\documentclass[11pt]{amsart} \usepackage[utf8]{inputenc} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage{booktabs} \usepackage{graphicx} \usepackage{xcolor} \usepackage[margin=1in]{geometry} \usepackage[colorlinks=true,linkcolor=blue!50!black,citecolor=blue!50!black,urlcolor=blue!50!black]{hyperref} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}} \newcommand{\li}{\operatorname{li}} \numberwithin{equation}{section} \title[The $6k\pm1$ race]{Cross-Energy of Dirichlet Partial Sums.\\ Part 37: The $6k\pm1$ race:\\ sieve structure, Chebyshev's bias and a GRH-equivalent inequality} \author{Pedro Caceres} \address{Universidad Europea de Valencia, Valencia, Spain} \email{pedrojesus.caceres@universidadeuropea.es} \date{November 2026 (formerly a standalone note). Part 37 of the book-series \emph{Cross-Energy of Dirichlet Partial Sums} (forty-seven parts, Parts 0--46; block VII: M\"obius cancellation and prime races). Part~0 is the guide. Version~2, 2~November 2026: Section~\ref{sec:dual}. Version~3, 2~November 2026: Section~\ref{sec:explicitC2}. Version~4, 10~November 2026: Hardy--Littlewood integrals recomputed, earlier work cited, title changed. In book version~1 (2 November 2026) this was Part~34; the present numbering is that of book version~2 (10 November 2026).} \begin{document} \begin{abstract} Every prime $p>3$ is $6k+1$ or $6k-1$. We collect what can be proved about the two resulting sequences and add three results. The races are those modulo $3$ (the prime $2$ counts for the class $2$); the race modulo $6$ differs by that one prime. (i) In the weighting $n^{-1/2}\log(N/n)$ of the Ces\`aro margin, the race between the prime powers $\equiv2$ and $\equiv1\pmod3$ is an exact identity. The $6k-1$ side leads by $2\frac{L'}L(\frac12,\chi_{-3})\log N-\frac12\psi'(\frac34)=1.1320\log N-1.2709$ plus the margin $M_\chi(N)$ of $L(s,\chi_{-3})$. The generalized Riemann hypothesis (GRH) for $L(s,\chi_{-3})$ is equivalent to $M_\chi\ge0$ for all large $N$, and under GRH the weighted race is never lost after $N=2$. This is a criterion of the type of Hardy--Littlewood and Landau (1918) for Chebyshev's assertion, here with the Ces\`aro weight and with explicit constants. (ii) We verify $0.1024\le M_\chi(N)\le0.3549$ for $2\le N\le2^{27}$, inside the window $[0,0.4536]$ that GRH predicts. We also check the formula against the first $271$ zeros of $L(s,\chi_{-3})$. (iii) For the unweighted race, the bias is $1/\sqrt{V_\chi}=2.97$ standard deviations, with $V_\chi=\sum_\rho\frac1{\rho(1-\rho)}=0.113230$ (equal to $\sum_\rho|\rho|^{-2}$ under GRH); under GRH and linear independence, a recomputation with $271$ zeros gives the density $0.99906$, in agreement with the $0.9990$ of Rubinstein and Sarnak. Along the way we prove the sieve characterization of $6k\pm1$ and of twin primes, explain why Legendre-type counting cannot give the error term, compare twin primes and Goldbach representations modulo $6$ with the Hardy--Littlewood predictions, and prove the formula for $\gcd_p(p^{2n}-1)$ (the sequence $24,240,504,480,\dots$). Version~2 adds the generating function $\frac12[\zeta_6(s)^2\pm L(s,\chi_6)^2]$ of the three forms $6xy\pm x\pm y$, and a duality theorem: averaged over the zeros of $\zeta$, the cross-energy $\mathcal C_2(N,\frac12,\gamma)$ of Part~1 returns $\log N/N$; averaged over the zeros of $L(s,\chi_{-3})$ it returns $\frac1N\sum_{d\mid N}\chi_{-3}(d)\Lambda(d)$; and M\"obius inversion separates the primes of the two lanes. Version~3 makes this exact at finite height and without RH: with a Gaussian weight on the zeros, the cross-energy satisfies an explicit formula whose pole term is $\mathcal C_2$ itself at $s=0$, whose prime term is $\log N$ exactly plus a nonnegative remainder from the near coincidences $nj\approx N$, and which returns every prime power up to $60$ from the $1517$ zeros below $2000$. Nothing here proves RH, GRH, or the twin prime or Goldbach conjectures. \end{abstract} \maketitle \section{Introduction} The integers prime to $6$ are exactly the numbers $6k\pm1$. So every prime $p>3$ lies in one of the two progressions \[ P^+=\{6k+1\},\qquad P^-=\{6k-1\}. \] In the author's notes \cite{Ca20}, these intertwined sequences are the starting point for primality tests, counting methods and several conjectures. One observation there is that $P^-$ is almost always ahead of $P^+$ \cite[Idea~7]{Ca20}. This paper puts these observations on a firm footing. It separates what is proved, what is conditional and what is open. For $p>3$, $p\equiv\pm1\pmod6$ exactly when $p\equiv\pm1\pmod3$. So the two series are distinguished by the real character $\chi=\chi_{-3}$ of conductor $3$ ($\chi(n)=0,1,-1$ for $n\equiv0,1,2\pmod3$). The race between them is governed by $L(s,\chi)$, and the Dedekind zeta function of $\mathbb Q(\sqrt{-3})$ factors as $\zeta(s)L(s,\chi)$. Our main tool is the Ces\`aro margin of Part~11 \cite{CE11}, transferred from $\zeta$ to $L(s,\chi)$. \emph{Earlier work.} Chebyshev asserted in 1853 that there are more primes $4n+3$ than $4n+1$, in the sense that $\sum_p(-1)^{(p-1)/2}e^{-p/x}\to-\infty$ as $x\to\infty$ \cite{Che}. Hardy and Littlewood proved this under GRH for the corresponding $L$-function \cite{HL18}, and Landau proved that the assertion is equivalent to that hypothesis \cite{Lan18}. Littlewood showed that the unweighted differences change sign infinitely often \cite{Li14}; Knapowski and Tur\'an began the systematic study of such races \cite{KT}, and Rubinstein and Sarnak computed their logarithmic densities under GRH and LI \cite{RuSa}. Theorems~\ref{thm:chilandau} and~\ref{thm:race} below belong to the Hardy--Littlewood--Landau line: a smoothed race whose one-sided behaviour is equivalent to GRH. What is added here is the choice of weight $n^{-1/2}\log(N/n)$, for which the race is an identity with an explicit line $1.1320\log N-1.2709$, an explicit window $[0,0.4536]$ for the margin, and a verification to $2^{27}$. The inequality is exact as an identity; as an inequality it is equivalent to GRH and is not proved. \emph{Which race.} The character $\chi_{-3}$ compares the classes $2$ and $1$ modulo $3$, and the prime $2$ belongs to the class $2$. All statements about $\pi(x;3,\cdot)$, $\psi(x,\chi)$, $C_1$, $C_2$, $M_\chi$ and the densities are for this race modulo $3$. The race modulo $6$ between $P^-$ and $P^+$ omits the prime $2$: $\pi(x;6,5)=\pi(x;3,2)-1$ and $\pi(x;6,1)=\pi(x;3,1)$ for $x\ge2$, so its lead is smaller by exactly one, and in the weighted race by the terms of the powers of $2$. The tables of counts (Table~\ref{tab:counts}) are for primes $>3$, that is, modulo $6$. \emph{Results.} Section~\ref{sec:sieve} proves the sieve characterization of $P^\pm$ and of twin primes, and shows why the error terms of Legendre-type counting are of the same order as the main term. Section~\ref{sec:bias} recalls Chebyshev's bias and computes it from the zeros of $L(s,\chi)$. Section~\ref{sec:margin} builds the margin $M_\chi$, proves that GRH for $L(s,\chi)$ is equivalent to $M_\chi\ge0$ (Theorem~\ref{thm:chilandau}), and verifies it to $2^{27}$. Section~\ref{sec:race} contains the exact form of the weighted race (Theorem~\ref{thm:race}) and two-sided bounds for each series. Section~\ref{sec:consec} treats consecutive primes and runs, and Sections~\ref{sec:twins}--\ref{sec:powers} treat twin primes, Goldbach modulo $6$ and the even powers of primes. Section~\ref{sec:dual} (new in version~2) writes the three forms as a Dirichlet series and proves that the cross-energy $\mathcal C_2$ of Part~1, averaged over the zeros of $\zeta$ or of $L(s,\chi_{-3})$, returns the primes of each lane (Theorem~\ref{thm:dual}); Section~\ref{sec:explicitC2} (new in version~3) gives the exact finite form of this duality, Proposition~\ref{prop:explicitC2}, and explains term by term how the cross-energy hears the primes through the zeros. Section~\ref{sec:status} summarizes the status. \emph{Place in the book.} This is Part~37 of the series, in block~VII (M\"obius cancellation and prime races). It builds on Part~11 (the Ces\`aro margin, transferred here from $\zeta$ to $L(s,\chi_{-3})$) and on the author's earlier notes on the primes $6k\pm1$. It is used by Part~38 (the energy of the same race), and it is cited by Parts~10 and~11. It also uses Part~10 \cite{CE10}. It is also used by Part~11 \cite{CE11}. Section~\ref{sec:dual} uses the cross-energy of Part~1 \cite{CE1}, and it is the dual of the zero detection by that cross-energy studied there; the localized Weil windows of Part~20 \cite{CE20} use the same functions $X_N$ in the other direction. \section{The sieve structure of \texorpdfstring{$6k\pm1$}{6k+-1}}\label{sec:sieve} \begin{lemma}[characterization]\label{lem:char} Let $k\ge1$ and $x,y$ range over the positive integers. \begin{enumerate} \item[(a)] $6k+1$ is composite if and only if $k=6xy+x+y$ or $k=6xy-x-y$ for some $x,y$. \item[(b)] $6k-1$ is composite if and only if $k=6xy+x-y$ for some $x,y$. \item[(c)] $(6k-1,6k+1)$ is a pair of twin primes if and only if $k$ is of none of the three forms. \end{enumerate} \end{lemma} \begin{proof} If $n=6k\pm1$ is composite, write $n=ab$ with $11$. If $n\equiv-1$, one factor is $\equiv1$ and the other $\equiv-1$, and $(6x-1)(6y+1)=6(6xy+x-y)-1$. Conversely, each form gives a nontrivial factorization. Part (c) combines (a) and (b). \end{proof} Status: elementary and complete; checked for $k\le10^5$. The three forms are the matrices $A_n,B_n,C_n$ of \cite{Ca20}, and (c) is \cite[Idea~8]{Ca20}. The ``duplicates'' counted there have an exact meaning: \begin{proposition}[multiplicities]\label{prop:mult} The number of ordered pairs $(x,y)$ with $6xy+x+y=k$ equals the number of ordered factorizations $6k+1=ab$ with $a,b>1$ and $a\equiv b\equiv1\pmod6$; similarly for the other two forms. In particular, the two forms $6xy+x+y$ and $6xy-x-y$ together represent $k$ in exactly $\tau(6k+1)-2$ ordered ways, and the form $6xy+x-y$ represents $k$ in $(\tau(6k-1)-2)/2$ ways. How the first number splits between the two forms depends on the residues of the prime factors modulo $6$, not only on how many there are. For example, $343=7^3$ has three prime factors, and $k=57$ has only the two representations $(1,8)$, $(8,1)$ by $6xy+x+y$ and none by $6xy-x-y$; and $175=5^2\cdot7$ gives $k=29$ two representations by each form, from $7\cdot25$ and from $5\cdot35$. \end{proposition} \begin{proof} The map $(x,y)\mapsto(6x+1,6y+1)$ is a bijection onto the factor pairs of the stated type. \end{proof} So a count of the composites in $P^\pm$ by ``elements minus duplicates'' is the inclusion--exclusion principle. Done exactly, it is Legendre's identity: for $z\ge5$ and $P(z)=\prod_{p\le z}p$, \begin{equation}\label{eq:legendre} \pi(N)-\pi(\sqrt N)+1=\sum_{d\mid P(\sqrt N)}\mu(d)\Big\lfloor\frac Nd\Big\rfloor , \end{equation} and the restriction to $P^\pm$ only removes the factors $2$ and $3$ from $P(z)$. The difficulty is not in the identity. It is in the error terms. \begin{proposition}[Legendre's main term is wrong by a constant factor]\label{prop:mertens} Write $\lfloor N/d\rfloor=N/d-\{N/d\}$ in \eqref{eq:legendre}. The main term is $N\prod_{p\le\sqrt N}(1-\frac1p)\sim2e^{-\gamma_E}\frac N{\log N}$, while $\pi(N)\sim\frac N{\log N}$. Hence the $2^{\pi(\sqrt N)}$ fractional parts contribute $-(2e^{-\gamma_E}-1+o(1))\frac N{\log N}$, a fixed proportion of the answer. \end{proposition} \begin{proof} Mertens' third theorem gives $\prod_{p\le z}(1-\frac1p)\sim e^{-\gamma_E}/\log z$ with $z=\sqrt N$. Compare with the prime number theorem. \end{proof} Table~\ref{tab:mertens} shows the effect. The main term exceeds $\pi(N)-\pi(\sqrt N)+1$ by $1.3\%$, $4.8\%$ and $5.7\%$ at $N=10^5$, $10^7$ and $10^8$, and the excess tends to $2e^{-\gamma_E}-1=12.3\%$, slowly because of the factor $(\log N-1)/\log N$. \begin{table}[ht] \centering \caption{Legendre's main term $N\prod_{p\le\sqrt N}(1-\frac1p)$ against the true count.}\label{tab:mertens} \begin{tabular}{rrrr} \toprule $N$ & main term & $\pi(N)-\pi(\sqrt N)+1$ & ratio\\ \midrule $10^4$ & $1\,203.2$ & $1\,205$ & $0.998$\\ $10^5$ & $9\,651.9$ & $9\,528$ & $1.013$\\ $10^6$ & $80\,965.3$ & $78\,331$ & $1.034$\\ $10^7$ & $696\,014.0$ & $664\,134$ & $1.048$\\ $10^8$ & $6\,088\,469.2$ & $5\,760\,227$ & $1.057$\\ \bottomrule \end{tabular} \end{table} This is why regressions for the duplicates, as in \cite[\S9--10]{Ca20}, fit well on finite ranges but cannot give an asymptotic error term. The fractional parts $\sum_d\mu(d)\{N/d\}$ are the M\"obius-weighted sawtooth sums that Part~11 \cite{CE11} identifies as the core of the problem. \begin{remark}[on primality tests and conjectures]\label{rem:tests} Characterizations like Lemma~\ref{lem:char} are exact, but they do not by themselves prove statements about gaps or representations, such as Legendre's, Grimm's or Goldbach's conjectures. Those require control of the error terms above. A related caution concerns the test ``$\gcd(2^n+n,3^n+n,n+1)=n+1$ iff $n+1$ is prime'' \cite[Idea~17]{Ca20}. The ``if'' direction follows from Fermat's little theorem. The converse fails for the Fermat pseudoprimes to bases $2$ and $3$, which include the Carmichael numbers prime to $6$: $n+1=1105,1729,2465,2701,2821,6601,8911,\dots$ \end{remark} \section{A smooth indicator and the near-factorization energy}\label{sec:tanh} Lemma~\ref{lem:char} can be packaged into a pair of smooth functions on the integers. Put, for $k\ge1$, \begin{equation}\label{eq:Fdef} F_+(k)=\prod_{x,y\ge1}\tanh^2(k-6xy-x-y)\,\tanh^2(k-6xy+x+y),\qquad F_-(k)=\prod_{x,y\ge1}\tanh^2(k-6xy+x-y), \end{equation} with the convention $\tanh^2(0)=0$; the infinite products converge, since $-\log\tanh^2 n\le4e^{-2n}/(1-e^{-2})$ and the number of lattice points with $6xy\pm(\cdots)=m$ is $O(\tau(6m\pm1))$, and truncating at $x,y\le k$, as one would compute them, changes nothing beyond $e^{-10k}$. \begin{proposition}[indicator and energy]\label{prop:tanh} \begin{enumerate} \item $F_+(k)>0$ if and only if $6k+1$ is prime, and $F_-(k)>0$ if and only if $6k-1$ is prime. \item Exactly, with $w(d)=-2\log\tanh|d|$, \begin{equation}\label{eq:energyF} -\log F_\pm(k)=\sum_{m\ge1,\,m\ne k}r_\pm(m)\,w(k-m),\quad r_+(m)=\tau(6m+1)-2,\quad r_-(m)=\frac{\tau(6m-1)-2}2. \end{equation} \end{enumerate} \end{proposition} \begin{proof} (1) is Lemma~\ref{lem:char}: a factor vanishes exactly at a representation. (2) By Proposition~\ref{prop:mult}, the number of pairs $(x,y)$ with $6xy+x+y=m$ or $6xy-x-y=m$ is the number of ordered factorizations $6m+1=ab$, $a,b>1$, which is $\tau(6m+1)-2$, both factors being automatically in matching classes mod $6$. For $6m-1=ab$ the classes are forced to be opposite, $a\equiv1$, $b\equiv-1$ or conversely, and the single form $6xy-x+y=m$ picks one orientation from each unordered pair, giving $(\tau(6m-1)-2)/2$ representations ($6m-1$ is never a square). Group the factors of \eqref{eq:Fdef} by the value $m$ and take logarithms. \end{proof} Since $w(d)=4e^{-2|d|}(1+O(e^{-4|d|}))$, identity \eqref{eq:energyF} says what $F$ is: an \emph{exponentially localized count of the factorizations of the neighbours}. $F(k)$ is not a useful primality test -- it is Lemma~\ref{lem:char} in disguise, at cost $k^2$ against $\sqrt{6k}$ for trial division, and the smoothness of $\tanh$ gives no analytic leverage, the arguments being integers. Its interest is the value it assigns to each \emph{prime}: a measure of how crowded the prime is by near-composites of its own progression. The nearest misses dominate: $F_+(3)=0.5799$ against $\tanh^2(1)=0.5800$, the single neighbour $k=4$ ($25=5^2$) accounting for all of it. Numerically, with \eqref{eq:energyF} checked against the defining product to $3\cdot10^{-10}$ and the divisor function sieved to $6\cdot10^6$ (\texttt{tanh\_f.py}): over the primes with $k\le10^6$, the mean of $-\log F_+$ grows like $0.47\log k$, and the mean of $-\log F_-$ like half of that -- the factor $2$ of \eqref{eq:energyF}, one representation per unordered factor pair instead of one per ordered pair, not any asymmetry of the progressions: indeed $\sum_{m\le10^6}\tau(6m+1)=6\,082\,216$ against $\sum_{m\le10^6}\tau(6m-1)=6\,082\,200$, equal to $2.6\cdot10^{-6}$. The divisor mass is blind, at this order, to the Chebyshev bias of Section~\ref{sec:bias}, which lives at the scale $\sqrt x/\log x$ of Proposition~\ref{prop:V}. The extremes of $F$ over the primes have arithmetic meaning (Figure~\ref{fig:tanh}(c)). The maximum on $100\le k\le10^6$ is $F_+=0.8246$ at $k=443\,567$, $p=2\,661\,403$: there $p-12$, $p-6$, $p$ and $p+6$ are all prime, a run in the progression, and the nearest factorization is three steps away. The minimum is $F_+=1.1\cdot10^{-20}$ at $k=846\,678$, $p=5\,080\,069$: its neighbour $6(k+1)+1=5\,080\,075=5^2\cdot7^2\cdot11\cdot13\cdot29$ has $\tau=72$, and the $70$ nontrivial ordered factorizations at distance $1$ each contribute $w(1)=0.5447$. So $\max F$ marks prime clusters and $\min F$ marks primes next to a highly divisible member of the progression: a graded structure on top of Lemma~\ref{lem:char}, with the hyperbola picture of Figure~\ref{fig:tanh}(b) classifying the points $e^{2\theta}=6k\pm1$ of $X^2-Y^2=1$ by it. Two further remarks. \emph{A real variable.} Replacing the lattice product \eqref{eq:Fdef} by \[ G_\pm(x)=\prod_{m\ge1}\tanh^{2r_\pm(m)}(x-m),\qquad x\in\mathbb R, \] gives smooth functions with the same values at the integers up to the truncation noted above, positive except for a zero of multiplicity $2r_\pm(m)$ at each composite index $m$; the exponents are even because $6m-1$ is never a square and $r_+$ counts ordered pairs, so no sign ambiguity arises. This is a clean interpolation, but it carries exactly the information of $\tau$ on the progressions and no more. \emph{Energy, gaps and twins.} Over the $205\,960$ primes $6k+1$ with $10^33$).}\label{tab:counts} \begin{tabular}{rrrr} \toprule $x$ & $\pi(x;6,1)$ & $\pi(x;6,5)$ & difference\\ \midrule $10^3$ & $80$ & $86$ & $6$\\ $10^4$ & $611$ & $616$ & $5$\\ $10^5$ & $4\,784$ & $4\,806$ & $22$\\ $10^6$ & $39\,231$ & $39\,265$ & $34$\\ $10^7$ & $332\,194$ & $332\,383$ & $189$\\ $10^8$ & $2\,880\,517$ & $2\,880\,936$ & $419$\\ \bottomrule \end{tabular} \end{table} The difference in Table~\ref{tab:counts} is positive throughout. It is not a fluctuation of the counts: it has a cause, the prime squares. Let $\psi(x,\chi)=\sum_{n\le x}\chi(n)\Lambda(n)$ and $\theta(x,\chi)=\sum_{p\le x}\chi(p)\log p=\theta(x;3,1)-\theta(x;3,2)$. \begin{proposition}[the explicit formula for the race]\label{prop:explicit} Since $\chi$ is primitive and odd, \[ \psi(x,\chi)=-\sum_\rho\frac{x^\rho}\rho-\frac{L'}L(0,\chi)+\sum_{m\ge1}\frac{x^{1-2m}}{2m-1} \] (sum over the nontrivial zeros, symmetrically), and \[ \frac{\theta(x;3,2)-\theta(x;3,1)}{\sqrt x}=1-\frac{\psi(x,\chi)}{\sqrt x}+o(1). \] \end{proposition} \begin{proof} The first formula is \cite[Ch.~19]{Da}. For the second, $\psi(x,\chi)-\theta(x,\chi)=\sum_{p^2\le x}\chi(p^2)\log p+O(x^{1/3}\log x)$, and $\chi(p^2)=1$ for $p\ne3$. So the difference is $\theta(\sqrt x)+O(x^{1/3}\log x)=\sqrt x(1+o(1))$ by the prime number theorem. \end{proof} Every square of a prime $p>3$ is $\equiv1\pmod 3$. In the normalized race, the squares therefore shift the $6k+1$ side down by exactly $1$. Under GRH, the remaining term $-\psi(x,\chi)/\sqrt x=\sum_\rho x^{i\gamma}/\rho+o(1)$ is a mean-zero almost periodic function in $\log x$ \cite{RuSa}. \begin{proposition}[the size of the fluctuation]\label{prop:V} Unconditionally, \[ V_\chi:=\sum_\rho\frac1{\rho(1-\rho)}=2\frac{\xi_\chi'}{\xi_\chi}(1)=\log\frac3\pi-\gamma_E+2\frac{L'}L(1,\chi)=0.113230 , \] where $\xi_\chi(s)=(3/\pi)^{(s+1)/2}\Gamma(\frac{s+1}2)L(s,\chi)$. Under GRH, $V_\chi=\sum_\rho|\rho|^{-2}$, and for simple zeros it is the mean square of $\psi(x,\chi)/\sqrt x$ in $\log x$. \end{proposition} \begin{proof} $\xi_\chi$ is entire of order $1$ with $\xi_\chi(s)=\xi_\chi(1-s)$. Its zeros are symmetric under $\rho\mapsto\bar\rho$ and $\rho\mapsto1-\rho$, so Hadamard's product gives $\frac{\xi_\chi'}{\xi_\chi}(s)=\sum_\rho\frac1{s-\rho}$, summed symmetrically. Since $\frac1{\rho(1-\rho)}=\frac1\rho+\frac1{1-\rho}$ and the zeros are symmetric under $\rho\mapsto1-\rho$, $V_\chi=2\sum_\rho\frac1{1-\rho}=2\frac{\xi_\chi'}{\xi_\chi}(1)$. On the line, $1-\rho=\bar\rho$, so $\frac1{\rho(1-\rho)}=\frac1{|\rho|^2}$. The value uses $\frac{L'}L(1,\chi)=0.368282$, computed from the Hurwitz zeta function. The mean square follows from Parseval in the Besicovitch space $B^2$. \end{proof} The $271$ zeros below height $400$ give $\sum1/|\rho|^2=0.108257$. Since the mean is $1$ and the spread is $\sqrt{V_\chi}=0.3365$, the bias is $2.97$ standard deviations. \begin{theorem}[Rubinstein--Sarnak \cite{RuSa}]\label{thm:rusa} Assume GRH and the linear independence (LI) of the ordinates of $L(s,\chi)$. Then the set of $x$ with $\pi(x;3,2)>\pi(x;3,1)$ has logarithmic density $\delta(3;2,1)=0.9990\ldots$ \end{theorem} Under GRH and LI, $\psi(x,\chi)/\sqrt x$ has the limiting distribution of $X=\sum_{\gamma>0}\frac{2}{|\rho|}\cos\vartheta_\gamma$ with independent uniform phases. So the density is $P(X<1)=\frac12+\frac1\pi\int_0^\infty\frac{\sin t}t\prod_{\gamma>0}J_0\big(\frac{2t}{|\rho|}\big)\,dt$. With the $271$ zeros and a Gaussian factor for the tail of $V_\chi$, we obtain $0.999063$, in agreement with the value $0.9990$ printed in \cite{RuSa}. On $10^4\le x\le2^{27}$, sampled uniformly in $\log x$, the normalized race has mean $0.967$ and variance $0.1124$, against $1$ and $V_\chi=0.1132$. Its histogram follows the limiting density (Figure~\ref{fig:dist}(a)). On the computed range, $\pi(x;3,2)>\pi(x;3,1)$ for all $2\le x\le2^{27}$. The first reversal of this race modulo $3$ is at $x=608\,981\,813\,029$ \cite{BH}. For the race modulo $6$ the lead is one less: $\pi(x;6,5)\ge\pi(x;6,1)$ on the whole range; for $x\ge5$ equality holds exactly from each of the primes $7,13,19,37,43,79,163,223,229$ up to the next prime, and the inequality is strict for $233\le x\le2^{27}$. Its first reversal is strictly earlier than the value of \cite{BH}, because a tie of the race modulo $3$ precedes its first reversal. The $\theta$-race fails on $0.13\%$ of $[10^4,2^{27}]$ in logarithmic measure (Figure~\ref{fig:races}). For prime races in general see \cite{GM}. \begin{figure}[ht] \centering \includegraphics[width=\textwidth]{fig/F1_races.png} \caption{The race between $P^-$ and $P^+$ up to $2^{27}$. (a) Raw counts. (b) The normalized $\theta$-race of Proposition~\ref{prop:explicit}: mean $1$ from the prime squares, band $1\pm\sqrt{V_\chi}$.} \label{fig:races} \end{figure} \section{The Ces\`aro margin of \texorpdfstring{$L(s,\chi_{-3})$}{L(s,chi-3)}}\label{sec:margin} For $\zeta$, Part~11 \cite{CE11} uses Suzuki's criterion \cite{Su1}: with\[C(N)=2\sum_{n1$, unconditionally, \[ M_\chi(N)=\sum_\rho\frac{2\,(1-N^{it_\rho})}{t_\rho^2}=\sum_\rho F_{\log N}(t_\rho),\qquad F_L(z)=\frac{2-e^{iLz}-e^{-iLz}}{z^2}, \] the sum over the nontrivial zeros with multiplicity, absolutely convergent. Under GRH, \[ M_\chi(N)=4\sum_{\gamma>0}\frac{1-\cos(\gamma\log N)}{\gamma^2}\in\bigl[0,\,4(\log\xi_\chi)''(\tfrac12)\bigr], \] where $(\log\xi_\chi)''(\tfrac12)=\tfrac14\psi'(\tfrac34)+(\tfrac{L'}L)'(\tfrac12,\chi)=0.1134022$. \end{proposition} \begin{proof} For $c>\frac12$, Perron's formula with the kernel $N^s/s^2$ gives $C_\chi(N)=\frac1{2\pi i}\int_{(c)}2(-\frac{L'}L)(s+\frac12,\chi)N^s\frac{ds}{s^2}$. Move the contour to the left; $\frac{L'}L$ grows at most like $\log^2|t|$ on suitable horizontal segments, and the factor $s^{-2}$ makes the integrals converge. The residues are as follows. The double pole at $s=0$ gives $-2\frac{L'}L(\frac12)\log N-2(\frac{L'}L)'(\frac12)$. A zero $\rho$ gives $-2N^{\rho-1/2}/(\rho-\frac12)^2=2N^{it_\rho}/t_\rho^2$. The trivial zero $-(2m-1)$ gives $-2N^{-(2m-1/2)}/(2m-\frac12)^2$. There is no pole term. By Hadamard's product and the symmetry of $\xi_\chi$, $\sum_\rho t_\rho^{-2}=(\log\xi_\chi)''(\frac12)=\frac14\psi'(\frac34)+(\frac{L'}L)'(\frac12)$. Adding $\sum_\rho2(1-N^{it_\rho})/t_\rho^2$ to $C_\chi$ therefore gives $\frac12\psi'(\frac34)-2\frac{L'}L(\frac12)\log N-2\sum_m N^{-(2m-1/2)}/(2m-\frac12)^2=R_\chi(N)$. As a check, both sides vanish at $N=1$. Under GRH, pair $\pm\gamma$ and use $0\le1-\cos\le2$. \end{proof} \begin{theorem}[GRH for $\chi_{-3}$ in one variable]\label{thm:chilandau} The following are equivalent: \begin{enumerate} \item[(1)] GRH for $L(s,\chi_{-3})$; \item[(2)] $M_\chi(N)\ge0$ for all large $N$; \item[(3)] $M_\chi(N)\le B$ for some $B$ and all large $N$; \item[(4)] for every $N>1$, $R_\chi(N)-0.45361\le C_\chi(N)\le R_\chi(N)$. \end{enumerate} \end{theorem} \begin{proof} (1)$\Rightarrow$(4) is Proposition~\ref{prop:chimargin}, and (4)$\Rightarrow$(2),(3) is immediate. For (2)$\Rightarrow$(1): for $\Re s>\frac12$, Fubini gives \[ \int_0^\infty M_\chi(e^L)e^{-sL}dL=\sum_\rho\frac{2}{s(s^2+t_\rho^2)}=:\mathcal G_\chi(s), \] meromorphic with poles at $0$ and at $\pm it_\rho$. A zero $\frac12+\delta+i\gamma_0$ with $\delta>0$ gives a pole at $\delta+i\gamma_0$, with nonzero residue (only $\pm\gamma$ meet at a given pole). It lies off the real axis because $L(\sigma,\chi_{-3})\neq0$ for $0<\sigma<1$; indeed, for $\sigma>0$ the series converges (its coefficients have bounded partial sums), its value is the limit of the partial sums up to $3k+2$, and so, pairing consecutive terms, $L(\sigma,\chi_{-3})=\sum_{k\ge0}\big[(3k+1)^{-\sigma}-(3k+2)^{-\sigma}\big]>0$ (the values at the ends are $L(0)=\frac13$ and $L(1)=\frac\pi{3\sqrt3}$). So $\mathcal G_\chi$ is analytic on $(0,\infty)$. By Landau's theorem \cite{Wi}, the Laplace integral of a function that is eventually $\ge0$ has a singularity at the real point of its abscissa of convergence. Hence the abscissa is $\le0$, and there is no pole with $\Re s>0$. For (3)$\Rightarrow$(1), apply the same argument to $B-M_\chi$. \end{proof} \emph{Why this margin is well conditioned.} For $\zeta$, the two sides $R$ and $C$ are of size $8\sqrt N$ and must stay ordered to $O(1)$. For $\chi$ there is no pole, and $R_\chi,C_\chi=O(\log N)$. \begin{lemma}[concavity]\label{lem:concave} In $L=\log N$, $M_\chi$ is continuous, smooth and strictly concave between consecutive prime powers. At a prime power $q$ its slope jumps by $-2\chi(q)\Lambda(q)/\sqrt q$. So the local minima of $M_\chi$ lie at the prime powers with $\chi(q)=-1$, namely $p\equiv2\pmod3$ and its odd powers. \end{lemma} \begin{proof} Between prime powers, $C_\chi$ is linear in $L$, and $R_\chi''(L)=-2\sum_me^{-(2m-1/2)L}<0$. The jump is the derivative of $2\chi(q)\Lambda(q)q^{-1/2}(L-\log q)_+$. \end{proof} \begin{proposition}[verification]\label{prop:chiverified} For $2\le N\le2^{27}$, \[ 0.10242\le M_\chi(N)\le0.35486 , \] inside the window $[0,0.45361]$ of Theorem~\ref{thm:chilandau}, and $M_\chi(N)=R_\chi(N)>0$ for $10}m_\gamma^2/\gamma^2\ge(\log\xi_\chi)''(\frac12)$. Unconditionally, GRH holds if and only if $\int_1^X\alpha_\chi(x)^2\frac{dx}x=X^{o(1)}$. \end{proposition} \begin{proof} The Laplace transform of $\alpha_\chi(e^u)$ is $\frac1s\bigl(\frac{L'}L(\frac12,\chi)-\frac{L'}L(s+\frac12,\chi)\bigr)$. It is regular at $s=0$, so the mean is $0$ by the Abelian theorem, and its residue at $i\gamma$ is $-\frac{m_\gamma}{i\gamma}$. Membership in $B^2$ follows from the explicit formula, as for $\zeta$ \cite{RuSa,ANS}. The variogram identity is Parseval, and the comparison uses $m^2\ge m$. For the energy criterion, if the energy is $X^{o(1)}$, Cauchy--Schwarz shows that the Laplace integral converges absolutely in $\Re s>0$, so $\frac{L'}L(w,\chi)$ is holomorphic in $\Re w>\frac12$, exactly as for $\zeta$ in \cite{CE11}. \end{proof} The block energies of $\alpha_\chi$ on $[2^k,2^{k+1})$, $12\le k\le26$, have mean $0.110$, against $0.1134$. The amplitude spectrum of $\alpha_\chi$ has its peaks at the zeros of $L(s,\chi)$ with heights $2/\gamma$, and nothing at the zeros of $\zeta$ (Figure~\ref{fig:lfun}). \begin{figure}[ht] \centering \includegraphics[width=\textwidth]{fig/F2_lfunction.png} \caption{(a) $Z_\chi(t)$ and the zeros of $L(s,\chi_{-3})$ below $52$. (b) The amplitude spectrum of $\alpha_\chi(e^u)$ on $2^8\le x\le2^{27}$: peaks at the ordinates of $L(s,\chi_{-3})$ with the predicted heights $2/\gamma$; the ordinates of $\zeta$ (grey ticks) are absent.} \label{fig:lfun} \end{figure} \begin{figure}[ht] \centering \includegraphics[width=\textwidth]{fig/F3_margin.png} \caption{(a) $M_\chi$ at every 50th prime power up to $2^{27}$, inside the window $[0,4(\log\xi_\chi)''(\frac12)]$, around its mean $2(\log\xi_\chi)''(\frac12)$. (b) $M_\chi$ from the primes and from the $271$ zeros, $16\le\log N\le17$.} \label{fig:margin} \end{figure} \section{The weighted race}\label{sec:race} For $a=1,2$ let \[ C_a(N)=2\sum_{\substack{n3$ these are the series $P^+$ and $P^-$. Since $\chi=+1$ on the first class and $-1$ on the second, $C_2-C_1=-C_\chi$. \begin{theorem}[the weighted $6k\pm1$ race]\label{thm:race} For every $N>1$, unconditionally, \begin{equation}\label{eq:race} C_2(N)-C_1(N)=2\frac{L'}L(\tfrac12,\chi_{-3})\log N-\tfrac12\psi'(\tfrac34)+2\sum_{m\ge1}\frac{N^{-(2m-\frac12)}}{(2m-\frac12)^2}+M_\chi(N). \end{equation} Hence GRH for $L(s,\chi_{-3})$ holds if and only if, for all large $N$, \[ C_2(N)-C_1(N)\ \ge\ 2\frac{L'}L(\tfrac12,\chi_{-3})\,\log N-\tfrac12\psi'(\tfrac34), \] where $2\frac{L'}L(\frac12,\chi_{-3})=1.1319784770\ldots$ and $\frac12\psi'(\frac34)=1.2709398\ldots$ The equivalence is for the exact constants: a line with a rounded slope differs from this one by an unbounded multiple of $\log N$. Under GRH, $C_2(N)>C_1(N)$ for every $N>2$, and the lead exceeds the line by at most $0.45361$ plus the trivial-zero terms. Unconditionally, $C_2(N)>C_1(N)$ for $20$ and $T=O(N^{-3/2})$. The inequality of the statement says $M_\chi+T\ge0$ for all large $N$. GRH gives $M_\chi\ge0$ (Theorem~\ref{thm:chilandau}), hence the inequality. Conversely, $\int_0^\infty T(e^L)e^{-sL}dL=2\sum_{m\ge1}(2m-\frac12)^{-2}(s+2m-\frac12)^{-1}$ is analytic in $\Re s>-\frac32$, so $M_\chi+T$ has the same Laplace singularities in $\Re s>0$ as $M_\chi$, and the Landau argument in the proof of Theorem~\ref{thm:chilandau}, (2)$\Rightarrow$(1), applies to it without change. Under GRH, $M_\chi\ge0$, and the lead is positive once $1.1319785\log N>1.2709398$, that is, for $N>3.0733$. For $20$. The unconditional range is Proposition~\ref{prop:chiverified}. \end{proof} This is the precise form of the observation of \cite[Idea~7]{Ca20}, for the weighted counts. The $6k-1$ prime powers lead at the constant rate $2\frac{L'}L(\frac12,\chi_{-3})=1.132$ per unit of $\log N$. The deviation from the line is $M_\chi$, which GRH confines to a window of width $0.454$ (Figure~\ref{fig:race}). The weight is what makes the bias total. It is the Fej\'er smoothing of Part~11 \cite{CE11}, which turns every zero into a nonnegative term. For the raw counts there is no such smoothing, and the bias holds only $99.9\%$ of the time (Theorem~\ref{thm:rusa}). \begin{figure}[ht] \centering \includegraphics[width=\textwidth]{fig/F4_race.png} \caption{The weighted race of Theorem~\ref{thm:race} at every 50th prime power up to $2^{27}$. (a) $C_2-C_1$ and the line $1.132\log N-1.271$; the shaded band is the window allowed by GRH. (b) The deviation from the line, which is $M_\chi$ up to the trivial-zero terms.} \label{fig:race} \end{figure} \emph{Both series separately.} Adding the margins of $\zeta$ and of $\chi$ gives the margin of the Dedekind zeta function of $\mathbb Q(\sqrt{-3})$. Let $C_{3^\bullet}(N)$ be the contribution of the powers of $3$ to $C$. Then $C+C_\chi=2C_1+C_{3^\bullet}$ and $C-C_\chi=2C_2+C_{3^\bullet}$. \begin{corollary}[a sandwich for each series]\label{cor:sandwich} With\[R(N)=8(\sqrt N-1)-2\frac{\zeta'}\zeta(\tfrac12)\log N+2\sum_{m\ge1}\frac{1-N^{-(2m+1/2)}}{(2m+\frac12)^2}\]as in \cite{CE11}, for every $N>1$, \[ C_1=\tfrac12\bigl(R+R_\chi-C_{3^\bullet}\bigr)-\tfrac12(M+M_\chi),\qquad C_2=\tfrac12\bigl(R-R_\chi-C_{3^\bullet}\bigr)-\tfrac12(M-M_\chi). \] Under RH and GRH for $\chi_{-3}$, $C_1$ lies in a window of width $\frac12(0.18484+0.45361)=0.31923$ below $\frac12(R+R_\chi-C_{3^\bullet})$, and $C_2-\frac12(R-R_\chi-C_{3^\bullet})\in[-0.09242,\,0.22681]$. \end{corollary} \begin{proof} Linear algebra on the two identities, with $M\in[0,0.18484]$ \cite{Su1,CE11} and $M_\chi\in[0,0.45361]$. \end{proof} \section{Consecutive primes and runs}\label{sec:consec} The bias of Section~\ref{sec:bias} concerns totals. A different and stronger effect concerns neighbours. Let $p_n3$, and count the transitions between the series. \begin{table}[ht] \centering \caption{Transitions between consecutive primes $p_n\to p_{n+1}$ ($p_{n+1}\le x$), by series ($1=P^+$, $5=P^-$).}\label{tab:trans} \begin{tabular}{rrrrrr} \toprule $x$ & $1\to1$ & $1\to5$ & $5\to1$ & $5\to5$ & same/different\\ \midrule $10^4$ & $233$ & $377$ & $378$ & $238$ & $0.624$\\ $10^6$ & $16\,394$ & $22\,837$ & $22\,837$ & $16\,427$ & $0.719$\\ $10^8$ & $1\,264\,047$ & $1\,616\,470$ & $1\,616\,470$ & $1\,264\,465$ & $0.782$\\ $2^{27}$ & $1\,671\,812$ & $2\,129\,608$ & $2\,129\,608$ & $1\,672\,522$ & $0.785$\\ \bottomrule \end{tabular} \end{table} A prime is followed by one in the same series only $44.0\%$ of the time below $2^{27}$, not $50\%$ (Figure~\ref{fig:dist}(b)). Lemke Oliver and Soundararajan \cite{LOS} discovered this phenomenon for general moduli and explained it with the Hardy--Littlewood $k$-tuple conjecture. In their model the deviation from independence decays only like $\log\log x/\log x$, which matches the slow drift towards $\frac12$ in the figure. The small excess of $5\to5$ over $1\to1$ is the bias of Section~\ref{sec:bias} again. Runs behave as the transitions suggest. Below $2^{27}$ the longest strings of consecutive primes in one series have length $15$ in $P^-$ and $16$ in $P^+$, and there are $171\,387$ and $171\,605$ maximal strings of length $\ge4$. The sequences A296011 and A296055 of \cite{Ca20} list the starting generators $k$ of four consecutive primes $6k-1,\dots,6k+17$ in $P^-$ (resp.\ $6k+1,\dots,6k+19$ in $P^+$) with no prime of the other series between them; our computation reproduces $42,897,1052,2107,2242,2457,2632,2912,\dots$ \begin{theorem}[Shiu \cite{Shiu}]\label{thm:shiu} For every $m$, each of $P^+$ and $P^-$ contains $m$ consecutive primes. \end{theorem} Shiu's theorem holds for every residue class $a\bmod q$ with $\gcd(a,q)=1$, and it is unconditional. Its proof uses the Maier matrix method, not the sieve structure of Lemma~\ref{lem:char}. It does not settle A296011 and A296055. Those sequences require the four primes to sit at consecutive generators, that is, four consecutive primes in arithmetic progression with difference $6$. Their infinitude follows from the Hardy--Littlewood prime $k$-tuples conjecture and is open. \begin{figure}[ht] \centering \includegraphics[width=\textwidth]{fig/F7_distribution.png} \caption{(a) Distribution of the normalized $\theta$-race over $10^4\le x\le2^{27}$ (uniform in $\log x$) and the limiting density of Theorem~\ref{thm:rusa}, computed from the $271$ zeros by Fourier inversion of $\prod J_0(2t/|\rho|)$. (b) Fraction of consecutive pairs of primes in the same series.} \label{fig:dist} \end{figure} \section{Twin primes}\label{sec:twins} By Lemma~\ref{lem:char}(c), every twin pair other than $(3,5)$ is $(6k-1,6k+1)$, with $k$ a twin generator (OEIS A002822). Twin centres $6k$ form A014574, and the representations of a twin centre as a sum of two smaller ones form A305825 \cite{Ca20}. \begin{conjecture}[Hardy--Littlewood \cite{HL}]\label{conj:hl} $\pi_2(x)\sim2C_2\int_2^x\frac{dt}{\log^2t}$, with $C_2=\prod_{p>2}\big(1-\frac1{(p-1)^2}\big)=0.6601618158\ldots$ \end{conjecture} \begin{table}[ht] \centering \caption{Twin prime pairs up to $x$ against Conjecture~\ref{conj:hl}. The integral is computed by Gauss--Legendre quadrature in $\log t$; at $10^8$ and $2^{27}$ the count differs from it by $56$ and by $3$.}\label{tab:twins} \begin{tabular}{rrrr} \toprule $x$ & $\pi_2(x)$ & $2C_2\int_2^x dt/\log^2t$ & ratio\\ \midrule $10^4$ & $205$ & $214.2$ & $0.957$\\ $10^5$ & $1\,224$ & $1\,248.7$ & $0.980$\\ $10^6$ & $8\,169$ & $8\,248.0$ & $0.990$\\ $10^7$ & $58\,980$ & $58\,753.8$ & $1.004$\\ $10^8$ & $440\,312$ & $440\,367.8$ & $1.000$\\ $2^{27}$ & $571\,313$ & $571\,316.0$ & $1.000$\\ \bottomrule \end{tabular} \end{table} Brun proved that $\sum(\frac1p+\frac1{p+2})$ over twin primes converges \cite{Br}. Its value $B_2\approx1.90216$ is an extrapolation; rigorous bounds are much wider \cite{PTb}. Our partial sum to $2^{27}$ is $1.7611$. Whether there are infinitely many twin primes, that is, infinitely many $k$ avoiding the three forms of Lemma~\ref{lem:char}, is open. \begin{figure}[ht] \centering \includegraphics[width=\textwidth]{fig/F5_twins.png} \caption{Twin primes $(6k-1,6k+1)$ up to $2^{27}$ against the Hardy--Littlewood prediction.} \label{fig:twins} \end{figure} \section{Goldbach modulo 6}\label{sec:goldbach} \begin{lemma}[residue classes]\label{lem:gold} Let $N$ be even and $N=p_1+p_2$ with primes $p_1,p_2>3$. If $N\equiv0\pmod6$, one prime is in $P^+$ and the other in $P^-$. If $N\equiv2$, both are in $P^+$. If $N\equiv4$, both are in $P^-$. Correspondingly, $N/6$, $(N-2)/6$ or $(N+2)/6$ is a sum of two generators $k_1+k_2$ \cite[Idea~30]{Ca20}. \end{lemma} \begin{proof} $(\pm1)+(\pm1)\pmod6$. \end{proof} Let $r(N)$ be the number of ordered representations $N=p_1+p_2$. The observation behind A322921 \cite{Ca20} is that among three consecutive even numbers, the multiple of $6$ usually has the most representations. The Hardy--Littlewood singular series explains it. \begin{conjecture}[Hardy--Littlewood \cite{HL}]\label{conj:gold} For even $N$, \[ r(N)\sim2C_2\prod_{\substack{p\mid N\\p>2}}\frac{p-1}{p-2}\int_2^{N-2}\frac{dt}{\log t\,\log(N-t)} . \] \end{conjecture} The factor for $p=3$ is $2$. So a multiple of $6$ expects twice as many representations as a neighbour with the same other prime factors. On average over $N\le2\cdot10^5$, $r(N)\log^2N/N$ is $3.63$ for $N\equiv0$, and $1.80$ and $1.83$ for $N\equiv2,4\pmod6$. The ratios are $2:1:1$, not the factors $e:1:\sqrt2$ suggested in \cite[Figs.~26--28]{Ca20}. Divided by the prediction, the three classes give distributions of the same shape with slightly different centres: for $N>2\cdot10^4$ the means are $0.991$, $0.984$ and $0.998$ for $N\equiv0,2,4\pmod6$, with standard deviations $0.013$, $0.020$ and $0.019$ (Figure~\ref{fig:goldbach}). The offsets are real, and they are Chebyshev's bias again. The Hardy--Littlewood main term is natural for prime powers $p^k$ counted with weight $1/k$, and the squares of primes are all $\equiv1\pmod3$: a representation $N=p^2+q$ with $q>3$ prime exists only for $N\equiv2$ or $0\pmod6$, never for $N\equiv4$. Counting primes only therefore leaves the class $4$ almost untouched and removes a part of order $N^{-1/2}$ from the classes $2$ and $0$ (twice as much from the class $2$): the mean for $N\equiv2$ is $0.976$, $0.985$, $0.988$ on $(2\cdot10^4,6\cdot10^4]$, $(6\cdot10^4,1.2\cdot10^5]$, $(1.2\cdot10^5,2\cdot10^5]$. When prime powers are counted with weight $1/k$, the three means become $0.9987$, $0.9999$ and $1.0002$ (\texttt{goldbach\_powers\_10nov.py}). That the multiple of $6$ has the most representations among $6m,6m+2,6m+4$ holds in $33\,322$ of the $33\,325$ triples with $48\le6m<2\cdot10^5$, but not always. In the sliding window $(80078,80080,80082)$ (the aligned triple is $80076,80078,80080$), $r=1005,2012,2010$, and $80080\equiv4\pmod6$ wins. This is because $80080=2^4\cdot5\cdot7\cdot11\cdot13$ has singular series $1.94$, against $2.00$ for $80082=2\cdot3^3\cdot1483$. There are three such events below $2\cdot10^5$, near $80080$, $170170$ and $190190$ ($170170=2\cdot5\cdot7\cdot11\cdot13\cdot17$ and $190190=2\cdot5\cdot7\cdot11\cdot13\cdot19$); they make seven of the sliding windows of three consecutive even numbers exceptional. Goldbach's conjecture itself is verified to $4\cdot10^{18}$ \cite{OHP} and is open. \begin{figure}[ht] \centering \includegraphics[width=\textwidth]{fig/F6_goldbach.png} \caption{Goldbach representations for even $N\le2\cdot10^5$. (a) $r(N)\log^2N/N$ by $N\bmod6$: the multiples of $6$ form a band at twice the height. (b) $r(N)$ divided by the Hardy--Littlewood prediction, by class.} \label{fig:goldbach} \end{figure} \section{Even powers of primes}\label{sec:powers} Every prime $p\ge5$ satisfies $p^2\equiv1\pmod{24}$. The table of \cite[Idea~32]{Ca20} lists the largest such modulus for each even exponent: $24,240,504,480,264,65520,24,16320,\dots$ \begin{theorem}\label{thm:gcd} For $n\ge1$, let $g_n=\gcd\{p^{2n}-1:\ p\ \text{prime},\ p>2n+1\}$. Then \[ g_n=2^{v_2(2n)+2}\prod_{\substack{\ell\ \text{odd prime}\\(\ell-1)\mid2n}}\ell^{\,v_\ell(2n)+1}. \] \end{theorem} \begin{proof} Fix a prime power $\ell^k$. By Dirichlet's theorem, the primes $p>2n+1$ meet every residue class mod $\ell^k$ prime to $\ell$. So $\ell^k\mid g_n$ if and only if $a^{2n}\equiv1\pmod{\ell^k}$ for every unit $a$, that is, if and only if the exponent of $(\mathbb Z/\ell^k)^\times$ divides $2n$. This requires $\ell\neq p$ for all the $p$ involved; that holds whenever the condition below is met, since then $\ell\le2n+1$. For odd $\ell$ the group is cyclic of order $\ell^{k-1}(\ell-1)$, so the condition is $(\ell-1)\mid2n$ and $k-1\le v_\ell(2n)$. If $(\ell-1)\nmid2n$, not even $\ell\mid g_n$. For $\ell=2$ and $k\ge3$, $(\mathbb Z/2^k)^\times\cong C_2\times C_{2^{k-2}}$ has exponent $2^{k-2}$, so the condition is $k\le v_2(2n)+2$; for $k\le2$ it always holds. \end{proof} \begin{table}[ht] \centering \caption{$g_n$ of Theorem~\ref{thm:gcd} for $n\le12$. The direct gcd over the first $400$ admissible primes, the formula, and the denominator of $B_{2n}/(4n)$ agree in every case.}\label{tab:gcd} \begin{tabular}{rrrrrrrrrrrr} \toprule $n=1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$ & $12$\\ \midrule $24$ & $240$ & $504$ & $480$ & $264$ & $65520$ & $24$ & $16320$ & $28728$ & $13200$ & $552$ & $131040$\\ \bottomrule \end{tabular} \end{table} The formula is the denominator of $B_{2n}/(4n)$, where $B_{2n}$ are the Bernoulli numbers. This is a classical consequence of the von Staudt--Clausen theorem and its refinement for $B_{2n}/2n$ \cite{IR,Ad}. Equivalently, it is the denominator of $\frac12\zeta(1-2n)=-B_{2n}/(4n)$, the order of the image of the $J$-homomorphism in Adams' work \cite{Ad}. The case $n=1$ is $p^2\equiv1\pmod{24}$, and $n=6$ gives $p^{12}\equiv1\pmod{65520}$ for $p>13$. \section{The cross-energy as a two-way detector: the zeros return the two lanes}\label{sec:dual} This section is new in version~2 (2~November 2026). It connects the sieve structure of Section~\ref{sec:sieve} with the zeros of $\zeta$ and of $L(s,\chi_{-3})$ through the cross-energy $\mathcal C_2$ of Part~1 \cite{CE1}. Part~1 uses $\mathcal C_2$ to detect zeros by running over the integers. Here the roles are reversed: averaged over the zeros, $\mathcal C_2$ returns $\log N$, and a M\"obius inversion returns each prime of each lane. \subsection{The three forms as a Dirichlet series} Let $\chi_6$ be the character modulo $6$ induced by $\chi_{-3}$ ($\chi_6(n)=1,-1,0$ for $n\equiv1,5$ and otherwise), so that $L(s,\chi_6)=(1+2^{-s})L(s,\chi_{-3})$, and let $\zeta_6(s)=(1-2^{-s})(1-3^{-s})\zeta(s)$. Write $r_+(k)$ for the number of ordered pairs $(x,y)$, $x,y\ge1$, with $k=6xy+x+y$ or $k=6xy-x-y$, and $r_-(k)$ for the number with $k=6xy+x-y$. \begin{proposition}[the forms as divisor counts]\label{prop:rdiv} For $k\ge1$, \[ r_+(k)=d(6k+1)-2,\qquad r_-(k)=\tfrac12\,d(6k-1)-1, \] and \[ \sum_{n\equiv\pm1\,(6)}\frac{d(n)}{n^s}=\frac12\Bigl[\zeta_6(s)^2\pm L(s,\chi_6)^2\Bigr]\qquad(\operatorname{Re}s>1). \] In particular $6k+1$ is prime if and only if $r_+(k)=0$, and $6k-1$ is prime if and only if $r_-(k)=0$. \end{proposition} \begin{proof} By Proposition~\ref{prop:mult}, $r_+(k)$ counts the ordered factorizations $6k+1=ab$ with $a,b>1$ and $a\equiv b\pmod 6$. Every factorization of a number $\equiv1\pmod6$ has $a\equiv b$, so $r_+(k)=d(6k+1)-2$. For $n=6k-1$ the divisors split equally between the classes $1$ and $5$, since $d\mapsto n/d$ exchanges them. The factorizations $(6x+1)(6y-1)$ with $x,y\ge1$ correspond to the divisors $\equiv1$ other than $1$, which gives $\frac12d(n)-1$. For the series, $\sum_{(n,6)=1}d(n)n^{-s}=\zeta_6(s)^2$ and $\sum\chi_6(n)d(n)n^{-s}=L(s,\chi_6)^2$, because $\chi_6$ is completely multiplicative. Half the sum and half the difference select the two classes. \end{proof} \noindent Status: elementary; checked for $k\le20\,000$. The "operator" behind the forms $6xy\pm x\pm y$ is multiplication in the monoid of integers prime to $6$, graded by $\chi_6$: the rule $(+)(+)=(-)(-)=+$, $(+)(-)=-$ is exactly the factor $L(s,\chi_6)^2$. The diagonal $x=y$ of the first two forms consists of the squares $(6x\pm1)^2$. It falls only in the lane $6k+1$, and it is the source of the bias of Proposition~\ref{prop:explicit}. \subsection{Duality} For $N\ge2$ and $t\in\mathbb R$ let \[ \mathcal C_2(N,\tfrac12,t)=\frac{2}{\sqrt N}\sum_{j1$, Landau's formula \cite{Lan} and its analogue for $L(s,\chi)$ with $\chi$ real give \[ \sum_{0<\gamma\le T}x^{\rho}=-\frac T{2\pi}\Lambda(x)+O_x(\log T),\qquad \sum_{0<\gamma_\chi\le T}x^{\rho_\chi}=-\frac T{2\pi}\chi(x)\Lambda(x)+O_x(\log T), \] with $\Lambda(x)=0$ unless $x$ is a prime power. Under RH and GRH, $x^\rho=\sqrt x\,x^{i\gamma}$, and taking real parts, $\sum_{0<\gamma\le T}\cos(\gamma\log x)=-\frac T{2\pi}\Lambda(x)/\sqrt x+O_x(\log T)$. Insert $x=N/j$ for each $j1}\sqrt{\frac dN}\Bigl(-\frac T{2\pi}\frac{\Lambda(d)}{\sqrt d}\Bigr)+O_N(\log T)=-\frac T{\pi N}\sum_{d\mid N}\Lambda(d)+O_N(\log T), \] and $\sum_{d\mid N}\Lambda(d)=\log N$. The same computation with $\chi(d)\Lambda(d)$ gives $A_L$. For the last identity, $N(A_\zeta\pm A_L)/2=\sum_{d\mid N}\Lambda(d)\frac{1\pm\chi(d)}2$ is the Dirichlet convolution of $\Lambda\frac{1\pm\chi}2$ with $1$, and M\"obius inversion undoes it. \end{proof} \noindent Status: proved under RH and GRH. Without them, the same proof gives the identity with $\mathcal C_2(N,\frac12,\gamma)$ replaced by $\frac2{\sqrt N}\sum_j j^{-1/2}\operatorname{Re}(N/j)^{\rho-1/2}$. The two versions differ only through zeros off the line. The theorem is Landau's formula read through $\mathcal C_2$. What it adds is the reading: the same function $\mathcal C_2$ detects zeros as $N$ runs over the integers (Part~1), and returns $\log N$ as $\gamma$ runs over the zeros. The lane of a prime is decided only by which family of zeros is used. \begin{table}[ht] \centering \caption{The two detectors of Theorem~\ref{thm:dual} with the $1517$ zeros of $\zeta$ and the $1865$ zeros of $L(s,\chi_{-3})$ below $T=2000$; the last column is the value of $\Lambda(N)$ in its lane.}\label{tab:dual} \begin{tabular}{rcrrc} \toprule $N$ & lane & detector $6k+1$ & detector $6k-1$ & expected\\ \midrule $5$ & $-$ & $-0.002$ & $1.600$ & $\log 5$\\ $7$ & $+$ & $1.928$ & $-0.013$ & $\log 7$\\ $11$ & $-$ & $-0.030$ & $2.368$ & $\log 11$\\ $13$ & $+$ & $2.518$ & $-0.020$ & $\log 13$\\ $25$ & $+$ & $1.636$ & $-0.076$ & $\log 5$\\ $29$ & $-$ & $0.002$ & $3.220$ & $\log 29$\\ $35$ & $-$ & $-0.046$ & $-0.083$ & $0$\\ $43$ & $+$ & $3.652$ & $0.001$ & $\log 43$\\ $49$ & $+$ & $1.834$ & $-0.155$ & $\log 7$\\ $55$ & $+$ & $0.172$ & $-0.331$ & $0$\\ $59$ & $-$ & $-0.284$ & $4.000$ & $\log 59$\\ $77$ & $-$ & $0.061$ & $-0.191$ & $0$\\ $79$ & $+$ & $4.133$ & $-0.285$ & $\log 79$\\ \bottomrule \end{tabular} \end{table} \emph{Numerical check.} We computed the zeros directly (Euler--Maclaurin and Hardy's $Z$-function). The counts below $2000$ close against the Riemann--von Mangoldt formulas, and the first zero of $L(s,\chi_{-3})$ is at $8.0397$. For $N=2,10,30,60$, $A_\zeta(N)=0.3472$, $0.2290$, $0.1116$, $0.0660$, against $\log N/N=0.3466$, $0.2303$, $0.1134$, $0.0682$. Table~\ref{tab:dual} shows the two detectors. Every prime $5\le p\le79$ appears in its lane and not in the other. The composites $35,55,77$ give values near $0$. The squares $25=5^2$ and $49=7^2$ give $\log5$ and $\log7$ in the lane $6k+1$, as they must, since $\Lambda(p^2)=\log p$. These are the diagonal terms $x=y$ of Proposition~\ref{prop:rdiv}, the same squares that cause the bias. The largest error up to $N=79$ is $0.57$ (at $N=67$). The error term is $O_N(\log T)/T$ with a constant that grows with $N$, so the table degrades slowly with $N$ at fixed $T$. \emph{The densities as waves.} Differentiating the explicit formula gives, for the density of each lane, \[ \frac{d}{dx}\psi(x;6,\pm1)=\frac12\Bigl[1-\frac2{\sqrt x}\sum_{\gamma>0}\cos(\gamma\log x)\mp\frac2{\sqrt x}\sum_{\gamma_\chi>0}\cos(\gamma_\chi\log x)\Bigr]+\dots \] in the sense of distributions, where the dots are the contributions of the trivial zeros and of the powers of $2$ and $3$. Figure~\ref{fig:waves} shows the truncation at height $400$ with Fej\'er weights $1-\gamma/400$. In the lane $6k-1$ every prime up to $200$ has a peak, and no composite does. In the lane $6k+1$ the only peaks at composites are at $25$, $49$, $121$ and $169$, the squares of primes. The zeros of $\zeta$ contribute equally to both lanes. Only the zeros of $L(s,\chi_{-3})$ separate them, with opposite signs. \begin{figure}[ht] \centering \includegraphics[width=\textwidth]{fig/F9_waves.png} \caption{(a), (b) The prime densities of the two lanes rebuilt from the $201$ zeros of $\zeta$ and the $271$ zeros of $L(s,\chi_{-3})$ below $400$, with Fej\'er weights; vertical lines at the integers of each lane (solid: primes, dotted: composites). (c) The two families of waves: the zeros of $\zeta$ (common to both lanes) and the zeros of $L(s,\chi_{-3})$ (opposite signs in the two lanes).}\label{fig:waves} \end{figure} \subsection{The duality at finite height: an exact explicit formula for the cross-energy}\label{sec:explicitC2} This subsection is new in version~3 (2~November 2026). Theorem~\ref{thm:dual} needs two things: a limit $T\to\infty$, and RH and GRH to turn $x^\rho$ into a cosine. Here we remove both. A Gaussian weight on the zeros gives an exact identity at every $N$ and every resolution. Its four terms (zeros, pole, Gamma factor and primes) can be computed separately, and each has a meaning in terms of $\mathcal C_2$. We first state the identity, then explain how the cross-energy hears the primes through the zeros. On the line, $\mathcal C_2$ is the restriction of the reflected cross-energy of Part~3 \cite{CE3}, \[ \mathcal C_2^{\rm FR}(N,s)=N^{-s}X_{N-1}(1-s)+N^{s-1}X_{N-1}(s),\qquad \mathcal C_2^{\rm FR}(N,\tfrac12+it)=\mathcal C_2(N,\tfrac12,t), \] an entire function of $s$, symmetric under $s\leftrightarrow1-s$. For a zero $\rho$ write $t_\rho=(\rho-\frac12)/i$. In this subsection $\sum_\rho$ runs over all nontrivial zeros, of both signs of the ordinate; under RH, $t_\rho=\gamma_\rho$ is real and $\mathcal C_2^{\rm FR}(N,\rho)=\mathcal C_2(N,\frac12,\gamma_\rho)$. \begin{proposition}[the explicit formula for the cross-energy]\label{prop:explicitC2} For every integer $N\ge2$ and every $\varepsilon>0$, unconditionally, \begin{multline*} \underbrace{\sum_\rho e^{-\varepsilon^2t_\rho^2}\,\mathcal C_2^{\rm FR}(N,\rho)}_{\text{zeros}} =\underbrace{2e^{\varepsilon^2/4}\Big(H_{N-1}+1-\frac1N\Big)}_{\text{pole}}\\ +\underbrace{\frac1{2\pi}\int_{\mathbb R}e^{-\varepsilon^2r^2}\,\mathcal C_2(N,\tfrac12,r)\Big(\Re\frac{\Gamma'}{\Gamma}\big(\tfrac14+\tfrac{ir}2\big)-\log\pi\Big)dr}_{\text{Gamma}}\\ -\underbrace{\frac1{\sqrt\pi\,\varepsilon\sqrt N}\sum_{n\ge2}\sum_{j1$, $j=N/d$) have $\sqrt{nj}=\sqrt N$, and together they give exactly $\sum_{d\mid N}\Lambda(d)=\log N$. All other terms are nonnegative, so $R_\varepsilon(N)\ge0$. For the upper bound, note that $\sqrt\pi\,\varepsilon N$ times the prime term is $\sqrt N\sum_{n,j}\Lambda(n)(nj)^{-1/2}(\dots)$. In the first family, group the terms with $nj\ne N$ by $m=nj$ and use $\sum_{nj=m}\Lambda(n)\le\sum_{d\mid m}\Lambda(d)=\log m$. In the second family, $j\log n$, and $\sum_{j