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\author{\textbf{Runbo Li}}
\address{Department of Mathematics, Imperial College London, London SW7 2AZ, UK}
\email{runbo.li26@imperial.ac.uk, runbo.li.carey@gmail.com}

%\date{\today}

\makeatletter
\@namedef{subjclassname@2020}{\textup{}2020 Mathematics Subject Classification}
\makeatother

%------------Title and keywords------
\title[]{Note on variable role-reversals}
\subjclass[2020]{\textbf{11N05}, \textbf{11N35}, \textbf{11N36}}
\keywords{\textbf{Prime}, \textbf{Sieve methods}, \textbf{Variable role-reversals}}

\begin{document}

\begin{abstract}
In \cite{LRB2602}, the author uses variable role-reversals only when $\theta < \frac{45}{89} - \varepsilon$. In this note, we shall discuss how to use variable role-reversals when $\frac{45}{89} - \varepsilon \leqslant \theta < \frac{11}{20} - \varepsilon$.
\end{abstract}

\maketitle

%\tableofcontents

In a recent paper \cite{LRB2602}, the author studied the average distribution of majorants and minorants of $\mathbbm{1}_p(x)$ in arithmetic progressions to moduli larger than $\sqrt{x}$. During the constructing process of those majorants and minorants, the author applied the \textit{variable role-reversal}, which is also a very powerful technique in many other applications of Harman's sieve. For the sake of simplicity, the author only uses role-reversals when $\theta < \frac{45}{89} - \varepsilon$. Here, we generalize the application of role-reversals in \cite{LRB2602} to a wider range $\theta < \frac{11}{20} - \varepsilon$. We mention that further extension to $\theta < \frac{4}{7} - \varepsilon$ may still be possible.

From here we suppose that $\frac{1}{2} \leqslant \theta < \frac{11}{20} - \varepsilon$. Almost all of the notations and definitions in this note are same as those in \cite{LRB2602}. We only change the definitions of $\kappa$ and $\tau$ in order to state our Lemma~\ref{l02} and Lemma~\ref{l03} below simpler. Note that $\kappa$ here equals to $\kappa$ in \cite{LRB2602} when $\theta \leqslant \frac{7}{13} - \varepsilon$ and becomes $\kappa^{\prime}$ in \cite{LRB2602} when $\theta > \frac{7}{13} - \varepsilon$. The definitions of some sets below are changed correspondingly.
\begin{align*}
\kappa = \kappa(\theta) =
\begin{cases}
\frac{5-8\theta}{6} - \varepsilon, & \theta \leqslant \frac{17}{32} - \varepsilon, \\
\frac{5-8\theta}{12} - 3 \varepsilon, & \frac{17}{32} - \varepsilon < \theta \leqslant \frac{7}{13} - \varepsilon, \\
\frac{11-20\theta}{6} - 2 \varepsilon, & \frac{7}{13} - \varepsilon < \theta \leqslant \frac{11}{20} - \varepsilon, \\
\end{cases}
\end{align*}
\begin{align*}
\tau = \tau(\theta) =
\begin{cases}
\frac{3(1-\theta)}{5} - \varepsilon, & \theta \leqslant \frac{11}{21}, \\
\frac{2}{7} - \varepsilon, & \frac{11}{21} < \theta \leqslant \frac{6}{11} - \varepsilon, \\
\frac{5-6\theta}{7} - \varepsilon, & \frac{6}{11} - \varepsilon < \theta. \\
\end{cases}
\end{align*}

\begin{align}
\nonumber \boldsymbol{S} =&\ \boldsymbol{S}(\theta) = \left\{(s, t): s \leqslant 1 - \theta - \varepsilon,\ s + 2t \leqslant 2 - 2 \theta - \varepsilon,\ s + 4t \leqslant 2 - \theta - \varepsilon \right\}, \\
\nonumber \boldsymbol{S}_{j} =&\ \boldsymbol{S}_{j}(\theta) = \left\{\boldsymbol{\alpha}_{j}: \boldsymbol{\alpha}_{j} \text{ partitions exactly into } \boldsymbol{S} \right\}, \\
\nonumber \boldsymbol{A}_{j} =&\ \boldsymbol{A}_{j}(\theta) = \left\{\boldsymbol{\alpha}_{j}: (\log \log x)^{-3} \leqslant \alpha_j \leqslant \cdots \leqslant \alpha_1 < \tau,\ \alpha_1 + \cdots + \alpha_j \leqslant 1 \right\}, \\
\nonumber \boldsymbol{A}_{j}^{\prime} =&\ \boldsymbol{A}_{j}^{\prime}(\theta) = \left\{\boldsymbol{\alpha}_{j}: (\log \log x)^{-3} \leqslant \alpha_j \leqslant \cdots \leqslant \alpha_1 < \frac{3}{7} + \varepsilon,\ \alpha_1 + \cdots + \alpha_j \leqslant 1 \right\}, \\
\nonumber \boldsymbol{U}_{j} =&\ \boldsymbol{U}_{j}(\theta) = \left\{\boldsymbol{\alpha}_{j}: \boldsymbol{\alpha}_{j} \in \boldsymbol{A}_{j},\ (\alpha_1, \ldots, \alpha_j, 2\theta-1+\varepsilon) \in \boldsymbol{S}_{j+1} \right\}, \\
\nonumber \boldsymbol{U}_{j}^{\prime} =&\ \boldsymbol{U}_{j}^{\prime}(\theta) = \left\{\boldsymbol{\alpha}_{j}: \boldsymbol{\alpha}_{j} \in \boldsymbol{A}_{j}^{\prime},\ (\alpha_1, \ldots, \alpha_j, 2\theta-1+\varepsilon) \in \boldsymbol{S}_{j+1} \right\}.
\end{align}

The next three lemmas come directly from \cite{LRB2602}.
\begin{lemma}\label{l01}
Let $\frac{1}{2} \leqslant \theta \leqslant \frac{45}{89} - \varepsilon$, $m_1 \sim M_1$ and $m_2 \sim M_2$. Then
$$
\sum_{\left(\frac{\log M_1}{\log x}, \frac{\log M_2}{\log x} \right) \in \boldsymbol{S}} a_{1, m_1} a_{2, m_2} S\left(\mathcal{A}^{q}_{m_1 m_2}, x^{\kappa}\right)
$$
has an asymptotic formula.
\end{lemma}
\begin{proof}
This lemma is [\cite{LRB2602}, Lemma 2.14].
\end{proof}
\begin{lemma}\label{l02}
Let $\frac{1}{2} \leqslant \theta \leqslant \frac{17}{32} - \varepsilon$. Then
$$
\sum_{\boldsymbol{\alpha}_{j} \in \boldsymbol{U}_{j}^{\prime}} S\left(\mathcal{A}^{q}_{p_1 \cdots p_j}, x^{\kappa}\right)
$$
has an asymptotic formula.
\end{lemma}
\begin{proof}
This lemma can be proved in the same way of [\cite{LRB2602}, Lemma 2.13].
\end{proof}
\begin{lemma}\label{l03}
Let $\frac{1}{2} \leqslant \theta \leqslant \frac{11}{20} - \varepsilon$. Then
$$
\sum_{\boldsymbol{\alpha}_{j} \in \boldsymbol{U}_{j}} S\left(\mathcal{A}^{q}_{p_1 \cdots p_j}, x^{\kappa}\right)
$$
has an asymptotic formula.
\end{lemma}
\begin{proof}
This lemma can be proved in the same way of [\cite{LRB2602}, Lemma 2.13].
\end{proof}

Next we shall briefly introduce the use of role-reversals in this problem. Suppose that we want to give a nontrivial estimate for the sum
\begin{equation}
\sum_{\boldsymbol{\alpha}_{j}} S\left(\mathcal{A}^{q}_{p_1 \cdots p_j}, p_j \right).
\end{equation}
Assume that the sum (1) has a ``right sign'' (that is, $-$ in the construction of a majorant or $+$ in the construction of a minorant) and satisfies $\boldsymbol{\alpha}_{j} \in \boldsymbol{S}_{j}$ (or $\boldsymbol{U}_{j}^{\prime}$ or $\boldsymbol{U}_{j}$, depending on the value of $\theta$). A direct application of Buchstab's identity gives
\begin{equation}
\sum_{\boldsymbol{\alpha}_{j}} S\left(\mathcal{A}^{q}_{p_1 \cdots p_j}, p_j \right) = \sum_{\boldsymbol{\alpha}_{j}} S\left(\mathcal{A}^{q}_{p_1 \cdots p_j}, x^{\kappa} \right) - \sum_{\substack{\boldsymbol{\alpha}_{j} \\ \kappa \leqslant \alpha_{j+1} < \min\left(\alpha_{j}, \frac{1}{2}(1 - \alpha_1 - \cdots - \alpha_{j}) \right) }} S\left(\mathcal{A}^{q}_{p_1 \cdots p_j p_{j+1}}, p_{j+1} \right).
\end{equation}
After applying Buchstab's identity once, sometimes we cannot use it again since $(\alpha_1, \ldots, \alpha_{j}, \alpha_{j}) \notin \boldsymbol{S}_{j+1}$ (or $\boldsymbol{U}_{j+1}^{\prime}$ or $\boldsymbol{U}_{j+1}$). In this case, we cannot give an asymptotic formula for the sum (with a ``wrong sign'')
\begin{equation}
\sum_{\substack{\boldsymbol{\alpha}_{j} \\ \kappa \leqslant \alpha_{j+1} < \min\left(\alpha_{j}, \frac{1}{2}(1 - \alpha_1 - \cdots - \alpha_{j}) \right) }} S\left(\mathcal{A}^{q}_{p_1 \cdots p_j p_{j+1}}, x^{\kappa} \right)
\end{equation}
after the second Buchstab iteration. The idea of a role-reversal is to apply the second Buchstab iteration on a different variable, usually the largest explicit prime variable (denoted by $p_1$):
$$
(p_1 p_2 \cdots p_{j+1}) \beta \Rightarrow (\beta p_2 \cdots p_{j+1}) p_1.
$$
Reversing the roles of $\beta$ and $p_1$ gives
\begin{align}
\nonumber &\ \sum_{\substack{\boldsymbol{\alpha}_{j} \\ \kappa \leqslant \alpha_{j+1} < \min\left(\alpha_{j}, \frac{1}{2}(1 - \alpha_1 - \cdots - \alpha_{j}) \right) }} S\left(\mathcal{A}^{q}_{p_1 \cdots p_j p_{j+1}}, p_{j+1} \right) \\
=&\ \sum_{\substack{\boldsymbol{\alpha}_{j} \\ \kappa \leqslant \alpha_{j+1} < \min\left(\alpha_{j}, \frac{1}{2}(1 - \alpha_1 - \cdots - \alpha_{j}) \right) }} S\left(\mathcal{A}^{q}_{\beta p_2 \cdots p_j p_{j+1}}, \left(\frac{2x}{\beta p_2 \cdots p_{j+1}}\right)^{\frac{1}{2}} \right).
\end{align}
Then, using the second Buchstab's identity (actually on the variable $p_1$ in (3)), we have
\begin{align}
\nonumber &\ \sum_{\boldsymbol{\alpha}_{j+1}} S\left(\mathcal{A}^{q}_{\beta p_2 \cdots p_j p_{j+1}}, \left(\frac{2x}{\beta p_2 \cdots p_{j+1}}\right)^{\frac{1}{2}} \right) \\
=&\ \sum_{\boldsymbol{\alpha}_{j+1}} S\left(\mathcal{A}^{q}_{\beta p_2 \cdots p_j p_{j+1}}, x^{\kappa} \right) - \sum_{\substack{\boldsymbol{\alpha}_{j+1} \\ \kappa \leqslant \alpha_{j+2} < \frac{1}{2} \alpha_1 }} S\left(\mathcal{A}^{q}_{\beta p_2 \cdots p_j p_{j+1} p_{j+2}}, p_{j+2} \right).
\end{align}
Note that $\beta \asymp x^{1 - \alpha_1 - \cdots - \alpha_{j+1}}$. Now we only need the condition $(1 - \alpha_1 - \cdots - \alpha_{j}, \alpha_2, \ldots, \alpha_{j}) \in \boldsymbol{S}_{j}$ (or $\boldsymbol{U}_{j}^{\prime}$ or $\boldsymbol{U}_{j}$) instead of $(\alpha_1, \ldots, \alpha_{j}, \alpha_{j}) \in \boldsymbol{S}_{j+1}$ (or $\boldsymbol{U}_{j+1}^{\prime}$ or $\boldsymbol{U}_{j+1}$). We can then discard parts of the last sum on the right-hand side of (5) with $(1 - \alpha_1 - \cdots - \alpha_{j+1}, \alpha_2, \cdots, \alpha_{j+2}) \notin \boldsymbol{G}_{j+2}$. More role-reversals are also possible to apply on other variables (e.g. $p_2$, $p_{j+2}$, ...) of this sum.

When $\theta \leqslant \frac{45}{89} - \varepsilon$, $\boldsymbol{\alpha}_{j} \in \boldsymbol{S}_{j}$ and $(1 - \alpha_1 - \cdots - \alpha_{j}, \alpha_2, \ldots, \alpha_{j}) \in \boldsymbol{S}_{j}$, by the definition of $\boldsymbol{S}_{j}$, we can give asymptotic formulas for the first sums on the right-hand side of (2) and (5) using Lemma~\ref{l01} directly. In \cite{LRB2602} the author indeed used Lemma~\ref{l01} to give an asymptotic formula for the first sum on the right-hand side of (5). When $\theta > \frac{45}{89} - \varepsilon$, however, the available results (Lemmas~\ref{l02}--\ref{l03}) require a descending order between prime variables in the definitions of $\boldsymbol{A}_{j}$ and $\boldsymbol{A}_{j}^{\prime}$. Thus, at least a splitting argument on the (possibly non-prime) variable $\beta$ is necessary.

Since we have $\left(\beta, P(p_{j+1})\right) = 1$ and $\alpha_1, \alpha_2, \ldots, \alpha_{j+1} \geqslant \kappa$, any prime factor of $\beta$ is $\geqslant x^{\kappa}$, and $\frac{\log \beta}{\log x} = 1 - \alpha_1 - \cdots - \alpha_{j+1} \leqslant 1 - (j+1) \kappa$. Hence, the number of prime factors of $\beta$ is no more than
\begin{equation}
\lceil \frac{1 - (j+1) \kappa}{\kappa} \rceil < \infty.
\end{equation}
Now, we can split $\beta$ into a product of finitely many prime variables. Write $\beta = p_1^{\prime} p_2^{\prime} \cdots p_l^{\prime}$, $p_i^{\prime} \asymp x^{\alpha_i^{\prime}}$ and $\alpha_1^{\prime} \geqslant \alpha_2^{\prime} \geqslant \ldots \geqslant \alpha_l^{\prime} \geqslant \alpha_{j+1} \geqslant \kappa$. Note that we have $1 - \alpha_1 - \cdots - \alpha_{j+1} = \alpha_1^{\prime} + \cdots + \alpha_l^{\prime}$.

\textbf{Case 1.} $\frac{45}{89} - \varepsilon < \theta \leqslant \frac{17}{32} - \varepsilon$. In order to give an asymptotic formula for the first sum on the right-hand side of (2), we need $\boldsymbol{\alpha}_{j} \in \boldsymbol{U}_{j}^{\prime}$, and thus $\alpha_1 < \frac{3}{7} + \varepsilon$. We also need to give an asymptotic formula for the first sum on the right-hand side of (5). Assume first that $\alpha_1^{\prime} < \alpha_1$. Since
$$
\alpha_1 > \alpha_2 > \ldots > \alpha_{j+1} \geqslant \kappa \quad \text{and} \quad \alpha_1 > \alpha_1^{\prime} \geqslant \alpha_2^{\prime} \geqslant \ldots \geqslant \alpha_l^{\prime} \geqslant \alpha_{j+1} \geqslant \kappa,
$$
we know that the $j+l$ variables $\alpha_2, \ldots, \alpha_{j+1}, \alpha_1^{\prime}, \ldots \alpha_l^{\prime}$ can be arranged into a descending order (denoted by $\boldsymbol{\alpha}_{j+l}^{\prime}$) with the largest variable (=$\max(\alpha_2, \alpha_1^{\prime})$) less than $\alpha_1$. Hence, we have $\boldsymbol{\alpha}_{j+l}^{\prime} \in \boldsymbol{A}_{j+l}^{\prime}$. Now, we only need the condition $(\boldsymbol{\alpha}_{j+l}^{\prime}, 2 \theta - 1 + \varepsilon) \in \boldsymbol{S}_{j+l+1}$, which can be satisfied if we have
$$
(\alpha_2, \ldots, \alpha_j, \alpha_{j+1}, 1 - \alpha_1 - \cdots - \alpha_{j+1}, 2 \theta - 1 + \varepsilon) \in \boldsymbol{S}_{j+2}.
$$
Again, this can be satisfied if we have
$$
(\alpha_2, \ldots, \alpha_j, 1 - \alpha_1 - \cdots - \alpha_j, 2 \theta - 1 + \varepsilon) \in \boldsymbol{S}_{j+1}.
$$

Assume that $\alpha_1^{\prime} \geqslant \alpha_1$. Then we have $1 - \alpha_1 - \cdots - \alpha_{j+1} = \alpha_1^{\prime} + \cdots + \alpha_l^{\prime} > \alpha_1^{\prime} \geqslant \alpha_1$. Now, if we have
$$
(\alpha_2, \ldots, \alpha_j, \alpha_{j+1}, 1 - \alpha_1 - \cdots - \alpha_{j+1}, 2 \theta - 1 + \varepsilon) \in \boldsymbol{S}_{j+2},
$$
we must have
$$
(\alpha_1, \alpha_2, \ldots, \alpha_j, \alpha_{j+1}, 2 \theta - 1 + \varepsilon) \in \boldsymbol{S}_{j+2},
$$
and we can give an asymptotic formula for the sum (3). Thus, a role-reversal is not necessary to apply in this case, and we can just perform another direct Buchstab iteration on the last sum on the right-hand side of (2) to get the sums (3) and
\begin{equation}
\sum_{\substack{\boldsymbol{\alpha}_{j} \\ \kappa \leqslant \alpha_{j+1} < \min\left(\alpha_{j}, \frac{1}{2}(1 - \alpha_1 - \cdots - \alpha_{j}) \right) \\ \kappa \leqslant \alpha_{j+2} < \min\left(\alpha_{j+1}, \frac{1}{2}(1 - \alpha_1 - \cdots - \alpha_{j} - \alpha_{j+1}) \right) }} S\left(\mathcal{A}^{q}_{p_1 \cdots p_j p_{j+1} p_{j+2}}, p_{j+2} \right).
\end{equation}
We can then discard parts of (7) with $\boldsymbol{\alpha}_{j+2} \notin \boldsymbol{G}_{j+2}$.

\textbf{Case 2.} $\frac{17}{32} - \varepsilon < \theta < \frac{11}{20} - \varepsilon$. The discussions are very similar to those in \textbf{Case 1}. The only difference is that we need $\boldsymbol{\alpha}_{j} \in \boldsymbol{U}_{j}$ instead of $\boldsymbol{\alpha}_{j} \in \boldsymbol{U}_{j}^{\prime}$ (Lemma~\ref{l02} is not available here), so that $\alpha_1 < \tau$.

In other similar cases where $2 \theta - 1 + \varepsilon$ is replaced by $2 \theta_1 + \theta_2 - 1 + \varepsilon$ or $\varepsilon$ (see \cite{LRB2602}), and when performing further role-reversals on a sum that a role-reversal has already been performed on it (which means that more almost-prime variables become explicit, see [\cite{LRBLPF3}, (34)] for example), the required conditions are similar.

As an application, we get the following better bounds for $C_0^{*}(\theta)$ using the \textbf{first method} in \cite{LRB2602} by applying role-reversals in the extended range of $\theta$. Many other bounds in \cite{LRB2602} can be improved using a similar approach.

\newpage

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline \boldmath{$\theta$} & \boldmath{$C_0^{*}(\theta)$} \textbf{(Old)} & \boldmath{$C_0^{*}(\theta)$} \textbf{(New)} & \boldmath{$\theta$} & \boldmath{$C_0^{*}(\theta)$} \textbf{(Old)} & \boldmath{$C_0^{*}(\theta)$} \textbf{(New)} \\
\hline $0.501$ & \boldmath{$0.8636$} & \boldmath{$0.8636$} & $0.517$ & $0.4455$ & \boldmath{$0.4936$} \\
\hline $0.502$ & \boldmath{$0.8455$} & \boldmath{$0.8455$} & $0.518$ & $0.4086$ & \boldmath{$0.4624$} \\
\hline $0.503$ & \boldmath{$0.8298$} & \boldmath{$0.8298$} & $0.519$ & $0.3744$ & \boldmath{$0.4261$} \\
\hline $0.504$ & \boldmath{$0.8135$} & \boldmath{$0.8135$} & $0.520$ & $0.3363$ & \boldmath{$0.3890$} \\
\hline $0.505$ & \boldmath{$0.7972$} & \boldmath{$0.7972$} & $0.521$ & $0.2973$ & \boldmath{$0.3510$} \\
\hline $0.506$ & $0.7299$ & \boldmath{$0.7730$} & $0.522$ & $0.2546$ & \boldmath{$0.3113$} \\
\hline $0.507$ & $0.7082$ & \boldmath{$0.7505$} & $0.523$ & $0.2148$ & \boldmath{$0.2714$} \\
\hline $0.508$ & $0.6894$ & \boldmath{$0.7334$} & $0.524$ & $0.1706$ & \boldmath{$0.2277$} \\
\hline $0.509$ & $0.6606$ & \boldmath{$0.7045$} & $0.525$ & $0.1218$ & \boldmath{$0.1785$} \\
\hline $0.510$ & $0.6369$ & \boldmath{$0.6808$} & $0.526$ & $0.0778$ & \boldmath{$0.1324$} \\
\hline $0.511$ & $0.6128$ & \boldmath{$0.6551$} & $0.527$ & $0.0267$ & \boldmath{$0.0779$} \\
\hline $0.512$ & $0.5907$ & \boldmath{$0.6373$} & $0.528$ & $0$ & \boldmath{$0.0256$} \\
\hline $0.513$ & $0.5567$ & \boldmath{$0.6033$} & $0.529$ & $0$ & $0$ \\
\hline $0.514$ & $0.5332$ & \boldmath{$0.5796$} & $0.530$ & $0$ & $0$ \\
\hline $0.515$ & $0.5040$ & \boldmath{$0.5496$} & $0.531$ & $0$ & $0$ \\
\hline $0.516$ & $0.4762$ & \boldmath{$0.5245$} & & & \\
\hline
\end{tabular} \\
\textbf{Table 1: Lower Bounds for }\boldmath{$C_0^{*}(\theta)$} \textbf{(First Method, }\boldmath{$\frac{1}{2} < \theta \leqslant \frac{17}{32} - \varepsilon$}\textbf{)}
\end{center}







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