)-\log p(n-1)$ d Subscript n Baseline colon equals log p left parenthesis n right parenthesis minus log p left parenthesis n minus 1 right parenthesis . Since $p(n)$ p left parenthesis n right parenthesis is log-concave for $n\geq 25$ n greater than or equals 25 , the sequence $d_n$ d Subscript n is nonincreasing. Consequently,

$$ \begin{align*} nd_n\geq\log p(2n)-\log p(n) \geq C(\sqrt2-1)\sqrt n-\log2-\frac{4+2\sqrt2}{\sqrt n}. \end{align*} $$ StartLayout 1st Row n d Subscript n Baseline greater than or equals log p left parenthesis 2 n right parenthesis minus log p left parenthesis n right parenthesis greater than or equals upper C left parenthesis StartRoot 2 EndRoot minus 1 right parenthesis StartRoot n EndRoot minus log 2 minus StartFraction 4 plus 2 StartRoot 2 EndRoot Over StartRoot n EndRoot EndFraction period EndLayout

Hence,

$$ \begin{align*} d_n\geq n^{-1/2}\bigg( C(\sqrt2-1)-\frac{\log2}{\sqrt n} -\frac{4+2\sqrt2}{n}\bigg)\geq\beta n^{-1/2}, \end{align*} $$ d Subscript n Baseline greater than or equals n Superscript negative 1 divided by 2 Baseline left parenthesis upper C left parenthesis StartRoot 2 EndRoot minus 1 right parenthesis minus StartFraction log 2 Over StartRoot n EndRoot EndFraction minus StartFraction 4 plus 2 StartRoot 2 EndRoot Over n EndFraction right parenthesis greater than or equals beta n Superscript negative 1 divided by 2 Baseline comma

because the expression in parentheses is increasing in n.

Set

$$ \begin{align*} \alpha:=3\cdot2^{-2/3}\zeta(3)^{1/3},\quad B:=\frac{2^{25/36}e^{\zeta'(-1)}\zeta(3)^{7/36}} {\sqrt{12\pi}} \end{align*} $$ StartLayout 1st Row alpha colon equals 3 dot 2 Superscript negative 2 divided by 3 Baseline zeta left parenthesis 3 right parenthesis Superscript 1 divided by 3 Baseline comma upper B colon equals StartFraction 2 Superscript 25 divided by 36 Baseline e Superscript zeta prime left parenthesis negative 1 right parenthesis Baseline zeta left parenthesis 3 right parenthesis Superscript 7 divided by 36 Baseline Over StartRoot 12 pi EndRoot EndFraction EndLayout

and

$$ \begin{align*} \gamma:=\alpha(2^{2/3}-1) -\frac{25\log2}{36\cdot 2829^{2/3}} -\frac{200(1+2^{-2/3})}{2829^{4/3}}> 1.1697. \end{align*} $$ gamma colon equals alpha left parenthesis 2 Superscript 2 divided by 3 Baseline minus 1 right parenthesis minus StartFraction 25 log 2 Over 36 dot 2829 Superscript 2 divided by 3 Baseline EndFraction minus StartFraction 200 left parenthesis 1 plus 2 Superscript negative 2 divided by 3 Baseline right parenthesis Over 2829 Superscript 4 divided by 3 Baseline EndFraction greater than 1.1697 period

We also use the following estimate of Iyer [Reference Iyer5, Lemma 2.3].

Lemma 3.3. For every integer $n\geq 2829$ n greater than or equals 2829 ,

$$ \begin{align*} \bigg| \log\mathrm{PL}(n)-\alpha n^{2/3} +\frac{25}{36}\log n-\log B \bigg| \leq\frac{200}{n^{2/3}}. \end{align*} $$ StartAbsoluteValue log upper P upper L left parenthesis n right parenthesis minus alpha n Superscript 2 divided by 3 Baseline plus StartFraction 25 Over 36 EndFraction log n minus log upper B EndAbsoluteValue less than or equals StartFraction 200 Over n Superscript 2 divided by 3 Baseline EndFraction period

Lemma 3.4. For every integer $n\geq 2829$ n greater than or equals 2829 ,

$$ \begin{align*} \log\mathrm{PL}(n)\leq\alpha n^{2/3},\quad \log\mathrm{PL}(n)-\log\mathrm{PL}(n-1) \geq\gamma n^{-1/3}. \end{align*} $$ StartLayout 1st Row log upper P upper L left parenthesis n right parenthesis less than or equals alpha n Superscript 2 divided by 3 Baseline comma log upper P upper L left parenthesis n right parenthesis minus log upper P upper L left parenthesis n minus 1 right parenthesis greater than or equals gamma n Superscript negative 1 divided by 3 Baseline period EndLayout

Proof. The first assertion follows from Lemma 3.3, since

$$ \begin{align*} -\frac{25}{36}\log n+\log B+\frac{200}{n^{2/3}}<0 \quad(n\geq 2829). \end{align*} $$ minus StartFraction 25 Over 36 EndFraction log n plus log upper B plus StartFraction 200 Over n Superscript 2 divided by 3 Baseline EndFraction less than 0 left parenthesis n greater than or equals 2829 right parenthesis period

For the second, set

$$ \begin{align*} d_n:=\log\mathrm{PL}(n)-\log\mathrm{PL}(n-1). \end{align*} $$ StartLayout 1st Row d Subscript n Baseline colon equals log upper P upper L left parenthesis n right parenthesis minus log upper P upper L left parenthesis n minus 1 right parenthesis period EndLayout

Log-concavity of $\mathrm {PL}(n)$ upper P upper L left parenthesis n right parenthesis for $n\geq 12$ n greater than or equals 12 gives $ nd_n\geq \log \mathrm {PL}(2n)-\log \mathrm {PL}(n) $ n d Subscript n Baseline greater than or equals log upper P upper L left parenthesis 2 n right parenthesis minus log upper P upper L left parenthesis n right parenthesis and Lemma 3.3 gives

$$ \begin{align*} \log\mathrm{PL}(2n)-\log\mathrm{PL}(n) \geq\alpha(2^{2/3}-1)n^{2/3} -\frac{25}{36}\log2 -\frac{200(1+2^{-2/3})}{n^{2/3}}. \end{align*} $$ StartLayout 1st Row log upper P upper L left parenthesis 2 n right parenthesis minus log upper P upper L left parenthesis n right parenthesis greater than or equals alpha left parenthesis 2 Superscript 2 divided by 3 Baseline minus 1 right parenthesis n Superscript 2 divided by 3 Baseline minus StartFraction 25 Over 36 EndFraction log 2 minus StartFraction 200 left parenthesis 1 plus 2 Superscript negative 2 divided by 3 Baseline right parenthesis Over n Superscript 2 divided by 3 Baseline EndFraction period EndLayout

Therefore,

$$ \begin{align*} d_n\geq n^{-1/3}\bigg( \alpha(2^{2/3}-1)-\frac{25\log2}{36n^{2/3}} -\frac{200(1+2^{-2/3})}{n^{4/3}}\bigg) \geq\gamma n^{-1/3}, \end{align*} $$ d Subscript n Baseline greater than or equals n Superscript negative 1 divided by 3 Baseline left parenthesis alpha left parenthesis 2 Superscript 2 divided by 3 Baseline minus 1 right parenthesis minus StartFraction 25 log 2 Over 36 n Superscript 2 divided by 3 Baseline EndFraction minus StartFraction 200 left parenthesis 1 plus 2 Superscript negative 2 divided by 3 Baseline right parenthesis Over n Superscript 4 divided by 3 Baseline EndFraction right parenthesis greater than or equals gamma n Superscript negative 1 divided by 3 Baseline comma

as the expression in parentheses is increasing in n.

Proof of Corollary 1.4.

For every integer $n\geq 100$ n greater than or equals 100 , apply Theorem 1.1 with $(f_n,n_0)=(p(n),25)$ left parenthesis f Subscript n Baseline comma n 0 right parenthesis equals left parenthesis p left parenthesis n right parenthesis comma 25 right parenthesis . Lemma 3.2 gives

$$ \begin{align*} 2\log_b p(n)\leq\frac{4\pi}{\sqrt6\log b}\sqrt n \end{align*} $$ 2 log Subscript b Baseline p left parenthesis n right parenthesis less than or equals StartFraction 4 pi Over StartRoot 6 EndRoot log b EndFraction StartRoot n EndRoot

and

$$ \begin{align*} \frac{\rho_{w,b}(1-\rho_{w,b})} {\log_b p(n)-\log_b p(n-1)} \leq\frac{\rho_{w,b}\log b}{\beta}\sqrt n <1.1\rho_{w,b}\sqrt n\log b. \end{align*} $$ StartLayout 1st Row StartFraction rho Subscript w comma b Baseline left parenthesis 1 minus rho Subscript w comma b Baseline right parenthesis Over log Subscript b Baseline p left parenthesis n right parenthesis minus log Subscript b Baseline p left parenthesis n minus 1 right parenthesis EndFraction less than or equals StartFraction rho Subscript w comma b Baseline log b Over beta EndFraction StartRoot n EndRoot less than 1.1 rho Subscript w comma b Baseline StartRoot n EndRoot log b period EndLayout

Since $p(25)>2$ p left parenthesis 25 right parenthesis greater than 2 and $\rho _{w,b}\leq \log _b2$ rho Subscript w comma b Baseline less than or equals log Subscript b Baseline 2 ,

$$ \begin{align*} \lambda_{w,b,p}(25) =27+2\rho_{w,b}-2\log_b p(25) \leq27. \end{align*} $$ StartLayout 1st Row lamda Subscript w comma b comma p Baseline left parenthesis 25 right parenthesis equals 27 plus 2 rho Subscript w comma b Baseline minus 2 log Subscript b Baseline p left parenthesis 25 right parenthesis less than or equals 27 period EndLayout

This proves the asserted estimate for $p(n)$ p left parenthesis n right parenthesis .

Likewise, for every integer $n\geq 2829$ n greater than or equals 2829 , apply Theorem 1.1 with $(f_n,n_0)=(\mathrm {PL}(n),12)$ left parenthesis f Subscript n Baseline comma n 0 right parenthesis equals left parenthesis upper P upper L left parenthesis n right parenthesis comma 12 right parenthesis . Lemma 3.4 gives

$$ \begin{align*} 2\log_b\mathrm{PL}(n) \leq\frac{2\alpha}{\log b}n^{2/3} = \frac{6\zeta(3)^{1/3}}{2^{2/3}\log b}n^{2/3} \end{align*} $$ 2 log Subscript b Baseline upper P upper L left parenthesis n right parenthesis less than or equals StartFraction 2 alpha Over log b EndFraction n Superscript 2 divided by 3 Baseline equals StartFraction 6 zeta left parenthesis 3 right parenthesis Superscript 1 divided by 3 Baseline Over 2 Superscript 2 divided by 3 Baseline log b EndFraction n Superscript 2 divided by 3

and

$$ \begin{align*} \frac{\rho_{w,\,b}(1-\rho_{w,\,b})} {\log_b\mathrm{PL}(n)-\log_b\mathrm{PL}(n-1)} \leq\frac{\rho_{w,\,b}\log b}{\gamma}n^{1/3} <0.9\rho_{w,\,b}n^{1/3}\log b. \end{align*} $$ StartLayout 1st Row StartFraction rho Subscript w comma b Baseline left parenthesis 1 minus rho Subscript w comma b Baseline right parenthesis Over log Subscript b Baseline upper P upper L left parenthesis n right parenthesis minus log Subscript b Baseline upper P upper L left parenthesis n minus 1 right parenthesis EndFraction less than or equals StartFraction rho Subscript w comma b Baseline log b Over gamma EndFraction n Superscript 1 divided by 3 Baseline less than 0.9 rho Subscript w comma b Baseline n Superscript 1 divided by 3 Baseline log b period EndLayout

Since $\mathrm {PL}(12)>2$ upper P upper L left parenthesis 12 right parenthesis greater than 2 and $\rho _{w,\,b}\leq \log _b2$ rho Subscript w comma b Baseline less than or equals log Subscript b Baseline 2 ,

$$ \begin{align*} \lambda_{w,\,b,\mathrm{PL}}(12) =14+2\rho_{w,\,b}-2\log_b\mathrm{PL}(12) \leq14. \end{align*} $$ StartLayout 1st Row lamda Subscript w comma b comma upper P upper L Baseline left parenthesis 12 right parenthesis equals 14 plus 2 rho Subscript w comma b Baseline minus 2 log Subscript b Baseline upper P upper L left parenthesis 12 right parenthesis less than or equals 14 period EndLayout

This proves the plane-partition estimate.

Before giving the proof of Corollary 1.5, we record the following elementary lemma.

Lemma 3.5. Let s be a positive integer, $a>0$ a greater than 0 , and $a_1,\ldots ,a_s\geq 0$ a 1 comma ellipsis comma a Subscript s Baseline greater than or equals 0 . If a real number x satisfies

$$ \begin{align*} x>a+\sum_{1\leq\nu\leq s}\frac{a_\nu}{a^\nu}, \end{align*} $$ x greater than a plus sigma summation Underscript 1 less than or equals nu less than or equals s Endscripts StartFraction a Subscript nu Baseline Over a Superscript nu Baseline EndFraction comma

then

$$ \begin{align*} x>a+\sum_{1\leq\nu\leq s}\frac{a_\nu}{x^\nu}. \end{align*} $$ x greater than a plus sigma summation Underscript 1 less than or equals nu less than or equals s Endscripts StartFraction a Subscript nu Baseline Over x Superscript nu Baseline EndFraction period

Proof. The hypothesis implies $x>a>0$ x greater than a greater than 0 . Hence, $a_\nu /x^\nu \leq a_\nu /a^\nu $ a Subscript nu Baseline divided by x Superscript nu Baseline less than or equals a Subscript nu Baseline divided by a Superscript nu for each $\nu $ nu and summing these inequalities gives the result.

Proof of Corollary 1.5.

Put $\rho =\rho _{w,b}$ rho equals rho Subscript w comma b and $L=\log b$ upper L equals log b . For partitions, Corollary 1.4 implies that

$$ \begin{align*} \frac{S_{w,\,b,p}(n)}{\rho\sqrt n}> \sqrt n- \bigg(\frac{4\pi}{\rho\sqrt6L}+1.1L\bigg) -\frac{27}{\rho\sqrt n}. \end{align*} $$ StartFraction upper S Subscript w comma b comma p Baseline left parenthesis n right parenthesis Over rho StartRoot n EndRoot EndFraction greater than StartRoot n EndRoot minus left parenthesis StartFraction 4 pi Over rho StartRoot 6 EndRoot upper L EndFraction plus 1.1 upper L right parenthesis minus StartFraction 27 Over rho StartRoot n EndRoot EndFraction period

Lemma 3.5, applied with

$$ \begin{align*} a=\frac{4\pi}{\rho\sqrt6L}+1.1L, \quad a_1=\frac{27}{\rho}, \end{align*} $$ a equals StartFraction 4 pi Over rho StartRoot 6 EndRoot upper L EndFraction plus 1.1 upper L comma a 1 equals StartFraction 27 Over rho EndFraction comma

therefore shows that $S_{w,b,p}(n)>0$ upper S Subscript w comma b comma p Baseline left parenthesis n right parenthesis greater than 0 for $n\geq 100$ n greater than or equals 100 , as soon as

$$ \begin{align*} \sqrt n> \frac{4\pi}{\rho\sqrt6L}+1.1L +\frac{27}{\rho} \bigg(\frac{4\pi}{\rho\sqrt6L}+1.1L\bigg)^{-1}. \end{align*} $$ StartRoot n EndRoot greater than StartFraction 4 pi Over rho StartRoot 6 EndRoot upper L EndFraction plus 1.1 upper L plus StartFraction 27 Over rho EndFraction left parenthesis StartFraction 4 pi Over rho StartRoot 6 EndRoot upper L EndFraction plus 1.1 upper L right parenthesis Superscript negative 1 Baseline period

Squaring the right-hand side and bounding its coefficients gives the sufficient condition

$$ \begin{align*} n\geq \frac{8\pi^2/3}{\rho^2L^2} +\frac{66}{\rho}+29L^2. \end{align*} $$ n greater than or equals StartFraction 8 pi squared divided by 3 Over rho squared upper L squared EndFraction plus StartFraction 66 Over rho EndFraction plus 29 upper L squared period

The right-hand side is greater than $100$ 100 . Thus, every integer n satisfying the displayed sufficient condition lies in the range of Corollary 1.4. It follows that $S_{w,b,p}(n)>0$ upper S Subscript w comma b comma p Baseline left parenthesis n right parenthesis greater than 0 and (1.1) yields the asserted bound for $N_p(w,b)$ upper N Subscript p Baseline left parenthesis w comma b right parenthesis .

For plane partitions, write $x=n^{1/3}$ x equals n Superscript 1 divided by 3 . Corollary 1.4 gives

$$ \begin{align*} \frac{S_{w,b,\mathrm{PL}}(n)}{\rho n^{2/3}}> x- \frac{3\cdot2^{1/3}\zeta(3)^{1/3}}{\rho L} -\frac{0.9L}{x}-\frac{14}{\rho x^2}. \end{align*} $$ StartFraction upper S Subscript w comma b comma upper P upper L Baseline left parenthesis n right parenthesis Over rho n Superscript 2 divided by 3 Baseline EndFraction greater than x minus StartFraction 3 dot 2 Superscript 1 divided by 3 Baseline zeta left parenthesis 3 right parenthesis Superscript 1 divided by 3 Baseline Over rho upper L EndFraction minus StartFraction 0.9 upper L Over x EndFraction minus StartFraction 14 Over rho x squared EndFraction period

Set

$$ \begin{align*} A:=\frac{3\cdot2^{1/3}\zeta(3)^{1/3}}{\rho L}. \end{align*} $$ upper A colon equals StartFraction 3 dot 2 Superscript 1 divided by 3 Baseline zeta left parenthesis 3 right parenthesis Superscript 1 divided by 3 Baseline Over rho upper L EndFraction period

Since

$$ \begin{align*} \frac{0.9L}{A}+\frac{14}{\rho A^2} = \bigg( \frac{0.9}{3\cdot2^{1/3}\zeta(3)^{1/3}} + \frac{14}{9\cdot2^{2/3}\zeta(3)^{2/3}} \bigg)\rho L^2 <1.1\rho L^2, \end{align*} $$ StartFraction 0.9 upper L Over upper A EndFraction plus StartFraction 14 Over rho upper A squared EndFraction equals left parenthesis StartFraction 0.9 Over 3 dot 2 Superscript 1 divided by 3 Baseline zeta left parenthesis 3 right parenthesis Superscript 1 divided by 3 Baseline EndFraction plus StartFraction 14 Over 9 dot 2 Superscript 2 divided by 3 Baseline zeta left parenthesis 3 right parenthesis Superscript 2 divided by 3 Baseline EndFraction right parenthesis rho upper L squared less than 1.1 rho upper L squared comma

Lemma 3.5 shows that it suffices to require $ x\geq A+1.1\rho L^2. $ x greater than or equals upper A plus 1.1 rho upper L squared period Cubing and using

$$ \begin{align*} (3\cdot2^{1/3}\zeta(3)^{1/3})^3=54\zeta(3) \end{align*} $$ StartLayout 1st Row left parenthesis 3 dot 2 Superscript 1 divided by 3 Baseline zeta left parenthesis 3 right parenthesis Superscript 1 divided by 3 Baseline right parenthesis cubed equals 54 zeta left parenthesis 3 right parenthesis EndLayout

shows that this is implied by

$$ \begin{align*} n\geq \frac{54\zeta(3)}{\rho^3L^3} +\frac{65}{\rho} +15\rho L^3 +2\rho^3L^6. \end{align*} $$ n greater than or equals StartFraction 54 zeta left parenthesis 3 right parenthesis Over rho cubed upper L cubed EndFraction plus StartFraction 65 Over rho EndFraction plus 15 rho upper L cubed plus 2 rho cubed upper L Superscript 6 Baseline period

The right-hand side is greater than $132$ 132 . Thus, after adding $2697$ 2697 , every integer n satisfying the preceding sufficient condition is at least $2829$ 2829 . Corollary 1.4 therefore applies and (1.1) yields the second assertion.

Acknowledgement

The authors used OpenAI’s ChatGPT for exploratory discussions concerning possible proof strategies and language editing. All mathematical arguments, verification and responsibility for the final manuscript remain with the authors.

Footnotes

This work was partially supported by Guangxi Natural Science Foundation (No. 2026GXNSFBA00640038).

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