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Math-635 · mathematics · Number theory

The Prime Number Race

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Statement

Definitions and mathematical statement

Shared notation.
Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

For positive modulus $b$, $\pi(x;a,b)$ counts primes at most $x$ in the residue class $a\pmod b$. For each fixed $a$ coprime to $b$, ask for arbitrarily large integer cutoffs at which its count is strictly larger than that of every other residue class. The comparison is simultaneous over the finitely many classes; the winning cutoff may depend on $a,b$. [S1] [S2]

Statement recorded in the supplied source.
Let $\pi(n; a, b)$ be the number of primes $p \le n$ with $p \equiv a \pmod b$. For every $a$ and $b$ with $a \perp b$, are there infinitely many values of $n$ for which $\pi(n; a, b) > \pi(n; a_1, b)$ for every $a_1 \not\equiv a \pmod b$? [S1]
Residual task reported by the archive.

The embedded review (2026-08-17) records: Prove the simultaneous-lead statement or clarify necessary hypotheses. [S1]

Short English statement

Does every residue class coprime to a modulus take an outright lead in its prime-counting race infinitely often?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-635-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

gpt 6 astra ultra attempt on Math-635

Every Coprime Residue Leading a Prime Race

Definitions and mathematical statement

Shared notation.
Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

For positive modulus $b$, $\pi(x;a,b)$ counts primes at most $x$ in the residue class $a\pmod b$. For each fixed $a$ coprime to $b$, ask for arbitrarily large integer cutoffs at which its count is strictly larger than that of every other residue class. The comparison is simultaneous over the finitely many classes; the winning cutoff may depend on $a,b$. [S1] [S2]

Statement recorded in the supplied source.
Let $\pi(n; a, b)$ be the number of primes $p \le n$ with $p \equiv a \pmod b$. For every $a$ and $b$ with $a \perp b$, are there infinitely many values of $n$ for which $\pi(n; a, b) > \pi(n; a_1, b)$ for every $a_1 \not\equiv a \pmod b$? [S1]
Residual task reported by the archive.

The embedded review $2026-08-17$ records: Prove the simultaneous-lead statement or clarify necessary hypotheses. [S1]

Short English statement

Does every residue class coprime to a modulus take an outright lead in its prime-counting race infinitely often?

Research attempt

Quantifiers and source check.
The desired lead is strict and simultaneous against every other residue class, for arbitrarily large cutoffs depending on the chosen class and modulus. The primary Rubinstein--Sarnak publication record [E1] was inspected on 1 October 2026; its analytic framework uses hypotheses on Dirichlet $L$-function zeros. They are not assumed in the requested unconditional statement. This attempt isolates an exact distributional sufficient condition and the distinction between pairwise and simultaneous leads.
Nonreduced residue classes are a finite issue.
If $\gcd(c,b)>1$ and a prime $p$ is congruent to $c$ modulo $b$, then some prime divisor of $\gcd(c,b)$ divides $p$, so $p$ itself divides $b$. Hence the counts in all nonreduced classes are bounded by the number of prime divisors of $b$. For each reduced class, Dirichlet's theorem on primes in progressions gives an unbounded count. Using that standard theorem, all nonreduced competitors are eventually irrelevant. The substantive competition is among the $\varphi(b)$ reduced classes.

For $b=1$ there is only one class and the comparison is empty, so the assertion is vacuous. For $b=2$ the sole reduced class is the odd class; it has at least two primes by cutoff five, while the even class contains only the prime $2$ and has count one. Thus the odd class leads thereafter. These boundary moduli are completely settled.

Pairwise reversals are not simultaneous leadership.
Consider three hypothetical discrepancy coordinates taking alternately the values $(0,-1,1)$ and $(0,1,-1)$. The first coordinate beats the second infinitely often and beats the third infinitely often, but never beats both at the same time. Thus even pairwise sign changes against every rival do not imply the requested simultaneous lead. These are illustrative vectors, not asserted prime counts. They refute a logical implication sometimes used to join separate oscillation theorems without controlling their joint phases.

Likewise equidistribution to a common first-order main term permits persistent differences of smaller order. Equal asymptotic densities therefore neither prove that every class leads nor rule out a strong bias in the times at which a class leads.

An exact sufficient condition from joint distributions.
List the reduced classes as $a_1,\ldots,a_r$, put $P(x)=\sum_{j=1}^r\pi(x;a_j,b)$, and define for $x>1$ \[ E_j(x)=\frac{\log x}{\sqrt x} \left(r\pi(x;a_j,b)-P(x)\right). \] The vector $E(x)$ lies in the hyperplane $H$ where the coordinate sum is zero. The normalization is positive and the same main count is subtracted from each coordinate, so $a_j$ leads exactly when $E_j(x)>E_i(x)$ for all $i\ne j$.

Suppose that the logarithmic-time distributions \[ \frac1T\int_0^T f(E(e^t))\,dt \] converge for all bounded continuous $f$ to integration against a probability measure $\mu$ having full support on $H$. Then every leader cone $U_j=\{v\in H:v_j>v_i\ (i\ne j)\}$ has positive measure: it is a nonempty relatively open set. Choose a smaller open ball inside it with positive measure. The open-set lower bound for weak convergence yields positive lower limiting frequency of visits to $U_j$. Therefore there are arbitrarily large real cutoffs at which $a_j$ leads.

Prime counts are constant between consecutive integers, so such real cutoffs give arbitrarily large integer cutoffs as required. Nonreduced classes have already been disposed of. Thus the full-support limiting distribution assumption is a precise sufficient condition for the catalogue statement. The analytic issue is proving enough of that joint distribution and its support unconditionally, not the elementary implication once it is available.

The role of density versus mere infinitude.
Positive logarithmic frequency is stronger than unbounded winning times. A set with all its winning times bounded contributes at most a fixed amount to the logarithmic-time integral, divided by $T$, so its limiting frequency is zero. Conversely zero frequency does not imply only finitely many wins. Thus a proof that a class is heavily disfavored would not by itself answer the existence-of-infinitely-many-wins question. No numerical estimate of a bias is treated as a proof of the full-support condition above.
Executed finite race checks.
The script counts all residue classes, including nonreduced ones, through $200000$ for moduli $3,4,5,8$. For modulus four, class three first leads strictly at seven and class one first leads strictly at $26861$. Some reduced classes in the other tested moduli never lead within this finite prefix. The result records these as absent in the prefix, not as permanent losers. Checking only prime cutoffs suffices because counts change only there. Files are checks_second.py and results_second.json under _Local/Erdos_problems/UnsolvedMath/attempt_audit_2026_10_01/number_theory/.
Remaining gap and outcome.
The nonreduced classes and boundary moduli are handled, and an exact joint-distribution sufficient condition is proved. That condition has not been established unconditionally for all moduli here. Outcome: UNRESOLVED. Neither finite race data nor separately oscillating pairwise differences resolves simultaneous leadership.

Sources

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