bibkey: pollack2017nonresidues authors: Paul Pollack year: 2017 title: Bounds for the first several prime character nonresidues doi: 10.1090/proc/13432 url: https://arxiv.org/pdf/1508.05035v2 claim: Theorem 1.1 supplies prime counts; its published real-character inputs also yield fixed reciprocal-prime mass with modulus zeros retained, usable for actual affine sources only when the enlarged cutoff reaches the Robin prefix. strata_touched: [] license: citation-only triage: anchor

Prime nonresidues with the actual exceptions retained

The author publication list identifies the article as Proceedings of the American Mathematical Society 145 (2017), 2815–2826, DOI:10.1090/proc/13432. The inspected sources are arXiv:1508.05035v2, dated 24 August 2015 in its submission history, and the author-hosted manuscript. Theorem 1.1 appears on p.1 of v2 and p.2 of the author manuscript; Propositions 2.1–2.2, the smooth-number input on pp.2–3, the coprime count in the proof of Lemma 2.6 on p.6, and Theorem 1.2 were also inspected. The weighted deduction below is an application of those published inputs, not a quoted statement of Theorem 1.1. Their complete original analytic proofs and the weighted/Robin applications were not independently formalized in Lean. The linked local mod-eight application has its own limited verification scope. No complete current literature survey or numerical analytic onset threshold is claimed.

The existing uniform supply

For every fixed $\epsilon >0$, Theorem 1.1 gives constants $m_0(\epsilon )$ and $\kappa (\epsilon )>0$ such that every integer $m>m_0$ and every nonprincipal Dirichlet character $\chi$ modulo $m$ have more than $m^{\kappa }$ primes $\ell$ satisfying

$$\ell \le m^{\beta },\chi (\ell )\notin \{0,1\},\beta =\frac{1}{4\sqrt{e}}+\epsilon .$$

For quadratic characters the nonresidue value is $-1$. The statement is not restricted to prime moduli or primitive characters. Its threshold and count exponent are not supplied here as numerical constants, so it is not a finite numerical certificate. The theorem is unconditional; GRH and a zero-free-region assumption are absent from its hypotheses.

For an explicitly stated reciprocal-prime weight supply, reuse Bourgain–Lindenstrauss, Theorem 5.1. Its cutoff is $Y\ge D^{1/4+\epsilon }$ for sufficiently large positive nonsquare $D$. The linked note records its exact weighted Robin budget and actual-source exceptions; the prime-count theorem here does not automatically supply that weight at $Y=\log N$.

Theorem 1.2 strengthens the cutoff for characters of higher order, using the Dickman-function parameter $u_k$. Quadratic characters have order two, so their parameter is $u_2=\sqrt{e}$; increasing the modulus does not increase their order. Proposition 2.1 records Norton’s Burgess estimate. Its extra factor is $R_k(m)=\min \{M(m{)}^{3/4},Q(k{)}^{9/8}\}$; for $k=2$, $Q(k)=1$. Thus the source already handles composite quadratic moduli. Reproving a prime-modulus Burgess estimate would not improve this interface.

A weighted real-character consequence of the published inputs

Put ${\beta }_0=1/(4\sqrt{e})$. Fix

$${\beta }_0<b<\frac{1}{4},0<c<\log (4\sqrt{e}b).$$

The source inputs give the following paper-level consequence: there is $M(b,c)$ such that every integer $m\ge M(b,c)$ and every real nonprincipal Dirichlet character $\chi$ modulo $m$ satisfy

$$W_{\chi }(m^b):=\sum _{\begin{array}{c}\ell \le m^b\\ \chi (\ell )=-1\end{array}}\frac{1}{\ell }\ge c.$$

The modulus need not be prime and the character need not be primitive. Primes dividing $m$ remain zeros. This conclusion is obtained from the following inputs; it is not attributed as a separately stated theorem in Pollack, nor asserted to be unpublished or original.

• Proposition 2.1 records K. K. Norton’s A character-sum estimate and applications, Acta Arithmetica 85 (1998), 51–78, Theorem 1.6. For order two, $R_2(m)=1$, so for fixed integer $r\ge 1$ and $\eta >0$ it gives $|\sum_{k\le x}\chi(k)|\ll_{r,\eta} x^{1-1/r}m^{(r+1)/(4r^2)+\eta}$.
• Proposition 2.2 records the $c=1$ case of G. Tenenbaum’s Cribler les entiers sans grand facteur premier, Philosophical Transactions of the Royal Society A 345 (1993), 377–384. With $u=\log x/\log y$, it gives $\Psi_q(x,y)=(\varphi(q)/q)\Psi(x,y)(1+O( \log(1+u)\log(1+\omega(q))/\log y))$when$x\ge y\ge2$, $P^{+}(q)\le y$ and $\omega (q)\le y^{1/\log (1+u)}$.
• The smooth-number asymptotic $\Psi (x,x^{1/u})\sim \rho (u)x$ for fixed $u$, and $\rho (u)=1-\log u$ for $1\le u\le 2$, are recorded on pp.2–3. These are additional inputs, not consequences of Proposition 2.2 alone.
• Inclusion–exclusion gives $C_m(z):=\#\{k\le z:(k,m)=1\}=z\phi (m)/m+O(2^{\omega (m)})$. The corresponding coprime-multiple count is used on p.6 in Lemma 2.6.

To check the deduction, choose fixed $a>1/4$ sufficiently close to $1/4$ that $1<u:=a/b<\sqrt{e}$ and $\rho (u)-1/2>c$. Set $x=m^a$, $y=m^b$, and $A_m=\phi (m)/m$. Choosing $r$ large and $\eta$ small in Norton’s estimate gives ${\sum }_{k\le x}\chi (k)=o(x{A}_m)$ uniformly. Indeed, its power relative to $x$ is $(1/4-a)/r+1/(4r^2)+\eta <0$, while $A_m=m^{-o(1)}$ and $2^{\omega (m)}=m^{o(1)}$. Thus

$$\#\{k\le x:\chi (k)=1\}=(1/2+o(1))x{A}_m.$$

For the coprime smooth count take $q={\prod }_{p\mid m,p\le y}p$. Proposition 2.2’s hypotheses hold uniformly for large $m$: $u$ is fixed, $\omega (q)=O(\log m)$, and its allowed upper bound is a positive power of $m$. Its remark gives ${\Psi }_m(x,y)={\Psi }_q(x,y)$, and $\phi (q)/q\ge A_m$. Consequently

$${\Psi }_m(x,y)\ge (\rho (u)-o(1))x{A}_m.$$

The number $B$ of $y$-smooth integers at most $x$ with $\chi (k)=-1$ is therefore at least $(\rho (u)-1/2-o(1))x{A}_m$. Each has a prime factor $\ell \le y$ with $\chi (\ell )=-1$, and such a prime is coprime to $m$. Counting its coprime multiples gives

$$\begin{array}{rl}B& \le \sum _{\begin{array}{c}\ell \le y\\ \chi (\ell )=-1\end{array}}{C}_m(x/\ell )\\ & \le x{A}_m{W}_{\chi }(y)+O(y{2}^{\omega (m)}).\end{array}$$

After division by $x{A}_m$, the error is $m^{b-a+o(1)}=o(1)$, proving the claimed weight. Counting all multiples as $x/\ell$ instead would retain an unnecessary factor $A_m$ in the lower bound. The coprimality bookkeeping is what preserves constant weight despite the additional modulus zeros.

There is no exceptional-real-character deletion in these inputs. No numerical value for $M(b,c)$ has been extracted, and no exact compiled application of this analytic deduction is claimed.

What is required at the same arithmetic source

The FIB theory volume already owns the norm identities and character Euler-product estimates in §§178,181,202. They are reused here.

For §181’s actual $N=Cu$, an odd prime with $\chi (\ell )=-1$ is excluded from the primitive factor $u$, but may still divide the multiplier $C$. To obtain a prime missing from $N$, the supplied prime must also satisfy $\ell \nmid C$ and $\ell \ne 2$. The multiple-prime statement can survive deleting a finite exceptional set $E$ when $|E|<m^{\kappa }$; the source’s full cutoff must still lie in the required actual prime prefix. A bound for the first nonresidue alone does not remove the multiplier cost. The actual-source examples in §§183,186 already show why that cost matters.

For §202’s affine $N=h+gU$, the induced character modulo $4|D|$, with $D=5h^2-4g^{2}Q$ nonsquare, takes only values $0,1$ at primes dividing $N$. A prime where that induced character is $-1$ is consequently missing from $N$. If instead one uses the primitive character of conductor $q$, all primes dividing $2D$ remain exceptions: primes in the square part of $D$ may disappear from $q$ without disappearing from the original certificate. Square $D$ gives a principal character and is outside Theorem 1.1’s scope. For odd affine offsets, the actual mod-eight certificate recovers the prime-two contribution when the primitive character is negative there; the actual square-part intersection still has to be retained.

One standard way to retain a finite exceptional set is to induce the primitive character to a modulus $M$ divisible by $q$ and by every prime in $E$. Theorem 1.1 applies to that nonprincipal induced character and its nonresidue primes automatically avoid $E$. The cutoff then involves $M^{\beta }$, not merely $q^{\beta }$. Alternatively, retaining modulus $q$ requires enough supplied primes to survive deletion of $E$. These are applications of the cited result, not new character estimates.

For a CA integer with support containing every prime through $P=P^{+}(N)$, exclusion by a missing prime requires the supplied prime to be at most that actual $P$. The Euler-product split $x=\log N$ in §§178,181,202 is a different parameter. A usable source application must establish its cutoff below $P$; it cannot substitute $x$ for $P$ without the required comparison.

The existing canonical norm estimate $q<4N/g^2$ in §182.2 does not ensure that $q^{\beta }\le \log N$, much less that the exceptions-adjusted cutoff is at most $P$. A fixed positive power of a modulus comparable to a positive power of $N$ exceeds $\log N$ asymptotically. This is a limitation of the available bound at those parameters, not a lower bound on the actual first nonresidue and not a counterexample to Robin. Restricted families with a sufficiently small actual modulus may use the theorem, while all-candidate conductor, exception and weighted-deficit control remains missing.

Finding one omitted prime is also not, for a general integer, a complete Robin certificate. Its contribution must pay the actual remaining Euler budget using the same source. No general Robin or RH conclusion, new analytic theorem, or new Lean declaration is claimed here.

The enlarged modulus at the actual Robin cutoff

For §202’s actual affine source, the original conservative modulus $m=4|D|$ already preserves every prime-two and square-part zero. Its negative primes all miss the same $n$, without an oddness assumption on $h$. If $h$ is odd, write $D=\Delta t^2$ with $\Delta$ signed squarefree. Then $\Delta \equiv 1(\mathrm{mod}4)$ is the fundamental discriminant, and either $m=|D|$ or $m=\mathrm{rad}|D|$ supports the induced character

$${\chi }_m(k)={\chi }_{\Delta }(k)1_{(k,m)=1}.$$

Both choices retain all raw-square-part zeros; the linked mod-eight certificate also excludes two whenever ${\chi }_m(2)=-1$. Thus, whenever $m\ge M(b,c)$ and $m^b\le Y=\log n$,

$$J_{\mathrm{miss}}(n;Y)=\sum _{\begin{array}{c}p\le Y\\ p\nmid n\end{array}}\log \frac{p}{p-1}\ge W_{{\chi }_m}(m^b)\ge c.$$

The supplied weight already avoids the enlarged modulus: no square-part subtraction is silently discarded. The benefit of a smaller faithful modulus is a smaller cutoff. However, $|D|\to \infty$ does not imply $\mathrm{rad}|D|\to \infty$; a radical-based application still needs its own $m\ge M(b,c)$ check. Fixed or bounded moduli can use earlier fixed-character results, not an unspecified large-modulus threshold.

In particular fix $0<\delta <4\sqrt{e}$, put $K=4\sqrt{e}-\delta$, and choose

$${\beta }_0<b<\min \{1/4,1/K\},0<c<\log (4\sqrt{e}b).$$

Along actual affine sources with $n\to \infty$, $D$ nonsquare, $|D|\to \infty$, and $|D|\le (\log n{)}^K$, use $m=4|D|$. The strict margin $Kb<1$ absorbs the factor four, so $m^b\le \log n$ eventually. The same-integer Euler budget then implies the paper-level consequence

$$\mathrm{limsup}\frac{Z(n)}{e^{\gamma }\log \log n}\le e^{-c}<1.$$

This admits signed nonsquare discriminants, composite conductors and raw square parts. Relative to the previously recorded positive-$D$ cutoff $D^{1/4+\epsilon }$, it also admits fixed raw-discriminant log-powers strictly below $4\sqrt{e}$ instead of strictly below four. Intersections with the old safe families are applications, not new safe-family claims.

A concrete canonical realization of the additional range reuses §205.1, not a new FIB construction. For $j\equiv 1(\mathrm{mod}6)$, take $g=j^3$, $h=1$, and the seed $v=\alpha =(1,0)$. Then

$$(A_j,B_j)=j^3{M}^{j}v,n_j=j^3{F}_{j+3}+1.$$

The contraction readout is $A_j+B_j\psi =j^3{\psi }^j\to 0$ from below. Eventually it lies in $(-1,\phi -1)$, so the existing canonical-window test permits the external unit bit without changing the legal window word or End. The same actual composition gives

$$Q(A_j,B_j)=-j^6,D_j=4j^6+5,(2j^3{)}^2<D_j<(2j^3+1{)}^2(j>1).$$

In particular $D_j$ is nonsquare and $D_j\sim 4(\log n_j/\log \phi {)}^6$. For every fixed ${\beta }_0<b<1/6$, the cutoff $(4D_j{)}^b$ is eventually at most $\log n_j$. Taking all $0<c<\log (4\sqrt{e}b)$ and then $b\uparrow 1/6$ gives the paper-level application

$$\underset{\begin{array}{c}j\to \infty \\ j\equiv 1(\mathrm{mod}6)\end{array}}{\mathrm{limsup}}\frac{Z(n_j)}{e^{\gamma }\log \log n_j}\le \frac{3}{2\sqrt{e}}<1.$$

No primality or squarefreeness of $D_j$ is used. Here $D_j\equiv 1(\mathrm{mod}8)$ and $n_j$ is even, so the prime-two shortcut supplies no missing weight. The raw discriminant lies outside every previously recorded $D\le (\log n{)}^{4-\delta }$ range. The variant with $g=j$ has log-power two and was already covered by that earlier interface; it is not a new safe family. The original character-product sufficient condition in §202.4 also does not cover this cubic-multiplier sequence: for $m=4D_j$, $(2+\log m)m/\phi (m)$ is at least order $\log \log n_j$, not little-o of it. The result here remains a source-derived asymptotic application, without a numerical onset or Lean verification of the analytic chain.

General candidates need not have nonsquare $D$ or satisfy this joint modulus/cutoff restriction. These applications do not replace that missing all-candidate estimate, and do not prove general Robin or RH.

Prime-power exceptions for the actual unit bit one

The conservative induction above retains every raw square-part zero. For an actual canonical source with unit bit $h=1$, a smaller faithful exception set can be read from the same prime-power exponents and the same five-window lift. The following is a paper-level local refinement; it supplies no new analytic nonresidue estimate or Lean verification.

Let $n>2$, $n=1+2A+3B$, $c=4A+7B$, and $D=5-4(A^2+AB-B^2)$ be nonsquare. Write

$$D=\delta s^2,s\ge 1,\delta \text{ signed squarefree},q=|\delta |.$$

Positive and negative discriminants are admitted. Since $D\equiv 1(\mathrm{mod}4)$, $s$ is odd and $\delta \equiv 1(\mathrm{mod}4)$, so $\delta$ itself is the fundamental discriminant. Let $\chi$ be its primitive nonprincipal quadratic character of conductor $q$. The existing FIB identity is

$$c^2-\delta s^2=5n(n-2).\text{\hspace{1em}(P1)}$$

If an odd prime has $\chi (p)=-1$, then

$$v_p(c^2-\delta s^2)=2\min \{v_p(c),v_p(s)\}.\text{\hspace{1em}(P2)}$$

Unequal valuations cannot cancel; equal valuations could cancel only if $\delta$ became a nonzero square modulo $p$. In particular, writing $e_p=v_p(n)>0$, a negative-character support prime $p\ne 2,5$ must have $e_p$ even and $v_p(c),v_p(s)\ge e_p/2$, since $p\nmid n-2$. For $p=5$, negativity instead forces $e_5$ odd and both valuations at least $(e_5+1)/2$. An odd-exponent support prime other than five cannot be a hidden negative prime after removal of the raw square part.

Define the potential-exception set and required depths by

$${\mathcal{P}}_1=\{p\mid n:p\ne 2,5,e_p\text{ even}\}\cup \{5:5\mid n,e_5\text{ odd}\},b_p=\{\begin{array}{ll}{e}_p/2,& p\ne 5,\\ (e_5+1)/2,& p=5.\end{array}$$

Use the actual lift, rather than a free residue, to set

$$H_1=\prod _{p\in {\mathcal{P}}_1}{p}^{b_p},{\mathcal{E}}_1=\{p\in {\mathcal{P}}_1:p^{b_p}\mid c\},T_1=\prod _{p\in {\mathcal{E}}_1}{p}^{b_p},R_1=\mathrm{rad}(T_1).$$

The set ${\mathcal{E}}_1$ contains every negative-character support prime; its other members need not be negative. All its membership tests are determined by $c\mathrm{mod}{H}_1$. The same-object divisibility relations are

$$H_1\mid 5\frac{n}{\mathrm{rad}(n)},T_1\mid \gcd (c,s),q{T}_1^2\mid |D|.\text{\hspace{1em}(P3)}$$

For $p\ne 5$, $b_p=e_p/2\le e_p-1$. For five this inequality holds when $e_5\ge 3$; $e_5=1$ is paid by the displayed extra factor five. For each $p\in {\mathcal{E}}_1$, both $c^2$ and $5n(n-2)$ are divisible by $p^{2b_p}$. Thus $p^{2b_p}\mid D$, so $p^{b_p}\mid s$, proving the remaining relations in (P3).

Set $m_1=\mathrm{lcm}(q,R_1)$ and induce the character by ${\tilde{\chi }}_1(k)=\chi (k)1_{(k,m_1)=1}$. Then

$${\tilde{\chi }}_1(p)=-1⟹p\nmid n,m_1\mid \mathrm{rad}|D|,m_1\le q{R}_1\le |D|\frac{R_1}{T_1^2}.\text{\hspace{1em}(P4)}$$

For odd support primes the first statement follows from (P2) and the mask. If $n$ is even, $c$ is odd and $8\mid n(n-2)$, so $D\equiv 1(\mathrm{mod}8)$ and $\chi (2)=1$; if $n$ is odd, two is already missing. This handles the prime two without transporting an odd-prime argument to it. The second statement holds because $q$ and $R_1$ are squarefree divisors of $|D|$.

The extra factor five is necessary in this construction. The canonical source $n=95$ has $(A,B)=(14,22)$, $c=210$, $D=-75=-3\cdot 5^2$ and $\chi (5)=-1$. Here $H_1=T_1=R_1=5$, $m_1=15$, while $n/\mathrm{rad}(n)=1$.

The exponent excess is exactly

$$B_{\exp }(n)=\log \frac{n}{\mathrm{rad}(n)}=\sum _{p\mid n}(e_p-1)\log p,\log H_1\le \log 5+B_{\exp }(n).$$

Under the additional SA support asymptotic $\mathrm{rad}(n)=n^{1-o(1)}$, this gives $H_1=n^{o(1)}$. It does not bound the primitive conductor $q$ or guarantee $m_1^b\le \log n$. Whenever the existing weighted supply can actually reach the cutoff for $m_1$, (P4) lets that supplied negative-prime weight contribute to $J_{\mathrm{miss}}$ without deleting the listed exceptions again. The missing estimate must still cross the full same-integer Euler budget; $D$ square, unit bit zero and uncontrolled $q$ remain separate branches.

A canonical family outside the sufficient conductor cutoff

The canonical local-residue construction uses the same actual unit-one source and has $D\asymp n$, $Y=\log n\sim 2z$, no negative primitive-character prime up to $z$, and $W_{\delta }^{\log }(Y)\to 0$. It retains the ramified zero at three. Its primitive conductor $q=\delta$ satisfies $q>z$: choose $1\le a<q$ with ${\chi }_{\delta }(a)=-1$. Its prime factors have nonzero character, and at least one has value $-1$, hence lies strictly between $z$ and $q$. In particular $q\to \infty$.

For each preassigned ${\beta }_0<b<1/4$, choose $0<c<\log (4\sqrt{e}b)$. The weighted consequence above gives $W_{{\chi }_{\delta }}(q^b)\ge c$ once $q$ reaches its stated threshold. If $q^b\le Y$, then

$$c\le W_{{\chi }_{\delta }}(q^b)\le W_{\delta }^{\log }(Y),$$

which contradicts the latter’s convergence to zero. Thus this canonical family eventually has

$$q^b>\log n,m_1^b\ge q^b>\log n.$$

The actual primitive conductor already exceeds this sufficient cutoff; an improved upper certificate for $m_1$ cannot change that fact on this family. The onset may depend on the fixed $b,c$. This is a paper-level application of the existing weighted estimate, not a new analytic theorem, an assertion $q\asymp n$, or an obstruction proved on SA/CA candidates.

The terminal CA prime band forces a conductor lower bound

The preceding canonical family is not established to be an asymptotic CA family. For actual CA sources, a different restriction follows by applying the published Thorner–Zaman prime count. This is a paper-level application of existing analytic and canonical interfaces, without a new analytic theorem, originality claim or Lean verification.

Let $n$ be any actual CA maximizer, including an intermediate tied maximizer, and put $P=P^{+}(n)$. Retain its own canonical unit bit $h$, coordinates $A,B$, $c=4A+7B$, and signed discriminant $D=5h^2-4(A^2+AB-B^2)$. In the branch $h=1$ with nonsquare $D$, write $D=\delta s^2$ as above and let $q$ be its primitive quadratic conductor. The existing prime-power exception result gives

$$p\mid n,p\ne 2,5,{\chi }_{\delta }(p)=-1⟹v_p(n)\text{ is even}.\text{\hspace{1em}(T3)}$$

This implication is reused at the same source; no canonical coordinates are transported from a larger representative. Primes dividing $q$ remain character zeros.

The terminal support consists of exponent-one primes

At the actual CA price $ϵ$, use the classical thresholds $F({\xi }_k,k)=ϵ$, with $\xi ={\xi }_1$. The Nicolas manuscript, equation (3.8) and Remark 3.1, printed p.11, give the prime-layer support and tied choices; (3.19)–(3.23), printed pp.12–13, give ${\xi }_2\le \sqrt{2\xi }$. Every prime below $\xi$ has strictly positive first-layer gain and is included. A first-layer tie can only affect the possible endpoint prime $\xi$. Hence $P\sim \xi$ by ordinary PNT, while ${\xi }_2=o(P)$.

Every sufficiently large actual CA integer therefore satisfies

$$v_p(n)=1\text{for every prime }P/2<p\le P.\text{\hspace{1em}(T4)}$$

Indeed $P/2>{\xi }_2$ eventually: each such prime is in the initial support and has strictly negative second-layer gain. A tied square activation at ${\xi }_2$ lies below this band. The argument permits every tied maximizer, rather than only the all-ties representative.

The primitive conductor is bounded below at the same candidate

Let $C,x_0$ be the constants in the linked interval consequence (T2). If $P>\max \{10,2{\xi }_2\}$ and $P\ge x_0$, (T3)–(T4) prohibit any prime $p\in (P/2,P]$ with ${\chi }_{\delta }(p)=-1$. If $q^C\le P$, (T2) instead supplies at least $P/(8\log P)$ such primes. Consequently, with a fixed $\eta =1/C>0$,

$$\boxed{q>P^{\eta }}\text{\hspace{1em}(T5)}$$

for every sufficiently large actual CA integer in this $h=1$, nonsquare-$D$ branch. No numerical CA onset or infinitude of this particular canonical branch is supplied.

This is a lower bound for the actual primitive conductor, rather than for the raw discriminant or an upper-bound certificate for the mask. Since $m_1=\mathrm{lcm}(q,R_1)\ge q$, it also gives $m_1>P^{\eta }$. On an unbounded branch of these sources, neither $q$ nor $m_1$ can be bounded by a fixed power of $\log P$.

For the existing sufficient cutoff $m_1^b\le Y=\log n$, the necessary range is now

$$P^{\eta }<q\le m_1\le Y^{1/b},P\sim Y.\text{\hspace{1em}(T6)}$$

The small fixed $\eta$ supplied here leaves that interval compatible; it does not exclude fixed-power-in-$\log n$ conductors. No reciprocal-prime deficit, exception-weight payment or signed Robin margin follows. Unit-bit-zero sources, square discriminants and the remaining larger conductors still require their own estimates. The raw-discriminant chain witnesses are not known to belong to this canonical branch, so their bound cannot be combined with (T5) by treating separately realized witnesses as one source. No actual Robin violation or RH proof is obtained.

A numerical exponent from the published progression interval

The independently published Haynes–White interval theorem supplies a prime in $(P/2,P]$ with ${\chi }_{\delta }(p)=-1$ whenever $q\le P^{\eta }$, for every fixed $0<\eta <5/67$ and sufficiently large $P$. Applying the same contradiction (T3)–(T4), without reconstructing either interface, gives the numerical version

$$\boxed{q>P^{\eta }\text{eventually for every fixed }0<\eta <5/67}\text{\hspace{1em}(T7)}$$

on the same actual $h=1$, nonsquare-$D$ CA branch. The eventual onset can depend on $\eta$ and is not claimed effective. For example $q>P^{1/14}$ eventually. The effective existential version (T5) is retained independently; no comparison of its unnamed exponent with $5/67$ is asserted.

This numerical conductor bound alone does not cross the faithful-mask cutoff. Its inequalities permit $m_1\le Y^5$ at $b=1/5$, while the lower bound is only $m_1>P^{1/14}$ and $P\sim Y$. A conductor lower bound $q>P^{\eta }$ would contradict $m_1^b\le Y$ by powers alone if $\eta b>1$; equality requires further constant or lower-order information. The supplied rate does not cross that scale in the stated ${\beta }_0<b<1/4$ regime. All same-source and unresolved-branch limitations above remain in force.

The progression count crosses the deep-layer capacity

A published quantitative count now gives a stronger numerical conductor restriction than (T7). Reuse Maynard, published Theorem 3.2 through (M2), without reconstructing its proof or the preceding canonical and CA layer interfaces. This is a paper-level same-source application, with no new analytic theorem, originality claim or Lean verification.

Take an actual CA maximizer $n$, including intermediate tied maximizers, whose own canonical unit bit is $h=1$ and whose own signed $D$ is nonsquare. Let $q$ be its primitive quadratic conductor and $P=P^{+}(n)$. Every prime up to $P$ belongs to its initial support. By (T3), every negative-character prime there, except possibly two and five, has even positive exponent and therefore is at most ${\xi }_2$. The existing ${\xi }_2\le \sqrt{2\xi }$ and $P\sim \xi$ give a fixed constant $C_A>0$ such that, eventually at every such source,

$${\pi }_{-}(P;\chi )\le \pi ({\xi }_2)+2\le C_A\frac{\sqrt{P}}{\log P}.\text{\hspace{1em}(T8)}$$

This is an upper bound for the same actual negative-prime population; ramified primes remain zeros. It does not assume every prime below ${\xi }_2$ has even exponent.

Suppose instead that $q\le P^{1/8}$ and $q\ge q_0$, where $q_0,c_0$ are Maynard’s effective constants. Then $P\ge q^8$ and (M2) supplies

$${\pi }_{-}(P;\chi )\ge \frac{c_0}{2}\frac{P\log q}{\sqrt{q}\log P}\ge \frac{c_0\log q_0}{2}\frac{P^{15/16}}{\log P}.\text{\hspace{1em}(T9)}$$

The ratio of (T9)’s lower bound to (T8)’s upper bound tends to infinity at least as a positive constant times $P^{7/16}$, uniformly throughout this conductor range. For the finitely many $q<q_0$, reuse (T2) from the earlier effective terminal-interval supplier with $P\ge \max \{x_0,q_0^C\}$; its negative primes in $(P/2,P]$ contradict (T4). Thus

$$\boxed{q>P^{1/8}}\text{\hspace{1em}(T10)}$$

for every sufficiently large actual CA source in this canonical $h=1$, nonsquare-$D$ branch. The exact endpoint is allowed here because Maynard’s supplied range is $P\ge q^8$. No numerical CA onset or infinitude of this branch is asserted.

For fixed ${\beta }_0<b<1/4$, any sufficiently large actual source in this branch that also satisfies the sufficient cutoff $m_1^b\le Y=\log n$ must have its faithful mask in the range

$$P^{1/8}<q\le m_1\le Y^{1/b},Y=\log n\sim P.$$

The conductor-only comparison does not rule out that cutoff by powers alone; the additional actual-mask constraint is supplied in (T22). The count comparison excludes a larger conductor range than (T7), but supplies neither a reciprocal-prime deficit nor the required signed Robin margin. Unit-bit zero, square $D$ and the larger remaining conductors stay unresolved.

All odd layers strengthen the signed-character restriction

Together with a fixed contradiction assumption $q\le P^{\eta }$, the numerical bound (T10) supplies the compact parameter range needed to use Szabó’s published Proposition 6 with the actual alternating CA layers. Reuse (T3), the classical thresholds and the literature estimate (S1); no new analytic theorem, originality claim or Lean verification is made.

Continue with the same actual CA integer, its canonical $h=1$, nonsquare signed $D$, primitive quadratic character $\chi$ of conductor $q$, and $P=P^{+}(n)$. Put $L=\log P$. All asymptotics below hold along every sequence of such actual maximizers with $P\to \infty$, including intermediate tied choices; this quantification asserts no branch infinitude. For every fixed $k\ge 1$, the classical activation formula gives

$$\frac{\log {\xi }_k}{L}⟶\frac{1}{k}.\text{\hspace{1em}(T11)}$$

Indeed, $F(x,k)\sim 1/(x^k\log x)$, and $F({\xi }_k,k)=F({\xi }_1,1)$ gives ${\xi }_k^k\sim k{\xi }_1$, with ${\xi }_1\sim P$. This is a fixed-layer consequence of the existing CA formula. A prime of exponent $k$ lies between ${\xi }_{k+1}$ and ${\xi }_k$, apart from the tied endpoints. In the logarithmic coordinate $u=\log p/L$, the odd layers therefore approach

$$\mathcal{O}=\bigcup _{j\ge 0}\left(\frac{1}{2j+2},\frac{1}{2j+1}\right),{\int }_{\mathcal{O}}du=\log 2,{\int }_{\mathcal{O}}udu=\frac{{\pi }^2}{24}.$$

For each fixed $r\ge 1$, Mertens’ first theorem and partial summation give

$$A_r(n):=\sum _{\begin{array}{c}p\le P\\ v_p(n)\mathrm{odd}\end{array}}\frac{\log p}{p}\left(r-\frac{\log p}{L}\right)=\left(r\log 2-\frac{{\pi }^2}{24}+o(1)\right)L.\text{\hspace{1em}(T12)}$$

Here $\mathrm{odd}$ means the exponent $v_p(n)$ is odd. First fix $K$ and let $P\to \infty$ on the finitely many retained layers, then let $K\to \infty$. Their tied endpoint primes contribute $o(L)$ regardless of the intermediate maximizing choice. All omitted layers lie below ${\xi }_{K+1}$ and have total weight at most $r\log {\xi }_{K+1}+O_r(1)=O_r(L/(K+1))+o(L)$. Letting $K$ grow after $P$ proves (T12), without a uniform-in-$k$ threshold asymptotic.

At every odd-exponent support prime other than two and five, (T3) prohibits $\chi (p)=-1$. Thus its value is $1$ unless $p\mid q$, when it is zero. The latter correction is bounded by

$$r\sum _{p\mid q}\frac{\log p}{p}=O_r(\log \log (q+3)).\text{\hspace{1em}(T13)}$$

For large $q$, split at $p=\log q$: the smaller-prime contribution is $O(\log \log q)$ by Mertens, and the remaining contribution is at most $(\log q{)}^{-1}{\sum }_{p\mid q}\log p\le 1$. The fixed primes two and five cost $O_r(1)$. In a range $q\le P^{\eta }$ with fixed $\eta >0$, these corrections are $o(L)$.

Define the same-character sum over all primes, including those beyond $P$,

$$S_r(\chi ;P)=\sum _p\frac{\chi (p)\log p}{p}{\left(r-\frac{\log p}{L}\right)}_{+}.$$

Its total unsigned weight is $(r^2/2+o(1))L$. Give all primes outside the protected odd layers the worst permissible value $-1$; keep the ramified corrections (T13). This yields

$$S_r(\chi ;P)\ge \left(2r\log 2-\frac{{\pi }^2}{12}-\frac{r^2}{2}+o(1)\right)L.\text{\hspace{1em}(T14)}$$

To compare (S1), put $t=L/\log q$ and $\alpha =rt$. Then $f_{\alpha }(\log p/\log q)=t(r-\log p/L{)}_{+}$. In a contradiction range $P^{1/8}<q\le P^{\eta }$, supplied by (T10), $\alpha$ lies in a fixed compact interval. For fixed order bound two, the error in (S1) is uniform over the varying characters and moduli at each fixed mesh value of $\alpha$, as noted in the source citation. To make it uniform over this compact interval, use a finite mesh, the bound $|f_{\alpha }-f_{\beta }|\le |\alpha -\beta |$, and ${\sum }_{p\le q^A}\log p/p=O_A(\log q)$. First take $q\to \infty$ for this finite mesh, then let its spacing tend to zero. Consequently

$$S_r(\chi ;P)\le \frac{r}{8}\log q+o_{r,\eta }(L).\text{\hspace{1em}(T15)}$$

Combining (T14)–(T15) gives

$$\frac{\log q}{\log P}\ge 16\log 2-\frac{2{\pi }^2}{3r}-4r-o(1).$$

The right side is maximized for $r\ge 1$ at $r=\pi /\sqrt{6}$. Hence, with

$$c_{\ast }:=16\log 2-\frac{8\pi }{\sqrt{6}}=0.829956247664\dots ,$$

one obtains the paper-level necessary condition

$$\boxed{q>P^{\eta }\text{eventually for every fixed }0<\eta <c_{\ast }}\text{\hspace{1em}(T16)}$$

for every actual CA source in this canonical branch, including intermediate ties. For $\eta \le 1/8$, (T10) already suffices; the compact-parameter contradiction proves the remaining range. No endpoint $\eta =c_{\ast }$, branch infinitude or effective onset is claimed. The constant is the optimum of this particular triangular weight family, not an optimum over all explicit-formula weights.

The later odd layers are material. Keeping only the terminal logarithmic band $(1/2,1]$ would instead give the lower coefficient $r-3/4-r^2/2=-(r-1{)}^2/2-1/4<0$, so this weight supplies no contradiction from that band alone. No square-depth distribution estimate is supplied here.

Bound (T16) alone does not cross the sufficient faithful-mask cutoff: $c_{\ast }b<1$ for ${\beta }_0<b<1/4$. The full mask has the additional restriction (T22). The conductor-only bound proves neither $q^b\le P$ nor the faithful-mask condition. No weighted missing-prime deficit, signed Robin margin, h=0 exclusion, square-$D$ exclusion or RH proof follows. Every character and layer in the comparison belongs to the same actual integer; no separately realized raw-$D$ witnesses are joined.

A quantified same-source eligibility condition that would cross the cutoff

The preceding bound alone does not cross the sufficient cutoff. The existing set ${\mathcal{E}}_1$, however, gives a specific missing FIB readout for this signed supplier. For fixed $r\ge 1$, define its actual moment

$$M_r(n)=\sum _{\begin{array}{c}p\le P,p\ne 2,5\\ v_p(n)\mathrm{even}\\ p^{v_p(n)/2}\mid c\end{array}}\frac{\log p}{p}\left(r-\frac{\log p}{L}\right).\text{\hspace{1em}(T17)}$$

This is the weight of the same source’s supported potential exceptions, not the weight of an independently selected residue class. Its membership is determined by the existing readout $c\mathrm{mod}{H}_1$. The total support weight is $(r-1/2+o(1))L$. Every negative support prime outside two and five belongs to this set. Keeping the conductor-zero cost (T13), one can therefore replace (T14) by

$$S_r(\chi ;P)\ge \left(2r-1-\frac{r^2}{2}+o(1)\right)L-2M_r(n).\text{\hspace{1em}(T18)}$$

In any fixed power range $q\le P^C$, (T15) then gives the conditional comparison

$$\frac{\log q}{L}\ge 16-\frac{8}{r}-4r-\frac{16M_r(n)}{rL}-o(1).\text{\hspace{1em}(T19)}$$

The range is available whenever the same source satisfies $m_1^b\le Y=\log n$, since $q\le m_1$ and $Y\sim P$. Fix $b=6/25$, which lies in ${\beta }_0<b<1/4$, and $r=\sqrt{2}$. For any fixed coefficient

$$0\le \mu <\frac{71\sqrt{2}}{96}-1=0.045928780505\dots ,$$

the two conditions

$$M_{\sqrt{2}}(n)\le \mu \log P,m_1^{6/25}\le \log n\text{\hspace{1em}(T20)}$$

are incompatible at every sufficiently large actual CA source in the $h=1$, nonsquare-$D$ branch. Indeed, (T19)’s lower coefficient exceeds $25/6$, whereas the cutoff gives $\log q/L\le 25/6+o(1)$. The strict fixed gap absorbs both errors. This is a conditional application of the same published signed estimate. Formula (T21) supplies its moment hypothesis whenever the actual cutoff holds; formula (T22) rules out that cutoff at sufficiently large sources in this branch.

The unrestricted even-layer envelope is only

$$M_r(n)\le \left(r(1-\log 2)-\frac{1}{2}+\frac{{\pi }^2}{24}+o(1)\right)L,$$

whose coefficient at $r=\sqrt{2}$ is $0.345188935617\dots$. It does not imply the required $0.045928\dots$ bound without further conditions. The unrestricted actual-source upper estimate for square-depth membership in $c=4A+7B$ remains missing. The full large-mask signed budget and remaining branches require their own estimates. No distribution law, h=0 or square-$D$ exclusion, reciprocal-prime deficit or RH proof is supplied.

The actual mask pays the moment and leaves the reachable cutoff

The modulus in (P4) is $m_1=\mathrm{lcm}(q,R_1)$, so both $q\le m_1$ and $R_1\le m_1$ hold for the same actual canonical source. A polynomial mask cap therefore controls the moment in (T17), in addition to the primitive conductor. This is an application of existing results, without a new analytic theorem, originality claim or Lean verification.

Continue on the actual CA source, whose initial support contains every prime through $P$. For fixed $C>0$, suppose $m_1\le P^C$. Then every prime in (T17) divides $R_1$; splitting at $L=\log P$ gives

$$0\le M_r(n)\le r\sum _{p\mid R_1}\frac{\log p}{p}\le r\sum _{p\le L}\frac{\log p}{p}+\frac{r}{L}\log R_1\le r\log L+O_{r,C}(1)=o(L).\text{\hspace{1em}(T21)}$$

This reuses the same two-range Mertens estimate as (T13), here split at $L=\log P$. It requires no independence, residue distribution or additional upper hypothesis about the same source’s square-depth membership. In particular the cutoff $m_1^{6/25}\le \log n\sim P$ would imply the moment condition in (T20) for any fixed $0<\mu <71\sqrt{2}/96-1$, at sufficiently large sources. The combined (T19) lower coefficient $16-8\sqrt{2}=4.686291\dots$ then exceeds $25/6$. This already rules out that joint cutoff.

A stronger existing supplier directly applies to the faithful mask. The induced character ${\tilde{\chi }}_1\mathrm{mod}{m}_1$ is nonprincipal and quadratic: reduction from units modulo $m_1$ to units modulo $q$ is surjective, so a negative unit of the primitive character has a unit lift. Formula (P4) says every prime with ${\tilde{\chi }}_1(\ell )=-1$ is missing from the actual integer. For every actual CA maximizer with largest prime $P$, including ties, all primes through $P$ are supported. Thus every such negative prime satisfies $\ell >P$.

Along any unbounded sequence in this actual branch, the mask tends to infinity with $P$. Indeed a negative unit represented by $1\le a<m_1$ has a negative prime factor $\ell \le a<m_1$; faithfulness gives $P<\ell <m_1$. Reuse the original published Theorem 1.1 above, whose character need not be primitive. For every fixed $\epsilon >0$ and sufficiently large masks, it supplies a negative prime

$$P<\ell \le m_1^{{\beta }_0+\epsilon },{\beta }_0=\frac{1}{4\sqrt{e}}.$$

Consequently, for every fixed $0<\eta <4\sqrt{e}$, choosing $\epsilon$ with $\eta ({\beta }_0+\epsilon )<1$ gives

$$\boxed{m_1>P^{\eta }\text{eventually}.}\text{\hspace{1em}(T22)}$$

The statement covers every actual $h=1$, nonsquare-$D$ CA source with its own faithful mask, including intermediate ties; it asserts no infinitude of this branch, endpoint $\eta =4\sqrt{e}$ or numerical onset. It is a lower bound for the full mask, not for the primitive conductor $q$ alone. This direct use of Pollack’s theorem is stronger than the mask bound obtainable from (T21) and the triangular signed estimate; neither analytic proof is repeated.

For any fixed $b>{\beta }_0$, choose $1/b<\eta <4\sqrt{e}$ in (T22). Since $P\sim \log n$ on these actual CA sources,

$$\frac{m_1^b}{\log n}⟶\infty \text{\hspace{1em}(T23)}$$

along every unbounded sequence in this branch. Thus the sufficient cutoff $m_1^b\le \log n$ has no sufficiently large realization there, including the weighted supplier’s entire ${\beta }_0<b<1/4$ regime. The conductor-only lower bounds in (T10) and (T16) remain valid; their numerical power comparison alone omitted this further mask constraint.

This excludes applicability of that small-mask route at large CA sources; it proves no Robin violation or safety there. Outside a polynomial mask cap, (T21) supplies no $o(L)$ bound for the actual moment. The full large-mask signed Robin estimate, unit bit zero and square discriminants remain unresolved. An average on free Beatty indices or independently selected residues cannot discharge this same-source obligation.

Small conductor forces a large actual exception radical

The same moment budget remains useful when the primitive conductor is polynomial but the full mask is not. The split at $z=\max \{2,\log R_1\}$, again using the existing Mertens estimate, gives the uniform elementary bound

$$M_r(n)\le r\log \log (R_1+3)+O_r(1).\text{\hspace{1em}(T24)}$$

For the large primes the contribution is at most $r(\log R_1)/z\le r$; the smaller primes contribute $r\log z+O_r(1)$. This uses the actual $R_1$, including any overlap with the conductor, without treating the two factors as coprime.

Fix $0<C<K:=16-8\sqrt{2}$ and suppose the same actual source has $q\le P^C$. Formula (T19) is applicable through its fixed-power range and the existing bootstrap (T10), whether or not $m_1$ is polynomial. Taking $r=\sqrt{2}$ and inserting (T24) yields

$$\frac{\log q}{\log P}+16\frac{\log \log (R_1+3)}{\log P}\ge K-o(1).\text{\hspace{1em}(T25)}$$

Therefore for every fixed $0<\tau <(K-C)/16$ one has, eventually at every actual $h=1$, nonsquare-$D$ CA source satisfying $q\le P^C$,

$$\boxed{\log R_1>P^{\tau },m_1\ge R_1>\exp (P^{\tau }).}\text{\hspace{1em}(T26)}$$

For example $C=25/6$ permits every fixed $0<\tau <0.0324765521467\dots$. This is a conditional tradeoff between the same integer’s conductor and actual exception radical; it does not assert existence of infinitely many sources in that conductor range. No endpoint exponent, effective onset or universal distribution of the canonical lift is claimed. It quantifies why retaining the primitive conductor alone can lose the dominant mask scale, without proving Robin.