\documentclass[11pt,a4paper]{article} \usepackage[utf8]{inputenc} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage[margin=2.35cm]{geometry} \usepackage{graphicx} \usepackage{booktabs} \usepackage{xcolor} \usepackage[colorlinks=true,linkcolor=blue!50!black,citecolor=blue!50!black,urlcolor=blue!50!black]{hyperref} \usepackage{tcolorbox} \usepackage{enumitem} \newtheorem{theorem}{Theorem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \newtheorem{observation}[theorem]{Numerical observation} \newcommand{\C}{\mathcal{C}} \newcommand{\K}{\mathcal{K}} \newcommand{\E}{\mathcal{E}} \renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}} \newcommand{\rad}{\operatorname{rad}} \title{\textbf{Cross-Energy of Dirichlet Partial Sums.\\ Part 10: The Caceres Prime Function}\\[4pt] \large A square-root damped prime weight, its exact Prime Zeta structure,\\ and a dual spectrum of the Riemann zeros} \author{Pedro Caceres\\[2pt] \small Universidad Europea de Valencia\\ \small \href{mailto:pedrojesus.caceres@universidadeuropea.es}{pedrojesus.caceres@universidadeuropea.es}} \date{October 2026; book edition, 2 November 2026.\\ \small Part 10 of the book-series \emph{Cross-Energy of Dirichlet Partial Sums} (forty-seven parts, Parts 0--46; block II: Primes and Euler products), formerly a standalone note. Part 0 is the guide and Part~1 \cite{CE1} gives the plan of the series.} \begin{document} \maketitle \begin{abstract}\sloppy We study the \emph{Caceres Prime Function} \[ \C(s)=\sum_{n\ge 2}\frac{a(n)}{n^{s}},\qquad a(n)=\sum_{p^k\,\|\,n}\frac{\log p}{\sqrt{p^{k}}}, \] the Dirichlet series of the additive function that weights each prime factor $p$ of $n$ by $\log p$ divided by the square root of the exact power of $p$ dividing $n$. \emph{Structure.} We prove that $\C=\zeta\K$, where the \emph{prime kernel}, $\K(s)=\sum_p\log p\,(1-p^{-s})/(p^{s+1/2}-1)$, is an exact telescoping series of shifted derivatives of the Prime Zeta function. So $\K$ is meromorphic in $\Re s>-\frac14$, and the line $\Re s=-\frac14$ is a natural boundary. Every zero $\frac12+i\gamma$ of $\zeta$ produces two singular points of $\K$, at $i\gamma$ and $i\gamma/2$, which are poles unless $2\gamma$ is again an ordinate with twice the multiplicity (respectively $\gamma/2$ an ordinate with half the multiplicity, for $i\gamma/2$); the exponent $\frac12$ is the normalization that puts both families on the imaginary axis. More precisely, $\K(s)=-\frac{\zeta'}\zeta(s+\frac12)+\frac{\zeta'}\zeta(2s+\frac12)-\frac1{12s}+H(s)$, with $H$ holomorphic in a neighbourhood of $\Re s\ge0$. \emph{RH criteria.} RH is equivalent to the holomorphy of $\K(s)-\frac1{s-1/2}+\frac1{2(s-1/4)}$ in $\Re s>0$. It is also equivalent to a bound $O(x^\varepsilon)$, or even a one-sided bound, for the remainder $R_b$ of the kernel's partial sums. Both are exact reformulations of RH, not proofs. \emph{Mean value and explicit formula.} Unconditionally, $\sum_{n\le x}a(n)=\K(1)x+2\zeta(\frac12)\sqrt x+O(\sqrt x\,e^{-c_\ast\sqrt{\log x}})$, with the explicit bound $5.51\sqrt x$ for the error after the first term when $x\ge10^6$. Under RH, the Riesz mean of $a$ has an explicit formula in which every zero contributes the pure phases $x^{i\gamma}$ and $x^{i\gamma/2}$ at constant amplitude. \emph{The Ces\`aro margin.} The kernel is Suzuki's one-variable margin $M$ seen at two scales. Unconditionally, $R_b(x)=\alpha(x)-\alpha(\sqrt x)+o(1)$, where $\alpha$ is the slope of the margin. Under RH, the kernel's Riesz margin is $M(N)-2M(\sqrt N)$. Its fluctuation has energy $2(\log\xi)''(\frac12)$, reduced by $2\gamma^{-2}$ for every pair of ordinates with $\gamma'=2\gamma$; the near pair $\gamma_{14}\approx2\gamma_4$ is visible in the data. \emph{Zeros of $\C$.} A computer-assisted proof shows that $\C$ has a zero near $1.379+8.639\,i$, in the half-plane of absolute convergence, and Bohr's theory gives $\gg T$ such zeros up to height $T$. So $\C$ is not an $L$-function. The theorems are proved here, except that Section~\ref{sec:margin} uses the formula for the margin and the almost periodicity of the slope from Part~11. Uncertified computations are labelled as numerical observations. \end{abstract} \begin{tcolorbox}[colback=gray!4,colframe=gray!45,title=\textbf{The Caceres Prime Function in five equations}] \begin{align} a(n)&=\sum_{p^k\|n}\frac{\log p}{\sqrt{p^k}} \tag{E1}\\ \C(s)&=\sum_{n\ge2}\frac{a(n)}{n^s}=\zeta(s)\,\K(s) \tag{E2}\\ \K(s)&=\sum_p\log p\;\frac{1-p^{-s}}{p^{s+1/2}-1} \tag{E3}\\ \K(s)&=\sum_{k\ge1}\Bigl[P'\bigl(k(s+\tfrac12)+s\bigr)-P'\bigl(k(s+\tfrac12)\bigr)\Bigr],\qquad P(s)=\sum_p p^{-s} \tag{E4}\\ \textstyle\sum_{n\le x}a(n)&=\K(1)\,x+2\zeta(\tfrac12)\sqrt x+o(\sqrt x),\qquad \K(1)=1.16046\ldots \tag{E5}\\ \K(s)&=-\frac{\zeta'}{\zeta}\bigl(s+\tfrac12\bigr)+\frac{\zeta'}{\zeta}\bigl(2s+\tfrac12\bigr)-\frac1{12s}+H(s) \tag{E6} \end{align} \end{tcolorbox} \section{Introduction} Three classical arithmetic functions measure the prime content of an integer $n=\prod p^{k_p}$: \[ \log n=\sum_{p^k\|n}k\log p,\qquad \log\rad(n)=\sum_{p\mid n}\log p,\qquad \omega(n)=\sum_{p\mid n}1 . \] Their Dirichlet series are $-\zeta'(s)=\zeta(s)\bigl(-\tfrac{\zeta'}{\zeta}(s)\bigr)$, $-\zeta(s)P'(s)$ and $\zeta(s)P(s)$, where $P(s)=\sum_p p^{-s}$ is the Prime Zeta function. In each case $\zeta(s)$ appears multiplied by a function that is either $\zeta'/\zeta$ or a Prime Zeta object. This paper studies the function obtained by \emph{damping} the contribution of each prime by the square root of its exact power, \begin{equation}\label{eq:a} a(n)=\sum_{p^k\|n}\frac{\log p}{p^{k/2}} . \end{equation} Thus $a(p)=\log p/\sqrt p$, $a(p^2)=\log p/p$, $a(6)=\frac{\log 2}{\sqrt2}+\frac{\log 3}{\sqrt3}$ and $a(12)=\frac{\log2}{2}+\frac{\log3}{\sqrt3}$. The damping makes $a$ distinguish primes, prime powers and composite numbers, while keeping the additive structure of $\log\rad$. \begin{definition}[Caceres Prime Function]\label{def:C} The \emph{Caceres Prime Function} is the Dirichlet series \[ \C(s)=\sum_{n\ge2}\frac{a(n)}{n^s},\qquad \Re s>1 , \] and its \emph{prime kernel} is $\K(s)=\C(s)/\zeta(s)$. We also write $A(x)=\sum_{n\le x}a(n)$ and $F(x)=\sum_{n\le x}a(n)\log(x/n)=\int_1^x A(t)\,\frac{dt}{t}$. \end{definition} \paragraph{Main results.} \begin{enumerate}[leftmargin=*,itemsep=1pt] \item \textbf{Structure} (Theorems~\ref{thm:fact}, \ref{thm:pz}, \ref{thm:poles}, \ref{thm:nb}). $\C=\zeta\K$; $\K$ is the Prime Zeta telescope (E4); it is meromorphic in $\Re s>-\frac14$, with natural boundary $\Re s=-\frac14$; each zero $\rho$ gives poles at $\rho-\frac12$ and $\frac12(\rho-\frac12)$. \item \textbf{Critical damping} (Proposition~\ref{prop:damping}). $\frac12$ is the only exponent that puts both images of the critical line on the imaginary axis. \item \textbf{RH criteria} (Theorems~\ref{thm:rh}, \ref{thm:Ctrans}, \ref{thm:landau}). RH $\Leftrightarrow$ $\K$ has no poles in $\Re s>0$ other than $\frac12,\frac14$ $\Leftrightarrow$ $R_b(x)=O(x^\varepsilon)$ $\Leftrightarrow$ $R_b(x)\ge-C_\varepsilon x^\varepsilon$. \item \textbf{Mean value and explicit formula} (Theorems~\ref{thm:mean}, \ref{thm:second}, \ref{thm:explicitF}). An explicit $\sqrt x$ bound, the second term $2\zeta(\frac12)\sqrt x$, and, under RH, a constant-amplitude explicit formula for the Riesz mean. \item \textbf{The Ces\`aro margin} (Section~\ref{sec:margin}). The decomposition (E6); the bridge $R_b(x)=\alpha(x)-\alpha(\sqrt x)+o(1)$; the kernel's margin $M(N)-2M(\sqrt N)$ under RH; its energy, and the pairs $\gamma'=2\gamma$ that reduce it. \item \textbf{Zeros of the kernel} (Theorem~\ref{thm:zero}, Corollary~\ref{cor:manyzeros}). $\C$ vanishes in its half-plane of absolute convergence, $\gg T$ times up to height $T$. \end{enumerate} \paragraph{What the paper does not claim.} Theorem~\ref{thm:rh} is an exact reformulation of RH, not a proof. Every singularity of $\K$ in $\Re s>0$ comes from $\zeta'/\zeta$, so the construction cannot locate the zeros of $\zeta$ better than $\zeta'/\zeta$ itself. Its value is that it turns the zeros into a clean, explicit, computable signal, which we exploit numerically. In particular, Section~\ref{sec:margin} shows that the kernel carries exactly the information of $\Lambda$, at two scales. All results labelled Theorem, Proposition, Lemma or Corollary are proved in full; Theorem~\ref{thm:zero} is computer-assisted, with every error term bounded explicitly. Results labelled \emph{Numerical observation} are computations without certified error bounds. Parts 11 and 37 of the series \cite{CE11,CE37} treat Suzuki's margin and the $6k\pm1$ race. \paragraph{Place in the book.} This is Part~10 of the series, in block II (Primes and Euler products). It builds on Part~9 \cite{CE9} (the prime zeta function $P(s)$, whose shifted derivatives make up the kernel $\K$), on Part~1 \cite{CE1} (the C-transform, used in Section~\ref{sec:ctl}) and on Part~11 \cite{CE11} (the Ces\`aro margin, used in Section~\ref{sec:margin}). It is used by Part~11, which reads the margin through the kernel, and by Part~37 \cite{CE37} (the $6k\pm1$ race). \section{The arithmetic function} \begin{lemma}[Elementary properties]\label{lem:basic}\leavevmode \begin{enumerate}[label=(\roman*),itemsep=1pt] \item $a$ is additive: $a(mn)=a(m)+a(n)$ whenever $\gcd(m,n)=1$. \item $01$, \[ \boxed{\;\C(s)=\zeta(s)\,\K(s),\qquad \K(s)=\sum_p\log p\;\frac{1-p^{-s}}{p^{s+1/2}-1}\;} \] and the series for $\K$ converges absolutely for $\Re s>\frac12$. \end{theorem} \begin{proof} Write each $n$ with $p^k\|n$ as $n=p^km$ with $p\nmid m$. Grouping the terms of $\C$ by the prime $p$ and the exponent $k$ gives \[ \C(s)=\sum_p\log p\sum_{k\ge1}p^{-k/2}\,p^{-ks}\sum_{p\nmid m}m^{-s} =\zeta(s)\sum_p\log p\,(1-p^{-s})\sum_{k\ge1}p^{-k(s+1/2)} . \] The inner geometric series equals $p^{-(s+1/2)}/(1-p^{-(s+1/2)})=1/(p^{s+1/2}-1)$. Rearranging is legitimate because all series converge absolutely for $\Re s>1$. For $\Re s>\frac12$, the general term of $\K$ is $O(\log p\,p^{-\Re s-1/2})$, which is summable. \end{proof} \begin{remark} Equivalently, $\K(s)=\sum_{n}b(n)n^{-s}$ with $b$ as in Lemma~\ref{lem:basic}(iii). The factor $1-p^{-s}$ is essential: it removes the multiples of $p$ already counted by $\zeta(s)$. A naive guess with an $s$-independent denominator, such as $\sum_p\log p\,p^{-(s+1/2)}/(1-p^{-1/2})$, ignores the exponent $v_p(n)$ and is off by a factor of about $3$. With all $n\le10^8$ and the tail $\K(1)N^{1-s}/(s-1)+\zeta(\frac12)N^{1/2-s}/(s-\frac12)$, the direct series agrees with $\zeta(s)\K(s)$ to relative error $3\cdot10^{-11}$, $1\cdot10^{-12}$ and $9\cdot10^{-13}$ at $s=1.5,2,3$. \end{remark} \section{Exact relation with the Prime Zeta function} Let $P(s)=\sum_p p^{-s}$, so that $P'(s)=-\sum_p\log p\,p^{-s}$. It is natural to embed $a$ in a one-parameter family: \[ a_\alpha(n)=\sum_{p^k\|n}\frac{\log p}{p^{\alpha k}},\qquad \C_\alpha(s)=\sum_{n\ge2}\frac{a_\alpha(n)}{n^s},\qquad \K_\alpha=\C_\alpha/\zeta, \] so that $a=a_{1/2}$, $\C=\C_{1/2}$ and $\K=\K_{1/2}$. \begin{theorem}[Prime Zeta structure]\label{thm:pz} Let $\alpha\ge0$ and $w=s+\alpha$. For $\Re s>1$, \[ \boxed{\;\C_\alpha(s)=\zeta(s)\sum_{k=1}^{\infty}\Bigl[P'(kw+s)-P'(kw)\Bigr]\;} \] In particular, for the Caceres Prime Function, \begin{equation}\label{eq:pz} \K(s)=\sum_{k\ge1}\Bigl[P'\bigl(k(s+\tfrac12)+s\bigr)-P'\bigl(k(s+\tfrac12)\bigr)\Bigr]. \end{equation} \end{theorem} \begin{proof} The proof of Theorem~\ref{thm:fact} with $p^{-k/2}$ replaced by $p^{-\alpha k}$ gives \[ \K_\alpha(s)=\sum_p\log p\,(1-p^{-s})\sum_{k\ge1}p^{-kw}=\sum_{k\ge1}\sum_p\log p\,\bigl(p^{-kw}-p^{-kw-s}\bigr). \] The inner sum is $-P'(kw)+P'(kw+s)$. The interchange of sums is justified by absolute convergence for $\Re s>1$. \end{proof} \begin{corollary}[Telescoping and the undamped case]\label{cor:alpha0}\leavevmode \begin{enumerate}[label=(\roman*),itemsep=1pt] \item $\displaystyle \K_\alpha(s)=\sum_{k\ge1}\int_0^{s}P''(kw+v)\,dv$: the kernel adds up the increments of $P'$ along the lattice $\{kw\}$ with step $s$. \item For $\alpha=0$ we have $w=s$, so $P'(ks+s)=P'((k+1)s)$ and the series telescopes completely, since $P'(\sigma)\to0$ as $\sigma\to+\infty$: \[ \C_0(s)=-\zeta(s)\,P'(s),\qquad a_0(n)=\log\rad(n). \] \end{enumerate} \end{corollary} So the Caceres Prime Function is the $\sqrt{\ }$-damped form of $\log\rad(n)$. Setting $\alpha=0$ collapses (E4) to a single term, while $\alpha=\frac12$ leaves a genuinely infinite telescope, whose first two terms carry all the singularities (Theorem~\ref{thm:poles}). \begin{corollary}[Form in terms of $\zeta$ only]\label{cor:mobius} Using $P'(s)=\sum_{m\ge1}\mu(m)\,\frac{\zeta'}{\zeta}(ms)$, \[ \C_\alpha(s)=\zeta(s)\sum_{k\ge1}\sum_{m\ge1}\mu(m)\left[\frac{\zeta'}{\zeta}\bigl(m(kw+s)\bigr)-\frac{\zeta'}{\zeta}\bigl(mkw\bigr)\right]. \] \end{corollary} \section{Analytic continuation and the dual spectrum}\label{sec:poles} Recall that $P'(w)$ is meromorphic in $\Re w>0$, where the line $\Re w=0$ is a natural boundary (Landau--Walfisz \cite{LW}). By Corollary~\ref{cor:mobius} with $k=1$, its poles in $\Re w>0$ are at $w=1/m$ for squarefree $m$ and at $w=\rho/m$ for the nontrivial zeros $\rho$ of $\zeta$. Moreover, $\frac{\zeta'}{\zeta}$ has residue $-1$ at $1$ and residue $m_\rho$ (the multiplicity) at $\rho$. \begin{theorem}[Continuation and poles]\label{thm:poles} $\K$ extends meromorphically to $\Re s>-\frac14$. In $\Re s>0$, \begin{equation}\label{eq:split} \K(s)=-P'\bigl(s+\tfrac12\bigr)+P'\bigl(2s+\tfrac12\bigr)+G(s), \end{equation} where $G$ is holomorphic in $\Re s>0$. Consequently, the poles of $\K$ in $\Re s>0$ are among the following: \begin{center} \begin{tabular}{lll} \toprule origin & pole & residue\\ \midrule pole of $\zeta$ at $1$, first term & $s=\tfrac12$ & $+1$\\ pole of $\zeta$ at $1$, second term & $s=\tfrac14$ & $-\tfrac12$\\ zero $\rho=\beta+i\gamma$, first term & $s=\rho-\tfrac12$ & $-m_\rho$\\ zero $\rho=\beta+i\gamma$, second term & $s=\tfrac12(\rho-\tfrac12)$ & $+\tfrac12 m_\rho$\\ \bottomrule \end{tabular} \end{center} The poles in the last two rows lie in $\Re s>0$ only if $\beta>\frac12$. Two poles in this list can coincide only if $\rho-\frac12=\frac12(\rho'-\frac12)$ for two zeros $\rho,\rho'$, that is, if $\rho'=2\rho-\frac12$. \end{theorem} \begin{proof} In \eqref{eq:pz}, the $k$-th pair of terms involves $P'$ at arguments with real parts $k(\sigma+\frac12)$ and $(k+1)\sigma+\frac k2$, where $\sigma=\Re s$. Both are positive as soon as $\sigma>-\frac14$, and they grow linearly in $k$. Since $P'(w)=O(2^{-\Re w})$ as $\Re w\to\infty$, the series converges locally uniformly away from the poles, which gives the continuation. For $\sigma>0$, write $\K(s)=\sum_p\log p\,(1-p^{-s})\sum_k p^{-k(s+1/2)}$ and separate the two terms $p^{-(s+1/2)}$ and $-p^{-(2s+1/2)}$ (both from $k=1$). Every remaining term $p^{-k(s+1/2)}$ or $p^{-s-k(s+1/2)}$ with $k\ge2$ has exponent with real part at least $\min(2\sigma+1,3\sigma+1)>1$. So the remaining sum $G(s)$ converges absolutely and locally uniformly, and $G$ is holomorphic in $\sigma>0$. This proves \eqref{eq:split}. For the poles, apply Corollary~\ref{cor:mobius} to $P'(s+\frac12)$ and $P'(2s+\frac12)$. Their arguments have real part greater than $\frac12$ when $\sigma>0$. Hence only the $m=1$ term can be singular: for $m\ge2$, the points $\rho/m$ and $1/m$ have real part at most $\frac12$. The table follows by evaluating the residues of $\pm\frac{\zeta'}{\zeta}$ at $1$ and at $\rho$, divided by the derivative of the inner argument ($1$ for the first term, $2$ for the second). \end{proof} \begin{theorem}[Natural boundary]\label{thm:nb} The line $\Re s=-\frac14$ is a natural boundary of $\K$ and of $\C$: neither function has a meromorphic continuation to any disc centred at a point of this line. \end{theorem} \begin{proof} Split \eqref{eq:pz} as $\K(s)=P'(2s+\frac12)+H(s)$, where $H$ collects the terms $-P'(k(s+\frac12))$ for $k\ge1$ and $P'((k+1)s+\frac k2)$ for $k\ge2$. Their arguments have real parts $k(\sigma+\frac12)$ and $k(\sigma+\frac12)+\sigma$. For $\sigma>-\frac13$ and $k\ge2$, both are at least $\min\bigl(\tfrac13,\ 3\sigma+1\bigr)>0$, and for $k=1$ the first is $\sigma+\frac12>\frac16$. All these real parts grow linearly in $k$. As in the proof of Theorem~\ref{thm:poles}, $H$ is therefore meromorphic in $\Re s>-\frac13$. Suppose $\K$ were meromorphic in a disc $D$ centred at $s_*$ with $\Re s_*=-\frac14$. Shrinking $D$, we may assume $D\subset\{\Re s>-\frac13\}$. Then $P'(2s+\frac12)=\K(s)-H(s)$ would be meromorphic in $D$, so $P'$ would be meromorphic in a disc $D'$ centred at $w_*=2s_*+\frac12$, a point of $\Re w=0$. Since $P'$ has only finitely many poles in $D'$, there is a point $w_1\in D'$ with $\Re w_1=0$ and a disc $D_1\subset D'$ around $w_1$ on which $P'$ is holomorphic. On $D_1$ the function $P(w_2)+\int_{w_2}^{w}P'(u)\,du$, with $w_2\in D_1$, $\Re w_2>0$, is holomorphic and agrees with $P$ on $D_1\cap\{\Re w>0\}$. So $P$ would continue analytically across the line at $w_1$. This contradicts the theorem of Landau and Walfisz \cite{LW} that $\Re w=0$ is a natural boundary of $P$. Finally, if $\C$ were meromorphic in such a disc, so would be $\K=\C/\zeta$, because $\zeta$ is meromorphic and not identically zero. \end{proof} \begin{proposition}[Critical damping]\label{prop:damping} Let $0<\alpha<1$, and assume RH unless $\alpha=\frac12$. In the expansion $\K_\alpha(s)=\sum_{k\ge1}\bigl[P'(k(s+\alpha)+s)-P'(k(s+\alpha))\bigr]$, a zero $\rho=\frac12+i\gamma_\rho$ on the critical line makes the two terms $-P'(s+\alpha)$ and $P'(2s+\alpha)$ singular at \[ s=\bigl(\tfrac12-\alpha\bigr)+i\gamma_\rho\qquad\text{and}\qquad s=\tfrac12\bigl(\tfrac12-\alpha\bigr)+\tfrac{i\gamma_\rho}{2}. \] For $0<\alpha<\frac34$, and in particular for $\alpha=\frac12$, no term other than these two families is singular at these points. For $\alpha=\frac12$ the point $i\gamma$ of the first family coincides with a point of the second exactly when $2\gamma$ is an ordinate; the residue there is $-m_\gamma+\frac12m_{2\gamma}$. So $i\gamma$ is a pole unless $m_{2\gamma}=2m_\gamma$, and $i\gamma/2$ is a pole unless $\gamma/2$ is an ordinate with $m_{\gamma/2}=\frac12m_\gamma$. This case is unconditional: a term of the expansion with indices $(m,k)$ is singular at a point of the imaginary axis only if $\zeta$ has a zero of real part $mk/2$, which is $\ge1$ unless $m=k=1$; so a zero off the line cannot produce or cancel a singularity at these points. For $\alpha\ne\frac12$, under RH, the two families lie on the different lines $\Re s=\frac12-\alpha$ and $\Re s=\frac12(\frac12-\alpha)$, they are disjoint, and all their points are poles, with residues $-m_\rho$ and $\frac12m_\rho$. Without RH a hypothetical zero off the line could cancel one of them (a zero at $1-\alpha+2i\gamma_\rho$ would cancel the first-family point through $P'(2s+\alpha)$, and a zero at $\frac14+\frac\alpha2+\frac{i\gamma_\rho}2$ the second-family point through $-P'(s+\alpha)$), which is why RH is assumed. Both families lie on the imaginary axis if and only if $\alpha=\frac12$; this is the statement $\frac12-\alpha=0$, the usual normalization $\Lambda(n)/\sqrt n$. \end{proposition} \begin{proof} By Corollary~\ref{cor:mobius}, the singularities of a term $\pm\mu(m)\frac{\zeta'}{\zeta}(m(as+b))$ coming from zeros $\rho'=\frac12+i\gamma'$ lie on the vertical line $\Re s=(\frac1{2m}-b)/a$. Here $(a,b)=(k,k\alpha)$ for $-P'(k(s+\alpha))$, and $(a,b)=(k+1,k\alpha)$ for $P'(k(s+\alpha)+s)$. The two families in the statement come from $m=1$ with $(a,b)=(1,\alpha)$ and $(2,\alpha)$, and lie on $\Re s=\frac12-\alpha$ and $\Re s=\frac12(\frac12-\alpha)$. We check every other $(m,k)$. \emph{Terms $-P'(k(s+\alpha))$.} The real part is $\frac1{2mk}-\alpha$. \begin{itemize}[itemsep=1pt] \item It equals $\frac12-\alpha$ only if $mk=1$, which is the first family itself. \item It equals $\frac12(\frac12-\alpha)$ only if $mk=2/(1+2\alpha)<2$, that is, $mk=1$. But then $\alpha=\frac12$ and this is again the first family. \end{itemize} \emph{Terms $P'((k+1)s+k\alpha)$.} The real part is $(\frac1{2m}-k\alpha)/(k+1)$. \begin{itemize}[itemsep=1pt] \item It equals $\frac12-\alpha$ iff $\alpha=\frac{k+1}2-\frac1{2m}$. For $k\ge2$ this forces $\alpha\ge1$. For $k=1$ it forces $m\ge2$ and $\alpha=1-\frac1{2m}\ge\frac34$. \item It equals $\frac12(\frac12-\alpha)$ iff $\alpha(1-k)=\frac{k+1}2-\frac1m$. For $k\ge2$ the left side is $\le0$ and the right side is positive. For $k=1$ this forces $m=1$, which is the second family itself. \end{itemize} Singularities coming from the pole of $\zeta$ and from the trivial zeros are real. So for $0<\alpha<\frac34$ only the two dominant terms are singular at the stated points, and the residues follow as in Theorem~\ref{thm:poles}. Finally, $\frac12-\alpha=0$ iff $\alpha=\frac12$. \end{proof} \begin{figure}[ht] \centering \includegraphics[width=\textwidth]{fig_damping.pdf} \caption{Where the zeros of $\zeta$ (first 10 ordinates, assuming RH) land as poles of $\K_\alpha$. Shading marks $\Re s>0$. For $\alpha=\frac12$ both images, $i\gamma$ and $i\gamma/2$, lie exactly on the boundary $\Re s=0$, and the half-plane is free of zero-induced poles. Crosses mark the poles $1-\alpha$ and $\frac{1-\alpha}{2}$ inherited from the pole of $\zeta$.} \label{fig:damping} \end{figure} Proposition~\ref{prop:damping} explains why $\frac12$ is the natural exponent. With $\alpha=\frac12$, the critical line of $\zeta$ is mapped onto the imaginary axis of $\K$, and the ordinates appear unchanged ($\gamma$) and halved ($\gamma/2$). We call $\{\gamma_n\}\cup\{\gamma_n/2\}$ the \emph{dual spectrum} of the Caceres Prime Function. \section{A criterion for the Riemann Hypothesis} \begin{theorem}\label{thm:rh} Let \[ \E(s)=\K(s)-\frac{1}{s-\frac12}+\frac{1}{2\,(s-\frac14)} . \] The Riemann Hypothesis holds if and only if $\E$ is holomorphic in $\Re s>0$. \end{theorem} \begin{proof} ($\Rightarrow$) Under RH every zero has $\beta=\frac12$, so by Theorem~\ref{thm:poles} the only poles of $\K$ in $\Re s>0$ are $\frac12$ (residue $1$) and $\frac14$ (residue $-\frac12$). Subtracting their principal parts leaves a holomorphic function. ($\Leftarrow$) Suppose there is a zero $\rho_0$ with $\beta_0>\frac12$. Define $\rho_{j+1}=2\rho_j-\frac12$ for as long as $\rho_{j+1}$ is again a zero of $\zeta$. Then $\Re\rho_j=\frac12+2^j(\beta_0-\frac12)$, which grows without bound, while every zero satisfies $\Re\rho<1$. So the chain stops at some $\rho_J$, and $2\rho_J-\frac12$ is not a zero. Look at the point $s_*=\rho_J-\frac12$. It satisfies $\Re s_*>0$ and $\Im s_*\ne0$, so $s_*\ne\frac12,\frac14$. By Theorem~\ref{thm:poles}, the first term contributes a pole with residue $-m_{\rho_J}\neq0$ at $s_*$. The only pole that could cancel it would come from the second term with $\rho'=2\rho_J-\frac12$, and that point is not a zero. Hence $\K$, and therefore $\E$, has a pole at $s_*$. \end{proof} \begin{remark} Theorem~\ref{thm:rh} is a reformulation of RH in the same family as the classical statement that $\zeta'/\zeta$ is holomorphic in $\Re s>\frac12$ except at $s=1$. Its interest lies in two features. First, the criterion is expressed through a function with explicit non-negative coefficients, $\C(s)=\zeta(s)\K(s)$. Second, the critical line becomes the imaginary axis, so the boundary of the criterion is the natural spectral line $s=it$. \end{remark} \section{The mean value of $a(n)$} Throughout we use the explicit Chebyshev-type bounds of Rosser and Schoenfeld \cite{RS}: \begin{equation}\label{eq:RS} \theta(u)<1.01624\,u\quad(u>0),\qquad \Bigl|\theta(u)-u\Bigr|<\frac{u}{2\log u}\quad(u\ge563),\qquad \sum_{p\le y}\frac{\log p}{p}<\log y\quad(y>1). \end{equation} \begin{theorem}[Mean value with an explicit constant]\label{thm:mean} For every $x\ge2$, \[ \bigl|A(x)-\K(1)\,x\bigr|\le B(x):=5.0813\sqrt x+3.7263\,x^{1/4}+\frac{\log^2x}{\log2}+1.955\log x+2, \] where \[ \K(1)=\sum_p\frac{\log p\,(1-p^{-1})}{p^{3/2}-1}. \] In particular $|A(x)-\K(1)x|\le5.51\sqrt x$ for all $x\ge10^6$. \end{theorem} \begin{proof} By Lemma~\ref{lem:basic}(iii), $A(x)=\sum_{d\le x}b(d)\lfloor x/d\rfloor$. Writing $\lfloor x/d\rfloor=x/d-\{x/d\}$ and using $\sum_d b(d)/d=\K(1)$ (an absolutely convergent series), we get \begin{equation}\label{eq:Rdecomp} R(x):=A(x)-\K(1)x=-x\sum_{d>x}\frac{b(d)}{d}-\sum_{d\le x}b(d)\Bigl\{\frac xd\Bigr\}, \qquad |R(x)|\le x\sum_{d>x}\frac{|b(d)|}{d}+\sum_{d\le x}|b(d)|. \end{equation} We use $|b(p)|=\log p\,p^{-1/2}$, $|b(p^2)|\le\log p\,p^{-1/2}$ and $|b(p^k)|\le\log p\,p^{-(k-1)/2}\le\log p\,p^{-1}$ for $k\ge3$. \emph{Head.} By partial summation and \eqref{eq:RS}, \[ \sum_{p\le y}\frac{\log p}{\sqrt p}=\frac{\theta(y)}{\sqrt y}+\frac12\int_2^y\frac{\theta(u)}{u^{3/2}}\,du\le 2.0325\sqrt y . \] The primes therefore give at most $2.0325\sqrt x$ and the prime squares at most $2.0325\,x^{1/4}$. For each $k\ge3$ with $2^k\le x$, the powers $p^k\le x$ give at most $\sum_{p\le x^{1/k}}\log p/p<\log x$, and there are fewer than $\log x/\log2$ such $k$. \emph{Tail.} By partial summation and $\theta(u)<1.01624u$, \[ x\sum_{p>x}\frac{\log p}{p^{3/2}}\le x\cdot\tfrac32\cdot1.01624\int_x^\infty u^{-3/2}du=3.0488\sqrt x, \] and in the same way \[ x\sum_{p>\sqrt x}\frac{\log p}{p^{5/2}}\le1.6938\,x^{1/4}. \] For $k\ge3$, put $y=x^{1/k}$. If $p>y$ then $x/p^k\le(y/p)^3$, so the powers $p^k>x$ with $k\le\log x/\log2$ contribute at most \[ \sum_k y^3\sum_{p>y}\log p\,p^{-4}\le\sum_k\tfrac43\cdot1.01624=1.355\cdot\frac{\log x}{\log 2}=1.955\log x . \] For $k>\log x/\log2$ every prime satisfies $p^k>x$. Using $p^{-(3k-1)/2}\le2^{-(3k-1)/2}(2/p)^4$ and $\sum_p\log p\,p^{-4}\le-\zeta'(4)<0.069$, these $k$ contribute at most $2.42\,x^{-1/2}\le2$. Adding the head and the tail gives $B(x)$. The last claim follows because $B(x)/\sqrt x$ is decreasing for $x\ge e^4$ and $B(10^6)=5.5035\cdot10^3$. \end{proof} \begin{proposition}[Rigorous value of $\K(1)$]\label{prop:K1} \[ 1.16045671\le\K(1)\le1.16047301 . \] The value obtained from the Prime Zeta representation \eqref{eq:pz}, \[ \K(1)=\sum_{k\ge1}\Bigl[P'\bigl(\tfrac{3k}2+1\bigr)-P'\bigl(\tfrac{3k}2\bigr)\Bigr]=1.1604648419\ldots, \] lies in this interval. This value of the series is computed numerically and is not certified. \end{proposition} \begin{proof} Write $\K(1)=\sum_{p\le P}T(p)+\sum_{p>P}T(p)$ with $T(p)=\log p\,(1-p^{-1})/(p^{3/2}-1)$ and $P=10^8$. For $p>P$ we have $T(p)=\log p\,p^{-3/2}\phi(p)$ with $1-P^{-1}\le\phi(p)\le1$. Moreover \[ \sum_{p>P}\log p\,p^{-3/2}=-\theta(P)P^{-3/2}+\tfrac32\int_P^\infty\theta(u)u^{-5/2}du , \] which lies in \[ -\theta(P)P^{-3/2}+3P^{-1/2}\,[1-\eta,\,1+\eta],\qquad \eta=\frac1{2\log P}, \] by \eqref{eq:RS}. The finite sum, of $5\,761\,455$ positive terms with total $1.160265$, and $\theta(10^8)=99\,987\,730.018\ldots$ were computed in double precision with sorted summation. The rounding error of the finite sum is at most $n\,u\sum|T|\le5.8\cdot10^6\cdot2^{-52}\cdot1.17<2\cdot10^{-9}$, including the relative error $\le2^{-52}$ of each logarithm and power. The error in $\theta(10^8)$ is at most $0.13$, which changes $\theta(P)P^{-3/2}$ by less than $10^{-12}$. Both are absorbed by widening the interval by $10^{-8}$, which the stated rounded endpoints already include. \end{proof} \begin{theorem}[The second term]\label{thm:second} There is an absolute constant $c_\ast>0$ such that, unconditionally, \[ \boxed{\;A(x)=\K(1)\,x+2\zeta(\tfrac12)\sqrt x+O\bigl(\sqrt x\,e^{-c_\ast\sqrt{\log x}}\bigr)\;} \] Under RH, the error term can be replaced by $O_\varepsilon(x^{1/3+\varepsilon})$. \end{theorem} \begin{proof} \emph{Unconditional part.} In \eqref{eq:Rdecomp}, the prime powers $p^k$ with $k\ge2$ contribute $O(x^{1/4})$, as shown in the proof of Theorem~\ref{thm:mean}. Hence \[ R(x)=-x\,S(x)-U(x)+O(x^{1/4}),\qquad S(x)=\sum_{p>x}\frac{\log p}{p^{3/2}},\qquad U(x)=\sum_{p\le x}\frac{\log p}{\sqrt p}\Bigl\{\frac xp\Bigr\}. \] By the prime number theorem with the de la Vall\'ee Poussin error term, $\theta(u)=u+E(u)$ with $E(u)\ll u\,e^{-c_2\sqrt{\log u}}$. Then \[ S(x)=\int_x^\infty u^{-3/2}\,du+\int_x^\infty u^{-3/2}\,dE(u)=2x^{-1/2}+O\bigl(x^{-1/2}e^{-c_2\sqrt{\log x}}\bigr). \] For $U(x)$, fix an integer $J\ge2$. The primes $p\le x/J$ contribute at most $2.0325\sqrt{x/J}$. On each interval $x/(m+1)0$ \cite[(2.1.5)]{Tit}, gives at $s=\frac12$ the identity $\zeta(\frac12)=-1-\frac12I$. Therefore $-(2+I)=2\zeta(\frac12)$. \emph{Under RH.} Let $c_x=1+1/\log x$ and $x^{1/2}\le T\le x$. Since $00$ apart from the poles $\frac12$ and $\frac14$ (Theorem~\ref{thm:poles}). Moreover, every argument of $\zeta'/\zeta$ appearing in Corollary~\ref{cor:mobius} has real part at least $\frac12+\delta$ when $\Re s\ge\delta$. Under RH, the partial fraction formula $\frac{\zeta'}{\zeta}(u)=\sum_{|\Im u-\gamma|<1}\frac1{u-\rho}+O(\log|\Im u|)$ \cite[Ch.~15]{Dav} gives $\frac{\zeta'}{\zeta}(u)\ll_\delta\log(|\Im u|+2)$ for $\frac12+\delta\le\Re u\le2$, since there $|u-\rho|\ge\delta$. For $\Re u\ge2$, $\zeta'/\zeta(u)$ is bounded by $\sum_n\Lambda(n)n^{-2}$. Hence $\K(s)\ll_\delta\log(|t|+2)$ for $\Re s\ge\delta$. RH also implies the Lindel\"of hypothesis \cite[\S14.2]{Tit}, so $\zeta(\sigma+it)\ll|t|^{\max(0,\frac12-\sigma)+\varepsilon}$ for $\sigma\ge\delta$. Move the segment to $\Re s=\delta$. The residues at $s=1,\frac12,\frac14$ are $\K(1)x$, $2\zeta(\frac12)\sqrt x$ and $-2\zeta(\frac14)x^{1/4}$. The new vertical segment contributes $\ll x^{\delta}T^{1/2-\delta+2\varepsilon}$, and the horizontal segments contribute $\ll x^{1+\varepsilon}/T$. With $T=x^{2/3}$ and $\delta=\varepsilon$, all error terms, including $x^{1/4}$, are $O(x^{1/3+3\varepsilon})$. \end{proof} Table~\ref{tab:mean} illustrates Theorem~\ref{thm:second}: the ratio $(A(x)-\K(1)x)/\sqrt x$ approaches $2\zeta(\frac12)=-2.92071$, and numerically, over $10\le x\le10^8$, the ratio $|R(x)|/B(x)$ never exceeds $0.562$. \begin{table}[ht] \centering \caption{Mean value of $a(n)$. The last column is $R_2(x)=A(x)-\K(1)x-2\zeta(\frac12)\sqrt x+2\zeta(\frac14)x^{1/4}$.} \label{tab:mean} \begin{tabular}{rrrr} \toprule $x$ & $A(x)$ & $(A(x)-\K(1)x)/\sqrt x$ & $R_2(x)$\\ \midrule $10^3$ & $1\,077.489$ & $-2.6239$ & $0.238$\\ $10^4$ & $11\,330.017$ & $-2.7463$ & $1.174$\\ $10^5$ & $115\,150.436$ & $-2.8336$ & $-1.363$\\ $10^6$ & $1\,157\,597.587$ & $-2.8673$ & $2.018$\\ $10^7$ & $11\,595\,503.533$ & $-2.8919$ & $-0.261$\\ $10^8$ & $116\,017\,439.601$ & $-2.9045$ & $-0.154$\\ \bottomrule \end{tabular} \end{table} \section{The explicit formula: the zeros at constant amplitude}\label{sec:explicit} We now prove the explicit formula for the Riesz mean $F(x)=\sum_{n\le x}a(n)\log(x/n)$ under RH. For $c>1$, Perron's formula \[ F(x)=\frac1{2\pi i}\int_{(c)}\C(s)\,x^s\,\frac{ds}{s^2} \] holds with an absolutely convergent integral, since $\frac1{2\pi i}\int_{(c)}y^s s^{-2}ds$ equals $\log y$ for $y\ge1$ and $0$ for $0-\frac1{12}$ are contained in the set \[ \tfrac12,\quad \tfrac14,\quad 0,\quad \pm i\gamma,\quad \pm\tfrac{i\gamma}{2}, \] where $\gamma$ runs over the positive ordinates of the zeros. All of them are simple. The residues are $1$ at $\frac12$, $-\frac12$ at $\frac14$, $-\frac1{12}$ at $0$, $-m_\gamma$ from the first family at $i\gamma$, and $+\frac12m_\gamma$ from the second family at $\frac{i\gamma}{2}$. Where the two families meet, at $\gamma=\gamma'/2$, the residues add, and the point is a pole unless $-m_\gamma+\frac12m_{\gamma'}=0$. That cannot happen if the zeros are simple. The points $\frac12,\frac14,0$ are always poles. \end{lemma} \begin{proof} By Corollary~\ref{cor:mobius}, $\K$ is a locally uniformly convergent sum of terms $\pm\mu(m)\frac{\zeta'}{\zeta}\bigl(m(\alpha s+\beta)\bigr)$, with $(\alpha,\beta)=(k,\frac k2)$ (sign $-$) or $(k+1,\frac k2)$ (sign $+$). Each term has simple poles where $m(\alpha s+\beta)\in\{1\}\cup\{\rho\}\cup\{-2,-4,\dots\}$, that is, at \[ s=\frac{1/m-\beta}{\alpha},\qquad s=\frac{\rho/m-\beta}{\alpha},\qquad s=\frac{-2j/m-\beta}{\alpha}. \] Under RH $\Re\rho=\frac12$. A direct check over $(m,k)$ shows the following. The real parts of these points that are $>-\frac1{12}$ are $\frac12$ (only $m=k=1$, first type), $\frac14$ (only $m=k=1$, second type) and $0$. The value $0$ arises from the pole of $\zeta$ for $mk=2$, and from the zeros only for $m=k=1$, at $i\gamma$ and $i\gamma/2$. The next values are $-\frac1{12}$ (real, $m=3$, $k=1$, second type), then $-\frac18$, $-\frac16$, and so on. The trivial zeros give real parts $\le-\frac14$. The residue of $\frac{\zeta'}{\zeta}(m(\alpha s+\beta))$ is $-1/(m\alpha)$ at a pole coming from $1$, and $m_\rho/(m\alpha)$ at a pole coming from $\rho$. At $s=0$ the four contributions are $-\frac12+\frac14+\frac12-\frac13=-\frac1{12}$, from $(m,k)=(2,1)$ and $(1,2)$ of each type. \end{proof} \begin{theorem}[Explicit formula for $F$ under RH]\label{thm:explicitF} Assume RH and let $-\frac1{12}<\sigma_0<0$. For $x\ge1$, \[ \begin{aligned} F(x)={}&\K(1)\,x+4\zeta(\tfrac12)\sqrt x-8\zeta(\tfrac14)\,x^{1/4}+\tfrac1{48}\log^2x+c_1\log x+c_0\\ &+2\Re\sum_{\gamma>0}\frac{m_\gamma}{\gamma^2}\Bigl[\zeta(i\gamma)\,x^{i\gamma}-2\,\zeta\bigl(\tfrac{i\gamma}{2}\bigr)\,x^{i\gamma/2}\Bigr]+O_{\sigma_0}\bigl(x^{\sigma_0}\bigr). \end{aligned} \] Here $c_1$ and $c_0$ are the coefficients in the Laurent expansion $\C(s)=\frac{1}{24s}+c_1+c_0s+O(s^2)$ at $s=0$, and the series over the zeros converges absolutely. In particular, $F(x)$ minus its smooth part is a uniformly almost periodic function of $\log x$, with frequencies in $\{\pm\gamma\}\cup\{\pm\gamma/2\}$, up to $O(x^{-1/12+\varepsilon})$. \end{theorem} \begin{proof} \emph{Step 1: bounds on vertical lines.} Let $\sigma=\sigma_0$. In \eqref{eq:pz} the arguments of $P'$ have real parts \[ \tfrac12+\sigma_0\in(\tfrac5{12},\tfrac12),\quad \tfrac12+2\sigma_0\in(\tfrac13,\tfrac12),\quad 1+2\sigma_0\in(\tfrac56,1),\quad 1+3\sigma_0\in(\tfrac34,1), \] and greater than $1$ for the terms with $k\ge3$. Expanding each $P'$ by Corollary~\ref{cor:mobius}, the terms whose argument has real part greater than $1$ are bounded uniformly, since $|\frac{\zeta'}{\zeta}(u)|\le|\frac{\zeta'}{\zeta}(\Re u)|$ there. They are summable in $m$ and $k$ because $|\frac{\zeta'}{\zeta}(u)|\le C\,2^{-\Re u}$ for $\Re u\ge2$. Only finitely many terms remain: $\frac{\zeta'}{\zeta}$ at real parts $\frac12+\sigma_0$, $\frac12+2\sigma_0$, $1+2\sigma_0$, $1+3\sigma_0$ and $1+4\sigma_0$, none of which equals $\frac12$. For a real part $\sigma_u\in(\frac12,2]$, RH and the partial fraction formula give $\frac{\zeta'}{\zeta}(u)\ll\log|\Im u|/(\sigma_u-\frac12)$. For $\sigma_u<\frac12$, the functional equation gives $\frac{\zeta'}{\zeta}(u)=-\frac{\zeta'}{\zeta}(1-u)+O(\log|\Im u|)$, which reduces to the previous case. Hence \[ \K(\sigma_0+it)\ll_{\sigma_0}\log(|t|+2). \] By convexity \cite[\S5.1]{Tit}, $\zeta(\sigma+it)\ll(|t|+2)^{1/2-\sigma_0}\log(|t|+2)$ uniformly for $\sigma_0\le\sigma\le2$. \emph{Step 2: horizontal lines.} Choose $T_j\in[j,j+1]$ with $|T_j-\gamma|\ge c/\log j$ and $|2T_j-\gamma|\ge c/\log j$ for every ordinate $\gamma$. This is possible for a small absolute $c>0$, since $[j-1,j+3]$ and $[2j-2,2j+4]$ contain $O(\log j)$ ordinates. On $\Im s=T_j$, $\sigma_0\le\Re s\le2$, only the two arguments $s+\frac12$ and $2s+\frac12$ cross the critical line. For them the partial fraction formula gives $\frac{\zeta'}{\zeta}\ll\log^2T_j$. All other arguments stay at distance at least $\frac16$ from $\Re u=\frac12$, which gives $\ll\log T_j$. Hence $\K(\sigma+iT_j)\ll\log^2T_j$. \emph{Step 3: residues.} Apply the residue theorem to the rectangle with vertices $\sigma_0\pm iT_j$ and $2\pm iT_j$. By Lemma~\ref{lem:polesneg} the poles of $\C(s)x^s/s^2$ inside are as follows. The first family is cut at $\gamma0}m_\gamma\Bigl[-\frac{\zeta(i\gamma)}{i\gamma}\,x^{i\gamma}+\frac{\zeta(i\gamma/2)}{i\gamma}\,x^{i\gamma/2}\Bigr], \end{equation} where the sum is not absolutely convergent. We use \eqref{eq:explicit} only as the model in the spectral comparison below. Rigorous statements about $A$ are Theorems~\ref{thm:mean} and~\ref{thm:second}. \paragraph{Why this matters.} In the explicit formula for $\psi(x)=\sum_{n\le x}\Lambda(n)$, a zero contributes $x^{\rho}/\rho=x^{1/2+i\gamma}/\rho$, an oscillation whose amplitude grows like $\sqrt x$. The square-root damping in (E1) removes exactly this growth. In Theorem~\ref{thm:explicitF} every zero contributes a \emph{pure phase} $x^{i\gamma}$, and a second copy $x^{i\gamma/2}$, at constant amplitude. By contrast, a zero off the line would make $\K$ have a pole in $\Re s>0$ (Theorem~\ref{thm:rh}), and the corresponding term would grow like a positive power of $x$. \subsection*{Numerical verification} We computed $a(n)$ for all $n\le10^8$ and used the 96 zeros with $\gamma<230$ (Appendix~\ref{app:num}). After removing the smooth terms: \begin{itemize}[itemsep=1pt] \item the fitted coefficients of $\log x$ in $A$ and of $\log^2x$ in $F$ are $0.0459$ and $0.0226$, about $10\%$ above the proved values $\frac1{24}=0.0417$ and $\frac1{48}=0.0208$ (no error bars); \item the residual of $F$ follows the zero sum of Theorem~\ref{thm:explicitF} with correlation $0.96$--$0.99$ in every decade between $10^3$ and $10^8$; \item the windowed Fourier transform $\Phi(t)$ of the residual of $A$ in the variable $\log x$ reproduces the predicted spectrum with correlation $0.985$ over $5\le t\le60$, and the main peaks agree to within a few percent (Figure~\ref{fig:spec}, Table~\ref{tab:peaks}). \end{itemize} \begin{figure}[ht] \centering \includegraphics[width=\textwidth]{fig_spectrum.pdf} \caption{Spectrum of the arithmetic oscillation of $A(x)=\sum_{n\le x}a(n)$ in the variable $\log x$, computed with all $n\le10^8$ (solid), against the spectrum of the explicit formula \eqref{eq:explicit} with 96 zeros (dashed). Upper ticks mark $\gamma_n$ and lower ticks mark $\gamma_n/2$.} \label{fig:spec} \end{figure} \begin{table}[ht] \centering \caption{Largest spectral peaks of the residual of $A$: measured height against the height predicted by \eqref{eq:explicit}. The last column gives the nearest frequency of the dual spectrum. With a frequency resolution of about $0.9$, nearby frequencies merge.} \label{tab:peaks} \begin{tabular}{ccccc} \toprule $t$ & measured & predicted & difference & nearest\\ \midrule $7.02$ & $0.1848$ & $0.1798$ & $2.8\%$ & $\gamma_1/2=7.07$\\ $12.53$ & $0.1075$ & $0.1047$ & $2.6\%$ & $\gamma_3/2=12.51$\\ $16.50$ & $0.1726$ & $0.1727$ & $0.1\%$ & $\gamma_5/2=16.47$\\ $18.75$ & $0.1798$ & $0.1759$ & $2.2\%$ & $\gamma_6/2=18.79$\\ $26.52$ & $0.1358$ & $0.1422$ & $4.6\%$ & $\gamma_{11}/2=26.49$\\ $36.02$ & $0.1244$ & $0.1224$ & $1.6\%$ & $\gamma_{18}/2=36.03$\\ $37.63$ & $0.1211$ & $0.1293$ & $6.4\%$ & $\gamma_6=37.59$\\ $44.25$ & $0.1204$ & $0.1205$ & $0.1\%$ & $\gamma_{25}/2=44.40$\\ \bottomrule \end{tabular} \end{table} \section{The C-transformation and the Laplace transform of the kernel}\label{sec:ctl} \subsection{The C-transformation} The C-transformation \cite{CP} measures the difference between a discrete sum and its continuous counterpart. For a weight $w$ with continuous density $m_w$ and a test function $f$, \[ C_N(f;w)=\sum_{n\le N}w(n)f(n)-\int_1^N m_w(x)f(x)\,dx . \] For $w=1$, $m_w=1$ and $f(x)=x^{-s}$, its limit is $\zeta(s)+\frac1{1-s}$. We apply it with $f(x)=x^{-s}$ to the kernel weight $b$ (Lemma~\ref{lem:basic}) and to $a$. For each weight we use as density the function whose Mellin transform is the principal part of $\K$ (respectively $\C$) at its real poles $\frac12,\frac14,0$ (respectively $1,\frac12,\frac14,0$): \[ m_b(x)=x^{-1/2}-\tfrac12x^{-3/4}-\tfrac1{12}x^{-1},\qquad m_a(x)=\K(1)+\zeta(\tfrac12)x^{-1/2}-\tfrac12\zeta(\tfrac14)x^{-3/4}+\tfrac1{24}x^{-1}. \] The three terms of $m_b$ correspond to the poles of $\K$ at $\frac12,\frac14,0$, with residues $1,-\frac12,-\frac1{12}$ (Theorem~\ref{thm:poles} and Proposition~\ref{prop:meanzero} below). We write \[ B(x)=\sum_{n\le x}b(n),\qquad M_b(x)=\int_1^xm_b=2\sqrt x-2x^{1/4}-\tfrac1{12}\log x,\qquad R_b(x)=B(x)-M_b(x), \] and similarly $M_a(x)=\int_1^xm_a$ and $R_a(x)=A(x)-M_a(x)$. \begin{proposition}[$\E$ is the C-transform of the kernel]\label{prop:Ctrans} For every $N\ge1$ and every $s$, \[ C_N(x^{-s};b)=R_b(N)N^{-s}+s\int_1^NR_b(x)\,x^{-s-1}dx . \] For $\Re s>1$, the limit as $N\to\infty$ exists and \[ \boxed{\;\lim_{N\to\infty}C_N(x^{-s};b)=\E(s)+\frac1{12s}=s\int_1^\infty R_b(x)\,x^{-s-1}\,dx\;} \] \end{proposition} \begin{proof} Since $B(1)=M_b(1)=0$, Stieltjes integration by parts gives \[ C_N=\int_{1}^{N}x^{-s}\,d(B-M_b)(x)=R_b(N)N^{-s}+s\int_1^NR_bx^{-s-1}dx . \] For $\Re s>1$, $\sum b(n)n^{-s}=\K(s)$ and \[ \int_1^\infty m_b(x)x^{-s}dx=\frac1{s-\frac12}-\frac1{2(s-\frac14)}-\frac1{12s}, \] and the difference is $\E(s)+\frac1{12s}$. \end{proof} \begin{lemma}\label{lem:RbRH} Under RH, $R_b(x)=O(\log^3x)$. \end{lemma} \begin{proof} Under RH, $\theta(u)=u+\Delta(u)$ with $\Delta(u)\ll\sqrt u\log^2u$ \cite[Ch.~18]{Dav}. Hence \[ \sum_{p\le x}\frac{\log p}{\sqrt p}=\int_{2^-}^xu^{-1/2}\,d\theta(u)=2\sqrt x+O(1)+\frac{\Delta(x)}{\sqrt x}+\frac12\int_2^x\frac{\Delta(u)}{u^{3/2}}\,du=2\sqrt x+O(\log^3x). \] The prime squares contribute $\sum_{p\le\sqrt x}\log p\,(p^{-1}-p^{-1/2})$. By Mertens' theorem and the previous line with $x$ replaced by $\sqrt x$, this equals $-2x^{1/4}+O(\log^3x)$. The powers $p^k$ with $k\ge3$ contribute $O(\log^2x)$, as in the proof of Theorem~\ref{thm:mean}. So $B(x)=2\sqrt x-2x^{1/4}+O(\log^3x)$, and the term $\frac1{12}\log x$ is absorbed in the error. \end{proof} \begin{theorem}[RH as a constant-scale statement]\label{thm:Ctrans} The following are equivalent: \begin{enumerate}[label=(\roman*),itemsep=1pt] \item the Riemann Hypothesis; \item $C_N(x^{-s};b)$ converges as $N\to\infty$ for every $s$ with $\Re s>0$; \item $R_b(x)=O_\varepsilon(x^\varepsilon)$ for every $\varepsilon>0$. \end{enumerate} \end{theorem} \begin{proof} (i)$\Rightarrow$(iii) is Lemma~\ref{lem:RbRH}. (iii)$\Rightarrow$(ii): by Proposition~\ref{prop:Ctrans}, $R_b(N)N^{-s}\to0$ and the integral converges absolutely when $\Re s>\varepsilon$. (ii)$\Rightarrow$(i): suppose $C_N(x^{-\varepsilon};b)$ converges, and put $G(y)=\int_1^yx^{-\varepsilon}d(B-M_b)$. Then \[ R_b(N)=\int_1^Nx^{\varepsilon}\,dG(x)=N^\varepsilon\bigl(G(N)-G(\infty)\bigr)-\varepsilon\int_1^Nx^{\varepsilon-1}\bigl(G(x)-G(\infty)\bigr)dx+O(1)=o(N^\varepsilon). \] So (iii) holds, and the integral of Proposition~\ref{prop:Ctrans} converges locally uniformly in $\Re s>0$. It therefore defines a holomorphic function there, which agrees with $\E(s)+\frac1{12s}$ for $\Re s>1$. Hence $\E$ is holomorphic in $\Re s>0$, and Theorem~\ref{thm:rh} gives RH. \end{proof} \begin{remark} By partial summation, \[ \sum_{p\le x}\frac{\log p}{\sqrt p}-2\sqrt x=\frac{\theta(x)-x}{\sqrt x}+\frac12\int_2^x\frac{\theta(u)-u}{u^{3/2}}\,du+O(1). \] Moreover $R_b(x)=D(x)-D(\sqrt x)+O(\log^2x)$ with $D(y)=\sum_{p\le y}\log p/\sqrt p-2\sqrt y$, and iterating this in $x\mapsto\sqrt x$ recovers $D$ from $R_b$. So Theorem~\ref{thm:Ctrans} is von Koch's criterion $\theta(x)-x=O(x^{1/2+\varepsilon})$ written in normalized form. The $\sqrt{\ }$-damping of $\C$ moves the RH scale from $x^{1/2}$ to the constant scale: RH says that the discrete-minus-continuous difference of the prime kernel stays $O(x^\varepsilon)$. \end{remark} \begin{proposition}[C-transform of $\C$]\label{prop:CtransC} $C_N(x^{-s};a)$ converges unconditionally for $\Re s\ge\frac12$, and under RH for $\Re s>\frac13$. The limit is \[ \widehat\C(s)=\C(s)-\frac{\K(1)}{s-1}-\frac{\zeta(\frac12)}{s-\frac12}+\frac{\zeta(\frac14)}{2(s-\frac14)}-\frac1{24s}. \] If $\K$ has a pole on $\Re s=0$ (in particular, unconditionally, at $s=i\gamma_1/2$, where no first-family pole can cancel it because $\gamma_1$ is the least ordinate), the C-transform of $\C$ converges in no half-plane $\Re s>\sigma$ with $\sigma<0$. \end{proposition} \begin{proof} As in Proposition~\ref{prop:Ctrans}, $C_N(x^{-s};a)=R_a(N)N^{-s}+s\int_1^NR_ax^{-s-1}dx$. By Theorem~\ref{thm:second}, $R_a(x)=O(\sqrt x\,e^{-c_\ast\sqrt{\log x}})$ unconditionally, with the constant $c_\ast>0$ of that theorem. Since $\int^\infty e^{-c_\ast\sqrt{\log x}}\,\frac{dx}x<\infty$, the integral converges absolutely for $\Re s\ge\frac12$. Under RH, $R_a(x)=O(x^{1/3+\varepsilon})$. The value of the limit follows by computing $\int_1^\infty m_ax^{-s}dx$ for $\Re s>1$, as before. Finally, a convergent C-transform defines a holomorphic function in its half-plane. But $\widehat\C=\zeta\K-\cdots$ has a pole wherever $\K$ does on $\Re s=0$, because $\zeta(i\gamma)\ne0$. \end{proof} \begin{observation}\label{obs:ctrans} With all $n\le10^8$: \begin{itemize}[itemsep=1pt] \item $R_b(x)$ stays in $[-1.56,\,0.77]$ for $10\le x\le10^8$ (infimum $-1.557$ as $x\to11897^-$, supremum $0.768$ at $x=113$; left and right limits at every prime power); its maximum modulus per decade stays below $1.56$ and shows no systematic growth (Table~\ref{tab:ctrans}, Figure~\ref{fig:ctrans}). \item The C-transform of the kernel at $s=0.3$ gives $-0.20291$, $-0.19737$ and $-0.19913$ at $N=10^6$, $10^7$ and $10^8$, against $\E(0.3)+\frac1{3.6}=-0.19737$ computed by analytic continuation: agreement to about $3\cdot10^{-3}$ (the five-digit match at $10^7$ is accidental). At $s=0.2+10i$ and $N=10^8$ it gives $0.230+1.158i$, against $0.223+1.134i$. \item In contrast, $R_a$ grows slowly: its rms over a decade goes from $0.98$ to $2.64$, and its maximum from $2.2$ to $9.5$, between $10^2$ and $10^8$. \end{itemize} The growth of $R_a$ comes from the integer layer. Indeed, $A(x)=\sum_{m\le x}B(x/m)$, so $R_a$ contains hyperbola sums such as $\sum_p\frac{\log p}{\sqrt p}(\{\frac xp\}-\frac12)$, which behave like the error term of a divisor problem. Such sums are independent of the zeros of $\zeta$. \end{observation} \begin{table}[ht] \centering \caption{Size of the two C-transform remainders, decade by decade.} \label{tab:ctrans} \begin{tabular}{lccccccc} \toprule decade ending at & $10^2$ & $10^3$ & $10^4$ & $10^5$ & $10^6$ & $10^7$ & $10^8$\\ \midrule $\max|R_b|$ & $1.00$ & $1.36$ & $1.54$ & $1.56$ & $1.40$ & $1.49$ & $1.31$\\ $\max|R_a|$ & $2.20$ & $3.08$ & $3.74$ & $6.10$ & $6.50$ & $7.74$ & $9.47$\\ rms $R_a$ & $0.98$ & $1.13$ & $1.37$ & $1.57$ & $1.97$ & $2.19$ & $2.64$\\ \bottomrule \end{tabular} \end{table} \begin{figure}[ht] \centering \includegraphics[width=0.9\textwidth]{fig_ctransform.pdf} \caption{The C-transform remainders $R_b(x)$ of the kernel (top), which is bounded on the computed range, and $R_a(x)$ of $\C$ (bottom), which grows slowly through the divisor-type integer layer.} \label{fig:ctrans} \end{figure} \subsection{The Laplace transform} In the variable $u=\log x$ the identities above are Laplace transforms. With the notation $\mathcal L[f](s)=\int_0^\infty f(u)e^{-su}du$ and $f_b(u)=R_b(e^u)$, partial summation gives, for $\Re s>1$, \[ \C(s)=s\,\mathcal L[A(e^u)](s),\qquad \K(s)=s\,\mathcal L[B(e^u)](s),\qquad \E(s)+\frac1{12s}=s\,\mathcal L[f_b](s). \] The poles of $\mathcal L[f_b]=(\E+\frac1{12s})/s$ on $\Re s=0$ are the possible frequencies of $f_b$, and the residues there are its Bohr--Fourier coefficients whenever these exist. \begin{proposition}[The kernel fluctuation has mean zero]\label{prop:meanzero} \[ \lim_{s\to0}\Bigl(\K(s)+\frac1{12s}\Bigr)=0 . \] Consequently, $\E(s)+\frac1{12s}$ is holomorphic at $s=0$ and vanishes there, and $\mathcal L[f_b]$ has no pole at $s=0$. If $f_b$ has a mean value $\lim_U\frac1U\int_0^Uf_b$, that mean is $0$. (The existence of such a mean, or merely $\int_0^Uf_b=O(U)$, already forces $\mathcal L[f_b]$ to converge in $\Re s>0$, hence RH; so this statement has content only under RH.) \end{proposition} \begin{proof} Near $w=1$, $P'(w)=-\frac1{w-1}+h_1(w)$. Near $w=\frac12$, $P'(w)=\frac{1/2}{w-1/2}+h_{1/2}(w)$, the pole coming from the term $\mu(2)\frac{\zeta'}{\zeta}(2w)$. Here $h_1$ and $h_{1/2}$ are holomorphic, since no other term of Corollary~\ref{cor:mobius} is singular at these points. In \eqref{eq:pz}, only $k=1$ and $k=2$ have arguments tending to a pole of $P'$ as $s\to0$: \[ P'(2s+\tfrac12)-P'(s+\tfrac12)=\frac1{4s}-\frac1{2s}+o(1),\qquad P'(3s+1)-P'(2s+1)=-\frac1{3s}+\frac1{2s}+o(1). \] For $k\ge3$, both arguments tend to $\frac k2\ge\frac32$, where $P'$ is holomorphic, so each bracket is holomorphic near $0$ and vanishes at $s=0$. The series of these brackets converges uniformly near $0$, so its sum is holomorphic and vanishes at $0$. Altogether $\K(s)=-\frac1{12s}+o(1)$. At $s=0$, the terms $-\frac1{s-1/2}$ and $\frac1{2(s-1/4)}$ equal $2$ and $-2$, so $\E(s)+\frac1{12s}\to0$. The last claim is the Abelian theorem for Laplace transforms \cite{Wid}: if the mean $c$ exists, then $\sigma\mathcal L[f_b](\sigma)\to c$ as $\sigma\to0^+$. \end{proof} Numerically, with the exact Euler--Maclaurin derivative of $\zeta$, $\K(s)+\frac1{12s}$ extrapolates to $-8\cdot10^{-8}$ at $s=0$. The negative offset of about $-0.4$ visible in Figures~\ref{fig:ctrans} and~\ref{fig:bohr} is not a mean. It is the real pole of $\K$ at $-\frac1{12}$, with residue $\frac16$, which contributes $\frac{1/6}{-1/12}x^{-1/12}=-2x^{-1/12}$ to $R_b$ and decays very slowly. \begin{proposition}[Bohr spectrum of the kernel]\label{prop:bohr} Assume RH. Then $\mathcal L[f_b]$ is holomorphic in $\Re s>0$ and meromorphic in $\Re s>-\frac14$. Its poles on $\Re s=0$ lie at $\pm i\gamma$ and $\pm\frac{i\gamma}2$, with residues \[ \operatorname*{Res}_{s=i\gamma}\mathcal L[f_b]=\frac{-m_\gamma}{i\gamma}=\frac{i\,m_\gamma}{\gamma},\qquad \operatorname*{Res}_{s=i\gamma/2}\mathcal L[f_b]=\frac{m_\gamma/2}{i\gamma/2}=-\frac{i\,m_\gamma}{\gamma}, \] the two being added where the families meet. Consequently, whenever the Bohr--Fourier coefficient $c(\omega)=\lim_U\frac1U\int_0^Uf_b(u)e^{-i\omega u}du$ exists at $\omega=\gamma$ or $\omega=\gamma/2$, it equals the corresponding residue. Formally, \[ \boxed{\;R_b(x)\ \sim\ \sum_{\gamma>0}\frac{2m_\gamma}{\gamma}\Bigl[\sin\bigl(\tfrac\gamma2\log x\bigr)-\sin\bigl(\gamma\log x\bigr)\Bigr]+\text{(decaying terms)}.\;} \] \end{proposition} \begin{proof} The residues of $\K$ at $i\gamma$ and $i\gamma/2$ are $-m_\gamma$ and $\frac12m_\gamma$ (Lemma~\ref{lem:polesneg}). The terms $-\frac1{s-1/2}+\frac1{2(s-1/4)}+\frac1{12s}$ add no poles on $\Re s=0$ apart from $0$, which is removable by Proposition~\ref{prop:meanzero}. Dividing by $s$ gives the residues. If $c(\omega)$ exists, the Abelian theorem gives $\sigma\mathcal L[f_b](\sigma+i\omega)\to c(\omega)$, and near a simple pole $\sigma\mathcal L[f_b](\sigma+i\omega)\to\operatorname{Res}$. Finally, $2\Re[\frac{i}{\gamma}e^{i\gamma u}]=-\frac2\gamma\sin\gamma u$ and $2\Re[-\frac{i}{\gamma}e^{i\gamma u/2}]=\frac2\gamma\sin\frac{\gamma u}2$. \end{proof} \begin{observation}\label{obs:bohr} We fitted $f_b(u)$, sampled at $2^{16}$ points on $[\log10^2,\log10^8]$ and with the analytically known decaying terms removed, by least squares on $\{\cos,\sin\}(\omega u)$ for the $192$ frequencies $\omega\in\{\gamma_j,\gamma_j/2\}$ with $\gamma_j<230$. The sine coefficients agree with $\mp2/\gamma$ and the cosine coefficients are close to $0$ (Table~\ref{tab:bohr}). The reconstruction with 96 zeros has correlation $0.91$ with $R_b$ (Figure~\ref{fig:bohr}). The visible misses ($\gamma_4$, $\gamma_9$, $\gamma_{10}/2$) are explained by near-coincident frequencies of the two families that a window of length $13.8$ cannot separate. The closest is $\gamma_{14}=60.8318$ against $2\gamma_4=60.8498$, a gap of $0.018$; Proposition~\ref{prop:pairs} predicts the merged coefficient $-2(\frac1{\gamma_4}-\frac1{\gamma_{14}})=-0.0329$ at $\gamma_4$. \end{observation} \begin{table}[ht] \centering \caption{Bohr coefficients of the kernel fluctuation $f_b(u)=R_b(e^u)$: fitted sine coefficients against the predictions of Proposition~\ref{prop:bohr}.} \label{tab:bohr} \begin{tabular}{rccccc} \toprule $j$ & $\gamma_j$ & $\sin$ at $\gamma_j$ & predicted $-2/\gamma_j$ & $\sin$ at $\gamma_j/2$ & predicted $+2/\gamma_j$\\ \midrule 1 & 14.1347 & $-0.1418$ & $-0.1415$ & $+0.1438$ & $+0.1415$\\ 2 & 21.0220 & $-0.0973$ & $-0.0951$ & $+0.0906$ & $+0.0951$\\ 3 & 25.0109 & $-0.0807$ & $-0.0800$ & $+0.0807$ & $+0.0800$\\ 4 & 30.4249 & $-0.0362$ & $-0.0657$ & $+0.0717$ & $+0.0657$\\ 5 & 32.9351 & $-0.0615$ & $-0.0607$ & $+0.0618$ & $+0.0607$\\ 6 & 37.5862 & $-0.0544$ & $-0.0532$ & $+0.0561$ & $+0.0532$\\ 7 & 40.9187 & $-0.0456$ & $-0.0489$ & $+0.0482$ & $+0.0489$\\ 8 & 43.3271 & $-0.0446$ & $-0.0462$ & $+0.0462$ & $+0.0462$\\ 9 & 48.0052 & $-0.0326$ & $-0.0417$ & $+0.0419$ & $+0.0417$\\ 10 & 49.7738 & $-0.0412$ & $-0.0402$ & $+0.0295$ & $+0.0402$\\ \bottomrule \end{tabular} \end{table} \begin{figure}[ht] \centering \includegraphics[width=\textwidth]{fig_bohr.pdf} \caption{The kernel fluctuation $R_b(x)$ computed from all primes up to $10^8$ (blue), against its Bohr series of Proposition~\ref{prop:bohr} with 96 zeros plus the analytically known decaying terms (orange). The dashed curve is the slowly decaying real-pole part, dominated by $-2x^{-1/12}$.} \label{fig:bohr} \end{figure} \begin{theorem}[A one-sided criterion]\label{thm:landau} The Riemann Hypothesis holds if and only if, for every $\varepsilon>0$, there is $C_\varepsilon$ with \[ R_b(x)\ge-C_\varepsilon\,x^\varepsilon\qquad(x\ge1). \] The same holds with the upper bound $R_b(x)\le C_\varepsilon x^\varepsilon$ in place of the lower bound. \end{theorem} \begin{proof} If RH holds, both bounds follow from Lemma~\ref{lem:RbRH}. Conversely, fix $\varepsilon>0$ and put $g(u)=f_b(u)+C_\varepsilon e^{\varepsilon u}\ge0$. For $\Re s>1$, \[ \mathcal L[g](s)=\frac{\E(s)+\frac1{12s}}{s}+\frac{C_\varepsilon}{s-\varepsilon}=:\Phi(s). \] By Theorem~\ref{thm:poles}, unconditionally, $\K$ has no real poles in $\Re s>0$ other than $\frac12$ and $\frac14$, because zeros of $\zeta$ are not real. Hence $\E$, and so $\Phi$, is holomorphic on a neighbourhood of every real point $\sigma>\varepsilon$. By Landau's theorem for Laplace transforms of nonnegative functions \cite[Ch.~II, Thm.~5b]{Wid}, the real point $\sigma_g$ of the abscissa of convergence of $\mathcal L[g]$ is a singularity of the function it defines. That function coincides with $\Phi$ wherever the integral converges, so $\sigma_g\le\varepsilon$. Thus $\mathcal L[g]$ converges in $\Re s>\varepsilon$, and $\E$ is holomorphic there. Since $\varepsilon$ is arbitrary, Theorem~\ref{thm:rh} gives RH. For the upper bound, use $g=C_\varepsilon e^{\varepsilon u}-f_b$. \end{proof} \begin{remark} Theorem~\ref{thm:landau} is the normalized analogue of Landau's classical one-sided theorem for $\psi(x)-x$. It is again a reformulation of RH. The case $\varepsilon=0$, a uniform one-sided bound, is not expected to hold: Littlewood's $\psi(x)-x=\Omega_\pm(\sqrt x\log\log\log x)$ suggests that $R_b$ is unbounded in both directions, though extremely slowly. On the computed range $R_b$ is asymmetric, with $\inf R_b=-1.56$ and $\sup R_b=0.77$, mainly because of the slowly decaying term $-2x^{-1/12}$. For $\C$ itself the Laplace picture is analogous, with Bohr coefficients, whenever they exist, $-m_\gamma\zeta(i\gamma)/(i\gamma)$ and $m_\gamma\zeta(i\gamma/2)/(i\gamma)$; here $\mathcal L[R_a(e^u)]$ is only known to converge in $\Re s>\frac13$ under RH. These are the amplitudes of the model \eqref{eq:explicit}, whose spectrum matches that of the residual of $A$ in Table~\ref{tab:peaks}. A one-sided criterion does not follow for $\C$, because zeros of $\zeta$ in $\C=\zeta\K$ could in principle cancel poles of $\K$. \end{remark} \section{The kernel and the Ces\`aro margin}\label{sec:margin} This section connects the kernel with Suzuki's one-variable form of RH, studied as the \emph{Ces\`aro margin} in Part~11 \cite{CE11}. We recall the objects. For $N>1$ and $L=\log N$, let \[ C(N)=2\sum_{n0}m_\gamma(1-\cos\gamma L)/\gamma^2\in[0,4(\log\xi)''(\frac12)]$, where $(\log\xi)''(\frac12)=\sum_\rho t_\rho^{-2}=0.0462100$. The \emph{slope} of the margin is \[ \alpha(x)=\sum_{n\le x}\frac{\Lambda(n)}{\sqrt n}-2\sqrt x+\frac{c_0}2 . \] Under RH, $u\mapsto\alpha(e^u)$ is a Besicovitch almost periodic function with mean $0$ and Fourier coefficients $\frac{im_\gamma}\gamma$ at the frequencies $\gamma$ \cite{RuSa,ANS,CE11}. \subsection{The decomposition of the kernel} By Lemma~\ref{lem:basic}(iii), $b(n)=\Lambda(n)n^{-1/2}+q(n)$, where $q$ lives on the prime powers $p^j$, $j\ge2$, with $q(p^j)=-\log p\,p^{-(j-1)/2}$. So \begin{equation}\label{eq:Ksplit} \K(s)=-\frac{\zeta'}\zeta\bigl(s+\tfrac12\bigr)+Q(s),\qquad Q(s)=\sum_nq(n)n^{-s}=\sum_{k\ge1}P'\bigl((k+1)s+\tfrac k2\bigr). \end{equation} \begin{proposition}[the echo term]\label{prop:E6} For $\Re s>0$, \[ \K(s)=-\frac{\zeta'}{\zeta}\bigl(s+\tfrac12\bigr)+\frac{\zeta'}{\zeta}\bigl(2s+\tfrac12\bigr)-\frac1{12s}+H(s), \] where $H$ extends holomorphically to a neighbourhood of the closed half-plane $\Re s\ge0$, and $H(0)=0$. Under RH, $H$ is holomorphic in $\Re s>-\frac1{12}$, and its first singularity is a simple pole at $s=-\frac1{12}$ with residue $\frac16$. \end{proposition} \begin{proof} Use $-\frac{\zeta'}\zeta(w)=-\sum_{m\ge1}P'(mw)$. In \eqref{eq:Ksplit}, the term $k=1$ of $Q$ is $P'(2s+\frac12)$, which is the term $m=1$ of $-\frac{\zeta'}{\zeta}(2s+\frac12)$ with the opposite sign. Hence \[ D(s):=Q(s)-\frac{\zeta'}\zeta\bigl(2s+\tfrac12\bigr)=\sum_{k\ge2}P'\bigl((k+1)s+\tfrac k2\bigr)-\sum_{m\ge2}P'\bigl(m(2s+\tfrac12)\bigr). \] At $s=0$ the arguments are $\frac k2\ge1$ and $\frac m2\ge1$. By Corollary~\ref{cor:mobius}, $P'$ is singular only at $w=1/j$ and $w=\rho/j$, $j$ squarefree. An argument equal to $1$ occurs only for $k=2$ ($3s+1$) and $m=2$ ($4s+1$). Since $P'(w)=-\frac1{w-1}+O(1)$ near $1$, we get $D(s)=-\frac1{3s}+\frac1{4s}+O(1)=-\frac1{12s}+O(1)$. Put $H=D+\frac1{12s}$. A zero $\rho=\beta+i\gamma$ makes a term singular where its argument equals $\rho/j$. For an argument $as+b$ with $b\ge1$, this happens at $\Re s=(\beta/j-b)/a<0$, since $\beta<1$. Near any point of $\Re s=0$ only finitely many terms are involved, so $H$ is holomorphic in a neighbourhood of $\Re s\ge0$. $H(0)=0$ is Proposition~\ref{prop:meanzero}, because the two $\frac{\zeta'}\zeta$ terms cancel at $s=0$. Under RH, the zero-induced singularities have real parts $(\frac1{2j}-b)/a\le-\frac18$. The real ones come from $as+b=1/j$. The largest of these below $0$ is $6s+\frac32=1$, from $m=3$, at $s=-\frac1{12}$; there $-P'(6s+\frac32)\sim\frac1{6s+1/2}$, which has residue $\frac16$. \end{proof} So the kernel is $-\frac{\zeta'}\zeta$ at two scales, $s+\frac12$ and $2s+\frac12$, with opposite signs and weights $1$ and $\frac12$ for the residues. Everything else is either the explicit pole $-\frac1{12s}$, which produces the term $-\frac1{12}\log x$ of $R_b$ and the $\frac1{48}\log^2x$ of Theorem~\ref{thm:explicitF}, or holomorphic across the imaginary axis. \subsection{The bridge} \begin{proposition}[the kernel's remainder is the slope at two scales]\label{prop:bridge} Without any hypothesis, \[ R_b(x)=\alpha(x)-\alpha(\sqrt x)+o(1)\qquad(x\to\infty). \] \end{proposition} \begin{proof} Let $A_\Lambda(x)=\sum_{n\le x}\Lambda(n)n^{-1/2}$. By \eqref{eq:Ksplit}, $B(x)=A_\Lambda(x)+\sum_{n\le x}q(n)$. The prime squares in $\sum q$ give $-\sum_{p\le\sqrt x}\log p/\sqrt p$, which is $-A_\Lambda(\sqrt x)$ up to the prime powers counted in $A_\Lambda(\sqrt x)$. Collecting terms, \[ R_b(x)-\bigl(\alpha(x)-\alpha(\sqrt x)\bigr)=\sum_{i\ge2}\Bigl(\sum_{p^{2i}\le x}-\sum_{p^{i+1}\le x}\Bigr)\frac{\log p}{p^{i/2}}+\frac1{12}\log x . \] For $i=2$ the bracket is $-\sum_{x^{1/4}x^{1/(2i)}}\log p\,p^{-i/2}$. \end{proof} With all prime powers below $10^{12}$, the difference in Proposition~\ref{prop:bridge} is at most $0.73$ per decade up to $10^5$, and decreases to $0.20$ on $[10^{11},10^{12}]$. The slow decay is the term $-2x^{-1/12}$ of Proposition~\ref{prop:E6}, followed by terms of order $x^{-1/8}$ ($O(x^{-1/8}\log^2x)$ under RH), (Figure~\ref{fig:kernelspec}(a)). \subsection{The kernel's margin} For a weight $w$, write $F_w(N)=\sum_{n0}\frac{m_\gamma\sin^2\frac{\gamma L}4\cos\frac{\gamma L}2}{\gamma^2}+O(N^{\sigma_0})\\ &\in[-0.3697,\,0.1848]+O(N^{\sigma_0}). \end{aligned} \] \end{theorem} \begin{proof} For $c>1$, Perron's formula with the kernel $N^t/t^2$ gives $F_q(N)=\frac1{2\pi i}\int_{(c)}Q(t)N^t\frac{dt}{t^2}$. The substitution $s=2t$ in the Perron integral of $C(X)$ at $X=\sqrt N$ gives $C(\sqrt N)=\frac1{2\pi i}\int_{(c)}\bigl(-\frac{\zeta'}{\zeta}\bigr)(2t+\frac12)N^t\frac{dt}{t^2}$. Hence $F_q(N)+C(\sqrt N)=\frac1{2\pi i}\int_{(c)}D(t)N^t\frac{dt}{t^2}$, with $D=-\frac1{12t}+H$ as in the proof of Proposition~\ref{prop:E6}. Under RH, every argument of $\frac{\zeta'}\zeta$ in the expansion of $D$ stays at distance at least $\frac16$ from the critical line when $\Re t\ge\sigma_0$. So $D(t)\ll\log(|t|+2)$ on $\Re t=\sigma_0$ and on horizontal segments, and the contour can be moved to $\Re t=\sigma_0$ with error $O(N^{\sigma_0})$. The only pole crossed is $t=0$, where $D(t)N^tt^{-2}$ has residue $-\frac1{24}\log^2N+H(0)\log N+H'(0)$; and $H(0)=0$. Since $D(t)=\K(t)+\frac{\zeta'}{\zeta}(t+\frac12)-\frac{\zeta'}{\zeta}(2t+\frac12)$, $D+\frac{1}{12t}$ has derivative $H'(0)$ at $0$, and the stated $d_1$ is $\frac{d}{dt}(\K(t)+\frac1{12t})|_{0}-(\frac{\zeta'}{\zeta})'(\frac12)$. For the margin, substitute $C(N)=R(N)-M(N)$ and $C(\sqrt N)=R(\sqrt N)-M(\sqrt N)$ in $2F_b=C+2F_q$. The bounds follow from $0\le M\le0.18484$ at $N$ and at $\sqrt N$. The last form uses $(1-\cos\theta)-2(1-\cos\frac\theta2)=4\sin^2\frac\theta4\cos\frac\theta2$. \end{proof} With the prime powers below $10^8$, a smooth fit gives $d_1\approx-10.34$, against $-10.254$ (here $(\K+\frac1{12s})'(0)=-6.5072$ and $(\zeta'/\zeta)'(\frac12)=3.7469$). Removing $-C(\sqrt N)$ from $F_q$ removes the echo: the correlation of the residual with $-\sum_\gamma\frac4{\gamma^2}\cos\frac{\gamma\log N}2$ drops from $0.83$ to $0.12$. Unlike $M$, the kernel's margin changes sign term by term, since $\cos\frac{\gamma L}2$ does. With the first $96$ zeros, $M(N)-2M(\sqrt N)$ lies in $[-0.183,-0.005]$ for $1.65\le N\le e^{25}$. This loss of positivity is why the one-sided criterion of Theorem~\ref{thm:landau} needs the factor $x^\varepsilon$. \subsection{Energy, and pairs of zeros with \texorpdfstring{$\gamma'=2\gamma$}{gamma'=2gamma}} \begin{proposition}[the kernel's energy]\label{prop:pairs} Assume RH and that all zeros are simple. Then $u\mapsto R_b(e^u)$ is Besicovitch almost periodic, and \[ \bigl\langle(R_b-\langle R_b\rangle)^2\bigr\rangle=2(\log\xi)''(\tfrac12)-2\sum_{\substack{\gamma>0\\ 2\gamma\text{ is an ordinate}}}\frac1{\gamma^2}. \] In particular, if no two ordinates satisfy $\gamma'=2\gamma$ (as the linear independence hypothesis implies), the energy is $2(\log\xi)''(\frac12)=0.092420$, and the variogram is $\langle|R_b(e^{u+L})-R_b(e^u)|^2\rangle=M(e^L)+M(e^{L/2})$. On a window of length $U$ in $\log x$, a pair with $|\gamma'-2\gamma|\ll1/U$ is not resolved, and the amplitude seen at $\gamma$ is $2(\frac1\gamma-\frac1{\gamma'})$ instead of $\frac2\gamma$. \end{proposition} \begin{proof} By Proposition~\ref{prop:bridge}, $R_b(e^u)=\alpha(e^u)-\alpha(e^{u/2})+o(1)$. The first term has the coefficients $\frac i\gamma$ at the frequencies $\pm\gamma$. The second has $-\frac i{\gamma'}$ at $\pm\frac{\gamma'}2$. If the frequency sets are disjoint, Parseval adds $2\sum_{\gamma>0}\gamma^{-2}=(\log\xi)''(\frac12)$ twice. If $\gamma'=2\gamma$, the frequency $\gamma$ receives $\frac i\gamma-\frac i{\gamma'}$, and $|\frac1\gamma-\frac1{\gamma'}|^2=\frac1{\gamma^2}+\frac1{\gamma'^2}-\frac2{\gamma\gamma'}$. Counting both signs, the defect is $\frac4{\gamma\gamma'}=\frac2{\gamma^2}$. The variogram is computed the same way, using the formula for $M$ at $L$ and at $L/2$. The finite-window statement is the same computation for two frequencies closer than the resolution $2\pi/U$. \end{proof} The linear independence hypothesis excludes exact pairs, but near pairs are unavoidable. Among the first zeros, $\gamma_{14}=60.8318$ and $2\gamma_4=60.8498$ differ by $0.018$, far below the resolution $2\pi/U\approx0.5$ of the range $2^8\le x\le2^{26}$. Proposition~\ref{prop:pairs} predicts the amplitude $0.0329$ at $\gamma_4$ instead of $\frac2{\gamma_4}=0.0657$. The fit finds $0.024$, while the amplitudes at $\gamma_1,\gamma_2,\gamma_3$ and at their halves are within $10\%$ of $\frac2\gamma$ (Figure~\ref{fig:kernelspec}(b); compare Table~\ref{tab:bohr}). The block energies of $R_b$ have mean $0.106$ for $2^{12}\le x\le2^{26}$. The excess over $0.0924$ matches the energy of the slowly decaying terms of Proposition~\ref{prop:E6} (Figure~\ref{fig:kernelspec}(c)). So the Caceres kernel is sensitive to additive relations $\gamma'\approx2\gamma$ among the ordinates, which $\alpha$ and $M$ do not see. \begin{figure}[ht] \centering \includegraphics[width=0.8\textwidth]{fig_kernelspec.png} \caption{The kernel and the margin. (a) $\alpha_b=R_b-\langle R_b\rangle$ against $\alpha(x)-\alpha(\sqrt x)$ (Proposition~\ref{prop:bridge}) and $\alpha(x)$, $10^6\le x\le4\cdot10^6$. (b) Amplitude spectrum in $\log x$ on $2^8\le x\le2^{26}$: $\alpha$ has peaks at $\gamma$; $\alpha_b$ at $\gamma$ and $\gamma/2$ (dotted), except at $\gamma_4$, which the dual copy of $\gamma_{14}\approx2\gamma_4$ cancels. (c) Energy per dyadic block: $\alpha_b$ around $2(\log\xi)''(\frac12)$ and $\alpha$ around $(\log\xi)''(\frac12)$.} \label{fig:kernelspec} \end{figure} \begin{remark}[which object is rigid, and what the kernel does not add]\label{rem:rigid} The coefficients of $R_b$, $\pm\frac i\gamma$, depend only on the ordinates. Those of $\sum_{n\le x}a(n)$ do not. Since $\C=\zeta\K$, the residue of $\C(s)x^s/s$ at $i\gamma$ is $-m_\gamma\zeta(i\gamma)x^{i\gamma}/(i\gamma)$. By the functional equation, $|\zeta(i\gamma)|=(\gamma/2\pi)^{1/2}|\zeta(1+i\gamma)|$, and $|\zeta(1+i\gamma)|\gg1/\log\gamma$. So these terms decay only like $\gamma^{-1/2}$, up to that factor. They depend on $\zeta$ at the edge of the strip, and the series over the zeros is not absolutely convergent. This is why Theorem~\ref{thm:explicitF} works with the Riesz mean $F$. Summing over all integers multiplies by $\zeta$ and destroys rigidity; the rigid object is $b=\mu*a$, which lives on the prime powers. Since $b(p)=\Lambda(p)/\sqrt p$ and $b(p^j)$ is a fixed function of $p$, the kernel carries exactly the information of $\Lambda$. A positive perturbation of $\Lambda$ that plants an off-line zero \cite{CE11} plants it in $b$ and $a$ as well. A true off-line zero $\frac12+\delta+i\gamma_0$ would appear in $R_b$ twice, as $x^\delta$ at $\gamma_0$ and as $-x^{\delta/2}$ at $\frac{\gamma_0}2$. These are distinct frequencies, so the two copies can neither cancel nor reinforce into a contradiction. Status: Propositions~\ref{prop:E6}, \ref{prop:bridge}, \ref{prop:pairs} and Theorem~\ref{thm:kernelmargin} are proved, the last two under RH. They give a dictionary between the kernel and the margin, not a new route to RH. \end{remark} \section{The critical line and the zeros of the kernel}\label{sec:zeros} \subsection{On the critical line} Theorem~\ref{thm:poles} lets us evaluate $\C(\frac12+it)=\zeta(\frac12+it)\K(\frac12+it)$ by analytic continuation. As an independent check, we used the smoothed Dirichlet sum \begin{multline*} \sum_n a(n)\,n^{-1/2-it}e^{-n/X}=\C(\tfrac12+it)+\K(1)\Gamma(\tfrac12-it)X^{1/2-it}\\ +\zeta(\tfrac12)\Gamma(-it)X^{-it}-\tfrac12\zeta(\tfrac14)\Gamma(-\tfrac14-it)X^{-1/4-it}+O(X^{-1/2}), \end{multline*} with $X=2\cdot10^5$. This expansion comes from the Mellin--Barnes formula $\sum a(n)n^{-w}e^{-n/X}=\frac1{2\pi i}\int\C(w+z)\Gamma(z)X^z\,dz$ by moving the contour past the poles of $\C$ at $1,\frac12,\frac14$. We use it only as an independent numerical check, and do not claim it with a proved error term. \begin{observation}\label{obs:line} On $5\le t\le65$ (step $0.1$), the two computations differ by $0.0065$ (median) and $0.028$ (maximum), on values of order $1$ (Figure~\ref{fig:crit}). $\C(\frac12+it)$ vanishes at every $\gamma_n$. On this grid $|\K(\frac12+it)|\ge0.005$, so $\C$ shows no other zeros on the line there. The minimum, at $t\approx45.30$, is explained by a zero of $\K$ at $0.4967+45.3037\,i$, just off the line. \end{observation} \begin{figure}[ht] \centering \includegraphics[width=0.88\textwidth]{fig_critical_line.pdf} \caption{Top: $|\C(\frac12+it)|$ from the smoothed Dirichlet sum (solid) and from $\zeta\K$ by analytic continuation (dashed). Bottom: $|\K(\frac12+it)|$. Dotted lines mark the ordinates $\gamma_n$.} \label{fig:crit} \end{figure} \subsection{The kernel has zeros of its own} Since $\C=\zeta\K$, the zeros of $\C$ are the zeros of $\zeta$ together with the zeros of $\K$. Newton's method applied to the analytic continuation of $\K$ locates the zeros in Table~\ref{tab:kzeros}. \begin{lemma}[Rigorous tail]\label{lem:tail} Let $N\ge10^6$ and $\Re s=\sigma>1$. Then \[ \Bigl|\C(s)-\sum_{n\le N}\frac{a(n)}{n^s}-\K(1)\frac{N^{1-s}}{s-1}\Bigr|\le5.51\,N^{1/2-\sigma}\Bigl(1+\frac{|s|}{\sigma-\frac12}\Bigr). \] \end{lemma} \begin{proof} By partial summation, \[ \sum_{n>N}a(n)n^{-s}=-A(N)N^{-s}+s\int_N^\infty A(t)t^{-s-1}dt . \] Inserting $A(t)=\K(1)t+R(t)$, the main part equals $\K(1)N^{1-s}/(s-1)$. By Theorem~\ref{thm:mean}, $|R(t)|\le5.51\sqrt t$ for $t\ge N$. Hence the remainder $-R(N)N^{-s}+s\int_N^\infty R(t)t^{-s-1}dt$ is at most the stated bound. \end{proof} \begin{theorem}[A zero of $\C$ in the half-plane of absolute convergence]\label{thm:zero} Let $s_0=1.379034+8.639155\,i$. The function $\C$, and hence $\K$, has exactly one zero in the disc $|s-s_0|<0.01$, and exactly one zero in $|s-s_0|<0.1$. \end{theorem} \begin{proof}[Proof (computer-assisted)] Let $N=10^6$ and \[ f(s)=\sum_{n\le N}a(n)n^{-s}+\kappa\,\frac{N^{1-s}}{s-1}, \] where $\kappa$ is the midpoint of the interval of Proposition~\ref{prop:K1}. By Lemma~\ref{lem:tail} and Proposition~\ref{prop:K1}, on a circle $|s-s_0|=r$ we have $|\C(s)-f(s)|\le E_r$, where \[ E_r=8.2\cdot10^{-6}\,\frac{N^{1-\sigma_r}}{|s_0-1|-r}+5.51\,N^{1/2-\sigma_r}\Bigl(1+\frac{|s_0|+r}{\sigma_r-\frac12}\Bigr),\qquad \sigma_r=\Re s_0-r . \] Let $L_r$ be the explicit bound for $|f'|$ on the closed disc given by $\sum_{n\le N}a(n)\log n\,n^{-\sigma_r}$ plus the derivative of the last term. We evaluated $f$ at $M$ equally spaced points $z_j$ of the circle. Let $h$ be the chord between consecutive points. The checks were: \[ \min_j|f(z_j)|-L_rh>E_r\qquad\text{and}\qquad\frac1{2\pi}\sum_j\arg\frac{f(z_{j+1})}{f(z_j)}=1 . \] The first inequality gives $|f|>E_r\ge|\C-f|$ on the whole circle, since $|f(z)|\ge|f(z_j)|-L_rh$ on the arc after $z_j$. It also gives $|f(z)-f(z_j)|<|f(z_j)|$ there, so each increment of the argument is determined by its endpoints and the sampled winding number equals the true one. By the argument principle $f$ has exactly one zero inside, and by Rouch\'e's theorem so does $\C$. Since $\zeta(s)\ne0$ for $\Re s>1$, it is a zero of $\K$. The certified numbers are listed in Table~\ref{tab:cert}. Every $f(z_j)$ was computed in double precision as a sum of at most $10^6$ terms whose absolute values add up to less than $2.61$, since $\sum_{n\le N}a(n)n^{-1.279}=2.6008$. There are three sources of floating-point error. \begin{itemize}[itemsep=1pt] \item \emph{The values $a(n)$.} Each is a sum of at most $20$ terms $b(p^k)$ with correctly rounded logarithms and powers, so its relative error is below $50\cdot2^{-52}$. \item \emph{Each term $a(n)e^{-s\log n}$.} The argument satisfies $|s\log n|\le9.9\cdot13.9<140$, and $\exp$, $\cos$ and $\sin$ are accurate to a few ulps. So the relative error of each term is below $140\cdot8\cdot2^{-52}<3\cdot10^{-13}$. \item \emph{The summation.} The rounding error is at most $10^6\cdot2^{-52}\cdot2.61<6\cdot10^{-10}$. \end{itemize} In total, every computed $f(z_j)$ is within $10^{-9}$ of its exact value. This is far below the smallest margin, $8.6\cdot10^{-4}$, and $L_r$ and $E_r$ are increased by an equally negligible amount. \end{proof} \begin{table}[ht] \centering \caption{Data of the computer-assisted proof of Theorem~\ref{thm:zero}, with $N=10^6$.} \label{tab:cert} \begin{tabular}{ccccccc} \toprule $r$ & $M$ & $\min_j|f(z_j)|$ & $L_r$ & $L_rh$ & $E_r$ & winding\\ \midrule $0.01$ & $4000$ & $1.3374\cdot10^{-3}$ & $6.67$ & $1.05\cdot10^{-4}$ & $3.73\cdot10^{-4}$ & $1$\\ $0.1$ & $6000$ & $1.2134\cdot10^{-2}$ & $11.46$ & $1.20\cdot10^{-3}$ & $1.44\cdot10^{-3}$ & $1$\\ \bottomrule \end{tabular} \end{table} \begin{corollary}[Infinitely many zeros with $\Re s>1$]\label{cor:manyzeros} There are constants $c>0$ and $T_0$ such that, for $T\ge T_0$, $\C$ has at least $cT$ zeros in the region $|\Re s-\Re s_0|<0.1$, $0<\Im s9\cdot10^{-3}$. Choose $M_0$ with $\sum_{n>M_0}a(n)n^{-\sigma_1}<\mu/4$, and put $Q(\tau)=\sum_{n\le M_0}w_n n^{-i\tau}$ with $w_n=a(n)n^{-\sigma_1}$. $Q$ is a trigonometric polynomial with distinct frequencies $\log n$, so it is almost periodic, and for every $\eta>0$ its $\eta$-almost periods form a relatively dense set \cite{Bes}. If $\tau$ is such an almost period, then for each $n\le M_0$ \[ w_n|n^{-i\tau}-1|=\Bigl|\lim_{U\to\infty}\frac1{2U}\int_{-U}^{U}\bigl(Q(t+\tau)-Q(t)\bigr)n^{it}\,dt\Bigr|\le\eta . \] Taking $\eta=\mu/(4M_0)$, for all $s\in D$ \[ |\C(s+i\tau)-\C(s)|\le\sum_{n\le M_0}w_n|n^{-i\tau}-1|+2\sum_{n>M_0}a(n)n^{-\sigma_1}<\tfrac\mu4+\tfrac\mu2<\mu\le|\C(s)|\quad(s\in\partial D). \] By Rouch\'e's theorem, $\C$ has a zero in $D+i\tau$. Relative density gives such $\tau$ in every interval of some fixed length $\ell$. Choosing one in each of the intervals $[k(\ell+1),k(\ell+1)+\ell]$, $k\ge0$, gives pairwise disjoint discs in $\Im s>8.5$, hence at least $T/(\ell+1)-2$ zeros with $0<\Im s\gamma_{96}}\gamma^{-2}=-0.013$ on average, so the true upper end is nearer $-0.018$. It cannot stay below a negative constant: under RH it is $\sum_\gamma\frac{8m_\gamma}{\gamma^2}c_\gamma(1-c_\gamma)$ with $c_\gamma=\cos\frac{\gamma L}2$, a uniformly almost periodic function of $L$ that vanishes at $L=0$, so its upper limit is $\ge0$; and if the first $649$ ordinates are linearly independent over $\mathbb Q$, Kronecker's theorem gives arbitrarily large $L$ where it comes arbitrarily close to, or exceeds, $2\sum_{\gamma<1000}\gamma^{-2}-16\sum_{\gamma>1000}\gamma^{-2}=0.0443-0.0155>0$. What is the first $N$ at which the kernel's margin is positive? \end{enumerate} \appendix \section{Numerical methods}\label{app:num} $a(n)$ for $n\le10^8$ was computed by sieving, adding $b(p^k)$ to every multiple of $p^k$ (Lemma~\ref{lem:basic}(iii)); it was checked against factorization for $n\le2\cdot10^4$ (error $\le9\cdot10^{-16}$), and cumulative sums were kept in 80-bit precision. $\zeta$ and $\zeta'$ were evaluated by Euler--Maclaurin summation with 12 Bernoulli corrections, differentiated term by term. The 96 zeros with $\gamma<230$ come from sign changes of Hardy's $Z(t)$, refined by Brent's method; their number agrees with $N(230)\approx96.06$, and the first 30 agree with tabulated values to $5\cdot10^{-10}$. For high-precision fits we used $\K(1)=1.1604648419$ from (E4). The spectrum is $\Phi(t)=\frac2{\sum_jw_j}|\sum_jR(e^{y_j})e^{-ity_j}w_j|$ with a Hann window on $2^{17}$ points $y_j\in[\log10^2,\log10^8]$. \section*{Data and code} \sloppy All numbers in this paper are produced by the Python (NumPy, SciPy, Matplotlib) scripts in the ancillary archive: \texttt{core.py}, \texttt{explicit.py}, \texttt{zeros.py}, the \texttt{step*} scripts (mean value, explicit formula, critical line, zeros of $\K$), \texttt{rigor\_meanvalue.py} and \texttt{rigor\_rouche.py} (Proposition~\ref{prop:K1}, Theorem~\ref{thm:zero}), \texttt{ctransform.py} and \texttt{laplace.py} (Section~\ref{sec:ctl}), \texttt{bridge\_M.py}, \texttt{bridge\_M2.py}, \texttt{bridge\_alpha.py}, \texttt{variogram.py}, \texttt{slope\_b.py} and \texttt{fig\_kernelspec.py} (Section~\ref{sec:margin}), and \texttt{paper\_figures.py}. \section*{Use of AI assistance} The function, the research programme and the questions are the author's. 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