bibkey: fabbian2026mertens authors: Giacomo Fabbian year: 2026 title: An explicit Mertens-product bound at the Morrill–Platt threshold, with applications to Robin’s inequality doi: 10.5281/zenodo.23025480 url: https://doi.org/10.5281/zenodo.23025480 claim: The preprint’s uniform Mertens-product bound implies Robin for integers with v₂ at most 25; the necessary interval constants were checked, while external analytic premises remain literature inputs. This filter is weaker than existing least-counterexample restrictions. strata_touched: [] license: citation-only triage: anchor

Explicit Mertens bound and the single-valuation Robin filter

The primary source is the Zenodo preprint, record 23025480, version 1.0, published 29 September 2026. Its PDF and supplement were inspected; no peer-reviewed publication or Lean verification is asserted. The analytic reduction below uses its stated external premises, not a new proof of those premises.

Exact interface

With $\vartheta (x)={\sum }_{p\le x}\log p$, Proposition 1 states, for every real $x\ge X_0$,

$$\sum _{p\le x}\log \frac{p}{p-1}-\gamma -\log \log \vartheta (x)\le B,X_0=29996208012611,B=1.3064373\cdot 10^{-8}.$$

Theorem 2 combines this bound with the existing Morrill–Platt finite verification: for a prime $p$ and an integer $J\ge 1$, the condition

$$-\log (1-p^{-J-1})>B$$

implies Robin’s inequality for every $n>5040$ with $v_p(n)\le J$. In particular, $2^{26}\nmid n$ suffices; being 26-free is a stronger restriction on $n$ and therefore a corollary. This does not exclude integers divisible by $2^{26}$.

The reduction uses ordered modified Euler factors (Lemma 3.2). If $N_k\le n<N_{k+1}$ for consecutive primorials and $p\le p_k$, it bounds $Z(n)=\sigma (n)/n$ by $(1-p^{-J-1})N_k/\phi (N_k)$. The argument also allows $p\nmid n$; it does not insert an absent prime into the actual Euler product. The displayed finite verification covers $5040<n\le X_0^{\#}$ by combining the two Morrill–Platt ranges. In arXiv:1809.10813v4, these are Theorem 13 and Corollary 14; the paper cites the published numbering 5 and 2. This imports an existing finite interval, without extending or rerunning it.

Verified constants and remaining dependencies

The supporting derivation in §§2–5 was checked for the selected $p=2,J=25$ consequence. In particular, the integrated explicit formula uses RH only at the already verified finite height $3\cdot 10^{12}$; its remaining zeros and infinite tail are bounded unconditionally. No use of global RH was found in that inspected chain. The external inputs are the Morrill–Platt verification, Platt–Trudgian finite-height zero verification, the integrated explicit formula and classical zero sum/count, Rosser–Schoenfeld bounds, and Büthe and Platt–Trudgian bounds for $\vartheta$. The original Morrill–Platt statements were checked; the other external original proofs and computations were not all re-audited.

The required C2–C3 checks and only the $p=2,J=25$ part of C4 were executed with python-flint 0.8.0, Python 3.12.13, 160-bit interval arithmetic and one thread. All selected strict comparisons passed, with exit code zero. The containing balls for the two final margins were

$$B-(A^{\ast }+P_2^{+}(X_0)+P_1^{+}(X_0))\in [1.13086113662967602691919\cdot 10^{-14}\pm 7.55\cdot 10^{-38}],$$

$$-\log (1-2^{-26})-B\in [1.83678830486995981542314977235185765273209096\cdot 10^{-9}\pm 1.57\cdot 10^{-54}].$$

Both lower endpoints are positive. The auxiliary sign conditions for the zero-count bound, the zeta logarithmic derivative at $-1$, and the $u,\kappa$ hypotheses were also checked. The full C4 prime catalogue, its maximality assertions, missing-prime threshold, and external finite Robin intervals were not rerun. These constant checks do not constitute a kernel proof or independent verification of every analytic premise.

Source integrity and reproduction boundary

The separate reproducibility.zip in the cited record contains the correct Robin certificate program; all 49 entries in its SHA-256 manifest matched. The archive SHA-256 is 258cca99fe0cc326d7ef325e07d6ba4d80502a4c3b4b0e7b930ed3901f22a44c. The checked robin_bounds.py has SHA-256 d38b6c86b742f4dd921bac593a92971d58612d8e15da75dde15bbfd66f3c8fd4.

The record’s source.zip, despite matching its advertised size and MD5, contains an unrelated prime-level GL(2) simple-zero manuscript. It was not used as the Robin source. This attachment mismatch does not imply that the separate certificate package is absent or invalid. Also, run_all.sh records individual child return codes without aggregating failure; its own successful exit alone is not a certificate. The selected checks called the relevant library functions and asserted each required interval comparison directly.

Reuse boundary for FIB and CA research

The valuation filter can be applied to an actual integer whose FIB source has already supplied its exact valuation. It supplies no new relation between a FIB address and divisibility. Its fixed bound on the joint signed prime quantity does not reach the finer scale needed to exclude the surviving candidates.

The preprint itself explains that its necessary conditions for arbitrary counterexamples are weaker than known conditions on a least counterexample, which must be superabundant. Its cited single-prime exchanges already give $v_2\ge 43$ and $v_3\ge 27$ there; these numbers are not the precise values of the stronger floor-log estimates. Thus the $2^{26}$ filter does not improve exclusion of the classical least-counterexample test set. Nor does it change the zero self-cutoff deficit at every CA integer or supply the missing actual-candidate signed estimate. It is a reusable source restriction, not progress from a proportion or finite filter to full RH.

Variable primitive allowance at surviving selected sources

Keeping the variable in the preprint’s primitive bound does not by itself reach the complete signed Robin target at the remaining selected sources. The following is a source-specific comparison of existing estimates, not a new prime-error bound, general RH criterion or originality claim. It is a paper derivation without Lean certification.

The primary allowance before the fixed constant

In the cited version-1 PDF, §4.1, printed p.8, let

$$w(t)=\frac{1+\log t}{t^2{\log }^{2}t},P_1(x)={\int }_x^{\infty }(t-\psi (t))w(t)dt=-I_{\psi }(x).$$

Use $H=3\cdot 10^{12}$ for the source’s verified height, called $T$ in the paper, and put

$$S=2+\gamma -\log (4\pi )>0,{\Sigma }^{\prime }=\frac{2(\log (H/(2\pi ))+1)}{\pi H},$$

$$M(t)=S{t}^{3/2}+{\Sigma }^{\prime }{t}^2+t\log (2\pi )+2+\frac{\log 2}{t}.$$

Lemma 4.4, printed p.9, bounds the absolute primitive error by $M(t)$. For zeros above $H$ it uses only $\beta \le 1$, replacing $t^{\beta +1}$ by $t^2$; the resulting positive allowance is ${\Sigma }^{\prime }{t}^2$. Lemma 4.5, printed pp.9–10, states, for $1<x\le Y$,

$$|P_1(x)|\le U(x,Y):=M(x)w(x)+M(Y)w(Y)+{\int }_x^{Y}M(t)|w^{\prime }(t)|dt+{\int }_Y^{\infty }ϵ(t)tw(t)dt.\text{\hspace{1em}(V1)}$$

Here $ϵ$ is the nonnegative pointwise envelope from the source’s (P6)(c). Every term of $U$ is nonnegative. The original analytic inputs and selected certificate checks are reused as described above; none is independently reproved or rerun.

Compare at the actual Robin source

Keep the conditionally selected least global Robin maximizer $N$ and its clock $A=\log N$, $L=\log A$, $T_A=\sqrt{A}\log A$. The existing selected-source restriction, (Z8), places every surviving such source at $A>3\cdot 10^{23}$. This is a restriction on that selected maximizer, not a least-counterexample bound or an all-integer finite verification. The effective core comparison, (G7)–(G9), supplies the sufficient signed target

$$T_A{I}_{\psi }(A)\ge -\mathcal{E}(L),\mathcal{E}(L)<2\sqrt{2}-2.00014<1(L\ge 26).\text{\hspace{1em}(V2)}$$

At this same clock, the first endpoint term in (V1) alone gives, for every $Y\ge A$,

$$T_{A}U(A,Y)\ge T_A{\Sigma }^{\prime }{A}^{2}w(A)={\Sigma }^{\prime }\sqrt{A}\left(1+\frac{1}{\log A}\right).\text{\hspace{1em}(V3)}$$

The elementary bounds $e<3$ and $\pi <4$ give

$$\frac{H}{2\pi }>\frac{H}{8}=375000000000>3^{24}=282429536481>e^{24}.$$

Thus $\log (H/(2\pi ))>24$ and ${\Sigma }^{\prime }>50/(12\cdot 10^{12})>4\cdot 10^{-12}$. Also $A>3\cdot 10^{23}>(5\cdot 10^{11}{)}^2$, so (V3) yields

$$\boxed{T_{A}U(A,Y)>2>\mathcal{E}(L)(Y\ge A).}\text{\hspace{1em}(V4)}$$

Consequently the direct absolute primitive allowance (V1), even with its handoff $Y$ optimized and its variable dependence retained, cannot pay (V2) at any surviving selected source. This is a lower bound on the allowance $U$, not on $|P_1|$ or the actual signed integral: (V1) still gives only $T_A{I}_{\psi }(A)\ge -T_{A}U(A,Y)$. No adverse sign of the actual error, Robin counterexample or impossibility of a sharper method follows.

The obstruction is the positive ${\Sigma }^{\prime }{t}^2$ envelope for unverified zeros, already present at the starting endpoint. A smaller majorant using additional zero information, or a genuinely signed estimate, would require a different supplier. The existing reflected-zero and zero-free-region estimates in the Polak note remain separate, stronger inputs; they are not repeated or added to this allowance. The complete same-source signed target and RH remain unproved.

A near-Euler contour for the complete negative-heat trace

The absolute primitive allowance (V1) remains insufficient at the selected source by (V4). A different interface is available from the complete signed zero trace already retained in the Nicolas comparison note, (NH1), (NH8)–(NH18). The result here improves its joint comparison error; it does not supply a signed main-term bound.

Reuse the cited Dobner negative-time convolution and positive Fourier kernel, the DLMF digamma sector estimate, Euler products, Cauchy estimates and the complete contour balance. These are literature-attested inputs. The parameter-dependent full-trace estimate below is a repo-derived paper-level application. Within the inspected existing notes and cached primary interfaces, no supplier of this exact joint estimate was identified; no exhaustive literature, mathematical-priority, numerical-constant or Lean-certification claim is made.

Keep $A\ge e^2$, $L=\log A$, and the complete principal continuation

$${\hat{F}}_A(z)=\frac{A^{z-1}}{zL}+E_1((1-z)L),Z_{ϵ}(A)=\sum _{{\xi }_{-ϵ}(w)=0}{\hat{F}}_A(w).$$

Both ordinate signs and every actual multiplicity are included. At zero time this is exactly the original trace. There are absolute constants $c>0$ and $C<\infty$ such that

$$\boxed{|Z_{ϵ}(A)-Z_0(A)|\le CϵA^{\eta }{\eta }^{-8},0<\eta \le \frac{1}{2},0<ϵ\le c{\eta }^4,ϵL\le 1.}\text{\hspace{1em}(VE1)}$$

In particular, choosing $\eta =1/L$ gives

$$\boxed{|Z_{ϵ}(A)-Z_0(A)|\le CϵL^8,0<ϵ\le c{L}^{-4}.}\text{\hspace{1em}(VE2)}$$

The constant $c$ may be reduced so that the small-time and $ϵL\le 1$ hypotheses hold in (VE2). The exponent $8$ is a paid upper-bound exponent, not an optimality assertion.

Pay normalization on a half-plane approaching the Euler boundary

Retain $J_{ϵ}(s)=s+(ϵ/4)\mathrm{Log}(s/(2\pi ))$, $\gamma (s)=s(s-1){\pi }^{-s/2}\Gamma (s/2)/2$ and $\gamma_{-\epsilon}(s)=\gamma(s) \exp((s-J_\epsilon(s))^2/\epsilon)$ from (NH7). Set

$$U_{ϵ}(s)=\frac{{\xi }_{-ϵ}(J_{ϵ}(s))}{{\gamma }_{-ϵ}(s)},R_s(v)=\frac{\gamma (s+iv)}{\gamma (s)}{e}^{-iv\mathrm{Log}(s/(2\pi ))/2}.$$

The exact Gaussian identity (NH8) extends to $\Re s>1$:

$$U_{ϵ}(s)=\frac{1}{\sqrt{\pi ϵ}}{\int }_{\mathbb{R}}{e}^{-v^2/ϵ}{R}_s(v)\zeta (s+iv)dv.\text{\hspace{1em}(VE3)}$$

Indeed, both sides are analytic there. The bound below makes the integral locally uniformly convergent, and the identity theorem extends the already retained equality; no new heat-convolution proof is needed.

For $\Re s\ge 1+\eta /2$, factor the two polynomial factors out of $\gamma$. The imaginary part of the log derivative of $\Gamma (s/2)$ is uniformly bounded on $\Re (s/2)\ge 1/2$, by the same fixed-sector DLMF expansion and its compact remainder. Consequently, for an absolute $C_0$,

$$|R_s(v)|\le C_0{\eta }^{-1}(1+|v|{)}^2{e}^{C_0|v|},|\zeta (s+iv)|\le C_0{\eta }^{-1}.\text{\hspace{1em}(VE4)}$$

The possible small denominator $s-1$ is paid by ${\eta }^{-1}$; it is not integrated into an $e^{C|v|/\eta }$ envelope.

Put $G_s(v)=R_s(v)\zeta (s+iv)$. On $|v|\le 1$,

$$|G_s^{\prime \prime }(v)|\le C_1{\eta }^{-3}.\text{\hspace{1em}(VE5)}$$

There are two complementary reasons. If $|s|>4$, the vertical segment stays uniformly away from both $0$ and $1$. The centered log-Gamma estimates and their Cauchy derivatives bound $R_s,R_s^{\prime },R_s^{\prime \prime }$ by an absolute constant. The absolutely convergent series bound ${\zeta }^{(j)}(s+iv)$ by $C_j{\eta }^{-j-1}$ for $j=0,1,2$. If $|s|\le 4$, instead use the exact cancellation

$$G_s(v)=\frac{{\xi }_0(s+iv)}{\gamma (s)}{e}^{-iv\mathrm{Log}(s/(2\pi ))/2}.$$

The numerator and its first two derivatives are bounded on this fixed compact set, whereas $|\gamma (s)|\ge c_1|s-1|\ge c_1\eta /2$. This gives $O({\eta }^{-1})$ there, hence (VE5). Treating the Gamma zero and zeta pole independently on this compact set would not justify the stated exponent.

Center the Gaussian before taking absolute values. Its linear moment vanishes and its second moment is $ϵ/2$. On $|v|>1$, (VE4) and the affine subtraction at zero pay the full tail by $O(ϵ{\eta }^{-3})$ for $ϵ\le 1/2$. Thus

$$U_{ϵ}(s)-\zeta (s)=O(ϵ{\eta }^{-3})(\Re s\ge 1+\eta /2).\text{\hspace{1em}(VE6)}$$

Euler inversion gives $|\zeta (s)|\ge 1/\zeta (\Re s)\ge c_2\eta$. Choosing $c$ sufficiently small in $ϵ\le c{\eta }^4$ ensures $|U_{ϵ}(s)|\ge c_2\eta /2$ on this whole half-plane. This pays the zero-free exterior for the deformed function, rather than applying an original-zeta zero-free region to different zeros.

Take logarithmic derivatives with their denominators paid

On $s=b+iy$, $b=1+\eta$, apply Cauchy’s estimate to (VE6) on radius-$\eta /4$ disks. It gives $U_{ϵ}^{\prime }-{\zeta }^{\prime }=O(ϵ{\eta }^{-4})$. The Euler logarithmic derivative satisfies

$$\left|\frac{{\zeta }^{\prime }}{\zeta }(b+iy)\right|\le \sum _{n\ge 2}\frac{\Lambda (n)}{n^b}=\frac{-{\zeta }^{\prime }(b)}{\zeta (b)}\ll {\eta }^{-1}.$$

Here $\zeta (b)\ge 1/\eta$ by integral comparison, and $-{\zeta }^{\prime }(b)\ll {\eta }^{-2}$. Use the identity

$$\frac{U_{ϵ}^{\prime }}{U_{ϵ}}-\frac{{\zeta }^{\prime }}{\zeta }=\frac{U_{ϵ}^{\prime }-{\zeta }^{\prime }}{U_{ϵ}}+\frac{{\zeta }^{\prime }}{\zeta }\frac{\zeta -U_{ϵ}}{U_{ϵ}}.$$

Both terms are $O(ϵ{\eta }^{-5})$. The exact normalization has log derivative $\gamma’/\gamma+ \epsilon\operatorname{Log}(s/(2\pi))/(8s)$. Therefore, with $H_{ϵ}={\xi }_{-ϵ}^{\prime }/{\xi }_{-ϵ}$ and $K_{ϵ}(s)=H_{ϵ}(J_{ϵ}(s))J_{ϵ}^{\prime }(s)$,

$$K_{ϵ}(s)-H_0(s)=O(ϵ{\eta }^{-5}),|H_0(s)|\ll {\eta }^{-1}+\log (|s|+2).\text{\hspace{1em}(VE7)}$$

All constants are independent of $A,\eta ,ϵ,y$.

Transport both complete coefficients near their singularities

Reuse the exact derivative (NH15):

$${\hat{F}}_A^{\prime }(z)=A^{z-1}{h}_L(z),h_L(z)=\frac{1}{z(1-z)}-\frac{1}{L{z}^2}.$$

Put $r=|b+iy|$, $\kappa =ϵL/4\le 1/4$ and $Q_\epsilon(s)=\widehat F_A(J_\epsilon(s))+ \widehat F_A(1-J_\epsilon(s))$. On the two unshifted lines, the rational brackets and their derivatives obey the joint bounds $C{\eta }^{-2}{r}^{-2}$ and $C{\eta }^{-3}{r}^{-3}$. Integrate the derivative to the corresponding imaginary infinity, where (NH2) gives zero, and integrate its $e^{\pm iLt}$ factor once by parts. Since $L\ge 2$, this gives

$$|Q_0(s)|\ll A^{\eta }{\eta }^{-3}{r}^{-2}.$$

Each ordinate sign uses its own branch; at $y=0$ take the one-sided limits. The reflected line has exponential factor $A^{-1-\eta }$, which is bounded by $A^{\eta }$.

After reducing $c$ if needed, every segment from $s$ to $J_{ϵ}(s)$ stays in $\Re z>1+\eta /2$. Its ordinate retains its sign, $|z|\asymp r$, and $|1-z|\ge c_3\eta r$; the reflected segment pays these same distances with the roles interchanged. The displacement is $O(ϵ\log (r+2))$, and the right-hand exponential factor is at most $C{A}^{\eta }{r}^{\kappa }$. The reflected exponential factor is at most $C{A}^{-1-\eta }$, using $ϵL\le 1$ when the real displacement is negative. Consequently

$$\begin{array}{rl}|Q_{ϵ}-Q_0|& \ll ϵA^{\eta }{\eta }^{-2}\log (r+2)r^{\kappa -2},\\ |Q_{ϵ}|& \ll A^{\eta }{\eta }^{-3}\log (r+2)r^{\kappa -2}.\end{array}\text{\hspace{1em}(VE8)}$$

The cancellation in ${\hat{F}}_A$ is retained; neither coefficient is replaced by its leading term.

Keep the all-height contour, slit and pole corrections

The injectivity, boundary-graph and properness argument in (NH17) applies unchanged to the line $\Re s=b$: $\Re s>1$ is enough for its positive secant real part. Equation (VE6) now pays its entire right-hand exterior. Functional reflection pays the left-hand exterior. With $b_{ϵ}=J_{ϵ}(b)>1$, the exact balance is

$$Z_{ϵ}(A)=\frac{1}{2\pi i}{\int }_{b-i\infty }^{b+i\infty }{Q}_{ϵ}(s)K_{ϵ}(s)ds+\log \frac{{\xi }_{-ϵ}(b_{ϵ})}{{\xi }_{-ϵ}(1)}-\frac{H_{ϵ}(0)}{AL}.\text{\hspace{1em}(VE9)}$$

The principal cut jump remains $+2\pi i$ on $x>1$, and the coefficient residue at zero remains $1/(AL)$. The real logarithms are valid by positivity of the original Fourier kernel. No cut or pole term is absorbed into an unspecified main trace.

Reuse the complete deformed zero count, Hadamard logarithmic derivative and good-height construction from (NH10), (NH17). At each fixed $A,\eta ,ϵ$, the two boundary curves still have real width $O(\log T)$, and the complete coefficients have the same $T^{\kappa -2}$ bound. Their horizontal ends therefore tend to zero with the retained $O_{A,\eta ,ϵ}(T^{\kappa -1}{\log }^{2}T)$ allowance. No new zero trajectories, simplicity hypothesis or finite spectrum truncation is introduced. The vertical integrals are absolutely convergent by (VE7)–(VE8).

Subtract (VE9) at zero time, on the same line. Equations (VE7)–(VE8) give

$$\left|\frac{1}{2\pi i}\int [Q_{ϵ}{K}_{ϵ}-Q_0{H}_0]ds\right|\ll ϵA^{\eta }{\eta }^{-8}{\int }_{\mathbb{R}}{r}^{\kappa -2}{\log }^2(r+2)dy\ll ϵA^{\eta }{\eta }^{-8}.$$

Here $1<b\le 3/2$ and $\kappa \le 1/4$, so the last integral has an absolute uniform bound. The slit and pole differences are $O(ϵ)$ uniformly in $\eta$: $b\in [1,3/2]$, $b_{ϵ}-b=O(ϵ)$, and the positive Fourier values, their first real derivatives and their time derivatives have common bounds on a fixed real compact interval. These are exactly the kernel-moment and denominator inputs already used in (NH18). Also $1/(AL)\le 1$. This proves (VE1) with all corrections paid; $A^{1/L}=e$ gives (VE2).

A broader common time, without a signed Robin conclusion

Choose one time for the whole spectrum at each clock,

$$ϵ(A)=\min \{(c/2)L^{-4},A^{-1/2}{L}^{-10}\}.$$

Then (VE2), with the unchanged positive elementary correction $r_A$ from (G9), gives

$$\boxed{-\sqrt{A}L{I}_{\psi }(A)=\sqrt{A}L{Z}_{ϵ(A)}(A)+r_A+O(L^{-1}).}\text{\hspace{1em}(VE10)}$$

The implicit constant is absolute. For sufficiently large $A$ the second branch is selected, so this comparison allows an asymptotically larger common negative time than $A^{-3}$. It never selects a different time per zero.

The original conditional least integer attaining the global Robin-ratio maximum under a violation remains the source, with $A=\log N>(7/2)10^{46}$; it is not replaced by a least counterexample. The complete original coefficients, actual real parts, both signs, all heights and multiplicities return through (VE1), and $r_A$ and the strict core remain required. No numerical $c,C$ or finite-clock slack has been certified. The signed upper bound for $\sqrt{A}L{Z}_{ϵ(A)}(A)$, the complete Robin target and RH remain unproved. The original response and its accepted (NH1)–(NH18) supplier are preserved verbatim in the Nicolas note.

The same negative-time prime filter has different absolute budgets on different cuts

The common-time comparison (VE10) does not restore coefficient positivity. The precise issue is the total negative coefficient budget, not only the already checked negative coefficient at index $6$. The joint estimate below keeps the same finite prime support, the same time and the same Dirichlet coefficients throughout.

Reuse Dobner’s damped Dirichlet series from (NH7)–(NH8) in the Nicolas note, and the finite Euler coefficient definition inspected in Planat, arXiv:2609.37164v2, Lemma 8.1. The latter’s nonnegative-head statement assumes nonnegative heat time and a finite head; neither assumption is extended here. Its divisor convolution and the previously checked two-prime sign identity are reused directly. Euler products, Dirichlet norm inequalities and the classical prime number theorem are literature-attested inputs. The growing-support aggregate budget below is a repo-derived paper application, without a priority, exhaustive literature, numerical-constant or Lean claim.

For $ϵ>0$, write $a_\epsilon(n)= \exp[-\epsilon(\log n)^2/4]$.Forafinitesetofprimes$P$, define

$${\mathcal{C}}_{ϵ,P}(s)=D_{ϵ}(s)\prod _{p\in P}(1-a_{ϵ}(p)p^{-s})=\sum _{n\ge 1}{c}_{ϵ,P}(n)n^{-s},$$

$$c_{ϵ,P}(n)=\sum _{d\mid (n,{\prod }_{p\in P}p)}\mu (d)\prod _{p\mid d}{a}_{ϵ}(p)a_{ϵ}(n/d).\text{\hspace{1em}(VF1)}$$

This coefficient formula agrees with the inspected finite filter when $P=\{2,3,5\}$. For fixed $ϵ>0$ and finite $P$, the series is absolutely convergent on every vertical line: Gaussian damping makes ${\sum }_n{a}_{ϵ}(n)n^{-\sigma }$ finite for every real $\sigma$, and the finite divisor convolution preserves that property. Thus the full negative mass

$${\mathcal{N}}_{ϵ,P}(\sigma )=\sum _{n\ge 1}[-c_{ϵ,P}(n){]}_{+}{n}^{-\sigma }\text{\hspace{1em}(VF2)}$$

is finite at each fixed time. No infinite positive-time Dirichlet series is asserted.

The right-of-Euler budget is uniformly small

For every finite $P$, every $0<\eta \le 1/2$ and every $ϵ>0$,

$$\boxed{{\mathcal{N}}_{ϵ,P}(1+\eta )\le Cϵ{\eta }^{-4}.}\text{\hspace{1em}(VF3)}$$

To pay the coefficient norm, use $\Vert \sum b_n{n}^{-s}{\Vert }_{\sigma }=\sum |b_n|n^{-\sigma }$, which is submultiplicative for absolutely convergent Dirichlet series. At zero time the coefficient of $\zeta (s){\prod }_{p\in P}(1-p^{-s})$ is the nonnegative indicator $1_{(n,\prod P)=1}$. The negative part in (VF2) is bounded by the coefficient norm of the difference from that zero-time series.

Let $E_{ϵ}={\prod }_{p\in P}(1-a_{ϵ}(p)p^{-s})$ and $E_0={\prod }_{p\in P}(1-p^{-s})$. On $\sigma =1+\eta$, Euler products and $1-e^{-x}\le x$ give

$$\begin{array}{rl}\Vert E_{ϵ}{\Vert }_{\sigma }& \le \prod _{p\in P}(1+p^{-\sigma })\le \zeta (\sigma )\ll {\eta }^{-1},\\ \Vert D_{ϵ}-\zeta {\Vert }_{\sigma }& \le (ϵ/4){\zeta }^{\prime \prime }(\sigma )\ll ϵ{\eta }^{-3},\\ \Vert E_{ϵ}-E_0{\Vert }_{\sigma }& \le (ϵ/4)\zeta (\sigma )\sum _p(\log p{)}^2{p}^{-\sigma }\ll ϵ{\eta }^{-3}.\end{array}$$

For the last bound, keep the Euler logarithmic derivative:

$$\sum _p(\log p{)}^2{p}^{-\sigma }\le (\log \zeta {)}^{\prime \prime }(\sigma )\le \frac{{\zeta }^{\prime \prime }(\sigma )}{\zeta (\sigma )}\ll {\eta }^{-2}.$$

The last inequality uses ${\zeta }^{\prime \prime }(1+\eta )\ll {\eta }^{-3}$ and $\zeta (1+\eta )\ge 1/\eta$. Subtract the two series as $(D_\epsilon-\zeta)E_\epsilon+ \zeta(E_\epsilon-E_0)$. Both products have norm $O(ϵ{\eta }^{-4})$, independently of $P$, proving (VF3). This uses a coefficient norm; a small function-value error alone would not pay it.

Every finite filter containing 2 has a large interior budget

Fix $0<\sigma <1$ and put $d=1-\sigma >0$. Suppose only that $2\in P$. For every prime $q\ne 2$, the already checked two-prime coefficient identity gives

$$c_{ϵ,P}(2q)=a_{ϵ}(2)a_{ϵ}(q)(e^{-ϵ\log 2\log q/2}-1)<0.\text{\hspace{1em}(VF4)}$$

If $q\notin P$, the two divisor terms are $a_{ϵ}(2q)-a_{ϵ}(2)a_{ϵ}(q)$. If $q\in P$, the four terms are $a_\epsilon(2q)-2a_\epsilon(2)a_\epsilon(q)+ a_\epsilon(2)a_\epsilon(q)$, giving the same coefficient. No other prime of $P$ divides $2q$. Adding more primes to the same filter therefore cannot remove this entire negative family. These are actual distinct integer indices, not independently chosen phases.

Let $X=e^{2d/ϵ}$ and $J=\lfloor {ϵ}^{-1/2}\rfloor$. Use the disjoint prime bins $[2^{j}X,2^{j+1}X)$ for $0\le j<J$. The classical prime number theorem gives one absolute $b>0$ such that every bin starting at $Q\ge X$, for sufficiently small $ϵ$, contains at least $bQ/\log Q$ primes. This is the uniform all-$Q$ consequence of that theorem, not a short-interval or RH-dependent estimate. Every one of these primes supplies (VF4), regardless of membership in $P$; no support deletion or bound on $\max P$ is needed.

Write $h=\log 2$, $u_j=\log (2^{j}X)=2d/ϵ+jh$. For every prime in that bin,

$$1-e^{-ϵh\log q/2}\ge 1-2^{-d},q^{-\sigma }{a}_{ϵ}(q)\ge 2^{-\sigma }{e}^{-\sigma u_j-ϵ(u_j+h{)}^2/4}.$$

Combining this with the bin count, $2^{-\sigma }{a}_{ϵ}(2)$ from (VF4), and the exact exponent identity

$$d{u}_j-\frac{ϵ}{4}(u_j+h{)}^2=\frac{d^2}{ϵ}-\frac{ϵ}{4}(jh{)}^2-\frac{ϵ}{2}{u}_{j}h-\frac{ϵ}{4}{h}^2$$

shows that every bin contributes at least $b_{\sigma }ϵe^{d^2/ϵ}$, for a constant $b_{\sigma }>0$ independent of $P,j,ϵ$. Indeed, $j<{ϵ}^{-1/2}$ bounds all subtracted exponents by constants, and $u_j\le (2d+h)/ϵ$ for $ϵ\le 1$. For $ϵ\le 1/4$, $J\ge 1/(2\sqrt{ϵ})$. Hence

$$\boxed{{\mathcal{N}}_{ϵ,P}(\sigma )\ge b_{\sigma }^{\prime }\sqrt{ϵ}{e}^{(1-\sigma {)}^2/ϵ}}\text{\hspace{1em}(VF5)}$$

for sufficiently small $ϵ$, uniformly over every finite $P$ containing $2$. Only a negative subfamily was used; all other coefficients remain in (VF2).

Apply both budgets at the same common Robin clock

At an actual integer $N$ with $2\mid N$, set $P=P_N=\{p:p\mid N\}$ and $A=\log N$. The same selected least global Robin-ratio maximizer has initial prime support, and hence $2\mid N$, by the retained selected-source reduction. No separate maximizing integer or artificial prime profile is substituted. The uniformity in (VF5) permits this actual support without a cardinality, largest-prime or additional resource assumption.

For the common time already chosen in (VE10), $ϵ(A)=\min \{(c/2)L^{-4},A^{-1/2}{L}^{-10}\}$, the time tends to zero. The two budgets for that same filtered series are

$$\begin{array}{rl}\sqrt{A}L{\mathcal{N}}_{ϵ(A),P_N}(1+1/L)& \le C{L}^{-5}⟶0,\\ {\mathcal{N}}_{ϵ(A),P_N}(\sigma )& \ge b_{\sigma }^{\prime }\sqrt{ϵ(A)}{e}^{(1-\sigma {)}^2/ϵ(A)}⟶\infty (0<\sigma <1\text{ fixed}).\end{array}\text{\hspace{1em}(VF6)}$$

Thus the full raw negative-coefficient cost cannot be treated as a vanishing soft correction on a fixed interior cut merely because the common negative time tends to zero. It can have a small coefficient budget on the paid right-of-Euler cut. The Gaussian saddle, prime density and full-filter invariance of the two-prime coefficient are the joint inputs beyond the isolated sign witness. These are uniform eventual estimates over the allowed integer family, without assuming an infinite sequence of selected global maximizers or certifying the bound at the standing finite clock.

This is an absolute coefficient budget, not a signed trace bound. For every real $\sigma >0$ the same filtered value is in fact $D_{ϵ}(\sigma ){\prod }_{p\in P}(1-a_{ϵ}(p)p^{-\sigma })>0$; large negative mass coexists with compensating positive mass. Finite-head statements, different groupings, extra kernels and paid signed cancellation are not excluded. In particular (VF5) does not refute the positive-time source, contradict (VE10), establish negative signed divergence, or give a Robin counterexample.

The original full $I_{\psi }(A)$ target, complete zero coefficients, actual real parts, both signs, infinite heights and multiplicities, elementary correction and strict core remain unchanged and unproved. No bound for the signed main trace in (VE10) or the actual $D_a$ lower allowance is obtained. The constants and asymptotic starting clocks have not been numerically certified; RH remains unproved.

A uniform shrinking-time cutoff for the complete signed trace

The fixed-parameter tail after (NH11) in the Nicolas note does not state a bound uniform as the clock and heat time change together. The following supplies that interface using the already paid near-Euler exterior in (VE6). The complete coefficient, heat convolution, positive Fourier kernel, Jensen theorem, contour balance and original response comparison are reused; none is reconstructed as a new source theorem. This is a repo-derived paper-level parameter application, without an exhaustive literature, mathematical-priority, numerical-constant or Lean claim.

Keep exactly the complete principal continuation ${\hat{F}}_A$ and actual multiset of ${\xi }_{-ϵ}$ zeros used in (VE1)–(VE10). Assume

$$A\ge e^2,L=\log A,0<\eta \le \frac{1}{2},0<ϵ\le c{\eta }^4,ϵL\le 1,\kappa =ϵL/4.$$

Use the absolute $c$ from (VE1), chosen small enough that $ϵ\le 1/2$ throughout this domain. There is an absolute $C<\infty$ such that, for every $T\ge 2$,

$$\boxed{\sum _{\begin{array}{c}{\xi }_{-ϵ}(w)=0\\ |\Im w|>T\end{array}}|{\hat{F}}_A(w)|\le \frac{C{A}^{\eta }}{L}{T}^{\kappa -1}\log (T+2).}\text{\hspace{1em}(VG1)}$$

The sum includes both ordinate signs, every actual real part and every multiplicity. Its constant is independent of $A,\eta ,ϵ,T$. This strengthens the parameter scope of the fixed-parameter tail; it gives no sign to the retained head.

The existing exterior confines every height uniformly

Put $b=1+\eta$ and write the right boundary from (VE9) as $J_{ϵ}(b+iu)$. Its ordinate and real part are

$$y=u+\frac{ϵ}{4}\arctan (u/b),x=b+\frac{ϵ}{4}\log \frac{\sqrt{b^2+u^2}}{2\pi }.$$

The ordinate is strictly increasing and has the sign of $u$, so $|u|\le |y|$. The zero-free exterior and reflected boundary already proved for (VE9) enclose all zeros, including bounded heights. As $1<b\le 3/2$, their actual coordinates $w=\beta +iy$ obey

$$\boxed{-\eta -\frac{ϵ}{4}\log (|y|+2)\le \beta \le 1+\eta +\frac{ϵ}{4}\log (|y|+2).}\text{\hspace{1em}(VG2)}$$

No asymptotic exceptional set or new zero trajectory is used. The weaker denominator-free upper envelope follows from $\sqrt{b^2+u^2}/(2\pi )\le |y|+2$; functional reflection gives the lower one. In particular $A^{\beta -1}\le A^{\eta }(|y|+2{)}^{\kappa }$.

The positive-kernel Jensen argument of (NH10) also has a common center lower bound on this small-time interval:

$${\xi }_{-ϵ}(1/2)\ge {\xi }_{-1/2}(1/2)>0,\underset{|s-1/2|\le R}{\max }|{\xi }_{-ϵ}(s)|\le 2{\xi }_0(1/2+R).$$

These are inequalities for the deformed function itself. Real Stirling growth and Jensen at radii $R$ and $2R$ therefore give a disk count $CR\log (R+2)$ with an absolute constant. By (VG2), for $T\ge 2$ and $|y|\le T$,

$$|\beta -1/2|\le 1+\frac{1}{8}\log (T+2)\le T.$$

Thus these zeros lie in $|w-1/2|<2T$, and the complete count satisfies

$$\boxed{\#\{w:{\xi }_{-ϵ}(w)=0,|\Im w|\le T\}\le CT\log (T+2),T\ge 2.}\text{\hspace{1em}(VG3)}$$

Multiplicity is retained in both Jensen and this count. No original-zeta zero-count theorem has been assigned to different zeros.

Keep the cancellation in the complete coefficient

Reuse (NH15), with

$${\hat{F}}_A^{\prime }(z)=A^{z-1}{h}_L(z),h_L(z)=\frac{1}{z(1-z)}-\frac{1}{L{z}^2}.$$

For every real $\beta$, $t\ge 2$ and $L\ge 2$,

$$|h_L(\beta +it)|\le \frac{1+1/L}{t^2},|h_L^{\prime }(\beta +it)|\le \frac{2(1+1/L)}{t^3}.$$

On either fixed ordinate sign, the complete coefficient tends to zero at the corresponding imaginary infinity by (NH2). Integrate the exact derivative along that vertical ray and integrate its $e^{\pm iLt}$ factor once by parts. Its endpoint and derivative integral give

$$|{\hat{F}}_A(\beta +iy)|\le \frac{A^{\beta -1}}{L}\left[\frac{1+1/L}{|y{|}^2}+{\int }_{|y|}^{\infty }\frac{2(1+1/L)}{t^3}dt\right]\le \frac{3A^{\beta -1}}{L|y{|}^2},|y|\ge 2.$$

This bound is uniform in $\beta$ and $L$. The two principal branches are kept separately; the positive-height ray crosses no cut, and conjugation supplies the same bound at negative height. Neither the $A^{z-1}/(zL)$ term nor $E_1((1-z)L)$ is estimated alone.

Combine this bound with (VG2)–(VG3) on the dyadic bands $2^{j}T<|y|\le 2^{j+1}T$. Since $0<\kappa \le 1/4$, their entire absolute mass is at most

$$\frac{C{A}^{\eta }}{L}\sum _{j\ge 0}(2^{j}T{)}^{\kappa -1}\log (2^{j+1}T+2)\le \frac{C{A}^{\eta }}{L}{T}^{\kappa -1}\log (T+2).$$

The geometric sums, including their $j$ factors, have common bounds because $2^{\kappa -1}\le 2^{-3/4}<1$. This proves (VG1) over all omitted heights, rather than on a selected subsequence.

One common time now gives a finite signed head

Use the same time already chosen for the full spectrum in (VE10):

$$ϵ(A)=\min \{(c/2)L^{-4},A^{-1/2}{L}^{-10}\},\eta =1/L,T(A)=\sqrt{A}{L}^3.$$

Here $A^{\eta }=e$. Moreover $ϵ(A)\le e^{-L/2}{L}^{-10}$, $\log T=L/2+3\log L$, and $\log L\le L$ for $L\ge 2$ give $\kappa \log T\le (7/8)e^{-L/2}{L}^{-8}<1$. Also $\log (T+2)\le 4L$. Hence (VG1) yields the joint bound

$$\boxed{\sqrt{A}L\sum _{\begin{array}{c}{\xi }_{-ϵ(A)}(w)=0\\ |\Im w|>T(A)\end{array}}|{\hat{F}}_A(w)|\le C{L}^{-2}.}\text{\hspace{1em}(VG4)}$$

It holds throughout the stated parameter domain with an absolute constant, without asserting that the second time branch is selected at one numerical clock. Define the finite, real signed head

$$Z_{ϵ(A)}^{\mathrm{head}}(A)=\sum _{\begin{array}{c}{\xi }_{-ϵ(A)}(w)=0\\ |\Im w|\le T(A)\end{array}}{\hat{F}}_A(w).$$

Positivity of the original Fourier kernel excludes real zeros at this negative time. The head is real by conjugation; its complete multiplicity count is at most $C\sqrt{A}{L}^4$ by (VG3). Finiteness is not a claim of a practical algorithm, computed zeros or certified numerical constants.

The exact signed tail has absolute normalized bound (VG4). Combining it with the unchanged complete comparison (VE10) gives

$$\boxed{-\sqrt{A}L{I}_{\psi }(A)=\sqrt{A}L{Z}_{ϵ(A)}^{\mathrm{head}}(A)+r_A+{\rho }_A,|{\rho }_A|\le C_1{L}^{-1}+C_2{L}^{-2}.}\text{\hspace{1em}(VG5)}$$

The $L^{-1}$ term is the existing whole-trace transport allowance; the $L^{-2}$ term pays the entire omitted deformed spectrum. The original positive elementary correction $r_A$ is still present. The principal cut and pole compensation remain paid by the reused (VE9)–(VE10); no additional correction is silently discarded by the head cutoff.

For the same conditional least integer attaining the global Robin-ratio maximum under a violation, $A=\log N>(7/2)10^{46}$, this supplies a finite head with a joint tail allowance. Its signed upper bound remains unproved. To reach the original sufficient target one still needs to control the head together with $r_A$ and ${\rho }_A$ against $\mathcal{E}(L)$; the strict core remains required. An asymptotic tail allowance is not a numerical finite-clock certificate or a uniform positive margin. The original real parts, both signs, all heights and multiplicities return through the complete response comparison. No RH conclusion or superiority to the already retained original-zero cutoffs is asserted.

Original density can be transported to a smaller negative-heat head

The common-time head in (VG5) need not retain the full height $\sqrt{A}{L}^3$. The additional interface below transports the existing original-zero density and zero-free tail to the actual negative-heat zeros on a growing finite band. No density theorem is assumed for a different function, and no simple-zero trajectory is used.

Reuse Lagarias, arXiv:math/0404394v4, Theorem 2.1(4), printed p.7, and Lemma 6.1, equation (6.8), printed p.27. The former gives the usual counting formula with an $O(\log T)$ remainder, hence $O(\log (y+3))$ original zeros in a fixed-length height interval, with multiplicities. The latter supplies the local logarithmic-derivative expansion in a fixed strip. Its Riemann specialization is used at heights at least $10$, away from the zeta pole and trivial zeros. The Gamma logarithmic derivative, vertical Stirling bounds, fixed-strip polynomial zeta growth, exact negative-time Gaussian convolution, Rouché theorem, and the complete (VE)–(VG) interfaces are also reused. The cached original Lagarias text and the already inspected Dobner convolution are the source inputs; their proofs are not reconstructed or independently certified here.

This is a repo-derived paper-level joint transport application. No exhaustive literature, mathematical-priority, Lean, computed-zero or numerical finite-clock claim is made.

Relative heat control away from the original zero multiset

Write $\xi ={\xi }_0$, and let ${\mathcal{Z}}_0$ be its actual zero multiset. There are absolute $c,C>0$ with the following property. For $Y\ge 10$, put $q=\log (Y+10)$. Assume

$$-1\le \Re s\le 2,10\le \Im s\le Y,\mathrm{dist}(s,{\mathcal{Z}}_0)\ge \delta ,$$

$$e^{-q}\le \delta \le 1/10,0<ϵ\le c{\delta }^2/q^2.$$

Then the actual, unshifted heat functions satisfy

$$\boxed{\left|\frac{{\xi }_{-ϵ}(s)}{\xi (s)}-1\right|\le Cϵq^2/{\delta }^2.}\text{\hspace{1em}(VH1)}$$

The denominator is paid below; an absolute function-value error is not substituted for this relative estimate.

Put $y=\Im s$. The cited local formula and the Gamma factor give $|{\xi }^{\prime }/\xi |\le Cq/\delta$ on the radius-$\delta /2$ neighborhood of $s$; on radius-$\delta /4$ disks there Cauchy gives $|({\xi }^{\prime }/\xi {)}^{\prime }|\le Cq/{\delta }^2$, with harmless changes to these radii. More explicitly, at a point $s+iu$, $|u|\le \delta /2$, the radius-$\delta /4$ disk stays at distance at least $\delta /4$ from every zero and in the fixed wider strip of Lemma 6.1. The local count is $O(q)$, and the remaining Gamma and local-formula terms are $O(q)$. Consequently, for $G(v)=\xi (s+iv)/\xi (s)$ and $|v|\le \delta /2$,

$$|G(v)-1-iv({\xi }^{\prime }/\xi )(s)|\le C{q}^2{\delta }^{-2}{v}^2{e}^{Cq|v|/\delta }.\text{\hspace{1em}(VH2)}$$

To pay the whole Gaussian tail, remove the original local factors

$$P_y(z)=\prod _{\begin{array}{c}\rho \in {\mathcal{Z}}_0\\ |\Im \rho -y|<2\end{array}}(z-\rho ),g_y(z)=\xi (z)/P_y(z).$$

Repeated factors retain multiplicities. Their number is at most $Cq$. In $-1\le \Re z\le 2$, $|\Im z-y|\le 1/2$, the source local formula, with the extra factors of ordinate distance at least one bounded separately, gives $g_y^{\prime }/g_y=O(q)$. This function is analytic and nonzero throughout the indicated rectangle. Compare its modulus along the horizontal segment from $2+iy$ to $s$. Every numerator local factor at $s$ has modulus at least $\delta$, and every factor at $2+iy$ has modulus at most $3$. Thus

$$|\xi (s)|\ge |\xi (2+iy)|\exp [-Cq\log (3/\delta )].$$

The Euler inverse bound at $\Re s=2$ and vertical Stirling supply a uniform lower bound for $|\xi (2+iy)|$. Fixed-strip polynomial zeta growth and vertical Stirling supply a uniform upper bound for $|\xi (\sigma +i(y+v))|$; compact heights include the removable Gamma singularities of the completed function. As $\log (3/\delta )\le q+\log 3$, these bounds combine, for every real $v$, to give fixed absolute $B,C$ with

$$|G(v)|\le C{e}^{C{q}^2}(1+|v|{)}^B{e}^{\pi |v|/4}.\text{\hspace{1em}(VH3)}$$

The existing exact vertical Gaussian identity is

$$\frac{{\xi }_{-ϵ}(s)}{\xi (s)}=\frac{1}{\sqrt{\pi ϵ}}{\int }_{\mathbb{R}}{e}^{-v^2/ϵ}G(v)dv.$$

Subtract $1+iv({\xi }^{\prime }/\xi )(s)$ before estimating: its Gaussian average is exactly one. On $|v|\le \delta /2$, (VH2) and square completion give $Cϵq^2/{\delta }^2$, since $ϵq^2/{\delta }^2\le c$. On the complement, (VH3) gives

$$C\exp [C{q}^2-{\delta }^2/(8ϵ)].$$

The polynomial and $e^{\pi |v|/4}$ factors are integrated against the remaining half-Gaussian, with a common bound for $ϵ\le 1/2$. Choose $c$ small enough to absorb $C{q}^2$ into half the negative exponent. Then $e^{-{\delta }^2/(16ϵ)}\le 16ϵ/{\delta }^2$ pays this part. The subtracted affine tail has the same bound using $|{\xi }^{\prime }/\xi (s)|\le Cq/\delta$. This proves (VH1) with constants independent of $Y,s,\delta ,ϵ$.

Small clusters transport multiplicities without separation assumptions

Return to $L=\log A$ and the same common $ϵ(A)$ from (VE10). Set

$$U=\sqrt{A}{L}^3,h=L^{-2},Y=U+2,q=\log (Y+10),\delta =\frac{h}{Kq},$$

where the fixed absolute $K$ is larger than a suitable multiple of the original local-count constant. For sufficiently large $A$, the domain conditions of (VH1) hold and

$$ϵ(A)q^2/{\delta }^2=K^2ϵ(A)q^4{h}^{-2}\le C{A}^{-1/2}{L}^{-2}⟶0.\text{\hspace{1em}(VH4)}$$

Here $q\asymp L$, $\delta \asymp L^{-3}$ and $\delta \ge e^{-q}$ eventually. No particular numerical starting clock is certified.

Take the union of radius-$r$ disks about every distinct original zero, with one common $\delta \le r\le 2\delta$. Repeated zeros are single centers but retain their multiplicities in counting. Choose $r$ to avoid tangencies and multiple boundary intersections for the finite collection of disks in the working band. The relevant union components are finite and have diameter at most $h$.

Here is the bound that prevents a long chain. For a component meeting an interior point of the working band, a chain of neighboring centers has successive distances at most $4\delta$. Before a chain could leave the unit neighborhood of its first center, its centers have ordinates in one fixed-length interval, containing at most $Cq$ original zeros. It cannot travel a unit distance because $4\delta Cq<1/2$ for large $A$. It therefore stays in that neighborhood. Its complete center count is at most $Cq$, and its disk-union diameter is at most $4\delta (Cq+1)\le h$, by the fixed choice of $K$. Thus no unproved minimum spacing or simplicity is needed. Components meeting the working band remain at least $10$ in height and below $Y$; there is no artificial truncation boundary through a cluster.

At the boundary of each such component, distance from every original zero is at least $r\ge \delta$. Equation (VH1), with (VH4), makes $|{\xi }_{-ϵ(A)}-\xi |<|\xi |$ on that complete boundary. Rouché, or the argument principle for the finite disk union including any hole boundaries, gives the same total zero multiplicity inside. Outside all these disks, (VH1) excludes deformed zeros in the working band. Hence the original and deformed multisets in each component can be matched, with

$$\boxed{|w-\rho |\le h=L^{-2}.}\text{\hspace{1em}(VH5)}$$

This is a multiplicity-preserving matching inside small clusters, not an assertion of simple trajectories or an inherited global density law. Conjugation supplies the lower-height version. The matching need not itself be used as a numerical algorithm.

Reuse the original weighted tail in the correct direction

The existing smaller-cut calculation in the Nicolas note uses the actual original spectrum, Chourasiya–Simonič’s inclusive density estimate and Johnston–Yang’s stated zero-free region. Reuse that already paid calculation, including its weighted intermediate quantity, rather than re-proving the density theorem or reassigning it to deformed zeros. With

$$R=57.54,\Omega =(L/\log L{)}^{1/3},{\kappa }_{\ast }=(28R{)}^{-1},T_0=A^{1/4}{e}^{-{\kappa }_{\ast }\Omega },$$

it supplies, for sufficiently large $A$,

$$B_2(A;T_0):=\sum _{\begin{array}{c}\rho =\beta +i\gamma \in {\mathcal{Z}}_0\\ \gamma >T_0\end{array}}\frac{A^{\beta -1/2}+A^{1/2-\beta }}{{\gamma }^2}\le Cb(A),$$

$$b(A)=L{A}^{-1/8}{e}^{{\kappa }_{\ast }\Omega }+L^3{e}^{-\Omega /(14R)}+L^3{A}^{-1/14}.\text{\hspace{1em}(VH6)}$$

This is the positive weighted-zero upper bound produced by the existing layer calculation, not a reverse inference from its signed-response upper bound. Reflection and conjugation preserve every multiplicity.

Put $S(A)=2T_0$. Work at sufficiently large $A$ that $S>12$, $U>S$ and $h<1$. The working band for (VH5) is $S<|\Im w|\le U$. By (VG2), its deformed real parts are in $[-1,2]$. Every relevant cluster stays in the high fixed strip and away from the outer height limits needed for (VH1). The matching assigns every such $w$ to a distinct original zero occurrence with $|\Im \rho |>T_0$, since $S-h>T_0$. The factor two in $S$ pays the lower-cut boundary without assuming a zero-free cutoff ordinate.

Use the complete-coefficient bound proved for (VG1), not a leading-term replacement. For each matched occurrence,

$$\sqrt{A}L|{\hat{F}}_A(w)|\le \frac{3A^{\Re w-1/2}}{|\Im w{|}^2}\le \frac{12A^h{A}^{\Re \rho -1/2}}{|\Im \rho {|}^2}.$$

Here $|\Im w|\ge |\Im \rho |-h\ge |\Im \rho |/2$. The matching is injective on the retained occurrences. For the complete original multiset, reflection followed by conjugation identifies the sum of $A^{\beta -1/2}/{\gamma }^2$ over both signs with $B_2(A;T_0)$. Since $A^h=e^{1/L}$, the whole matched middle costs at most $Cb(A)$. Beyond $U$, the unchanged (VG4) costs $C{L}^{-2}$. Consequently the entire deformed suffix obeys

$$\boxed{\sqrt{A}L\sum _{\begin{array}{c}{\xi }_{-ϵ(A)}(w)=0\\ |\Im w|>S(A)\end{array}}|{\hat{F}}_A(w)|\le Cb(A)+C{L}^{-2}=O(L^{-2}),S(A)=2A^{1/4}{e}^{-\Omega /(28R)}.}\text{\hspace{1em}(VH7)}$$

For the last equality, $\Omega /\log L\to \infty$ and $\Omega /L\to 0$ give $b(A)=o(L^{-2})$. All constants are independent of $A$; the bound is eventual, with no certified numerical threshold. The actual common time is unchanged and is not chosen separately for a zero or a cluster.

The reduced head still needs its signed bound

Let $Z_{ϵ(A)}^{\mathrm{small}}(A)$ be the same complete finite signed zero sum as in (VG5), with inclusive height $S(A)$ in place of $U$. It is real by conjugation. The unchanged uniform count (VG3) gives $O(A^{1/4}L{e}^{-\Omega /(28R)})$ zero occurrences in this head. Combining the complete suffix bound with (VE10) yields

$$\boxed{-\sqrt{A}L{I}_{\psi }(A)=\sqrt{A}L{Z}_{ϵ(A)}^{\mathrm{small}}(A)+r_A+{\tilde{\rho }}_A,|{\tilde{\rho }}_A|\le C_1{L}^{-1}+C_{2}b(A)+C_3{L}^{-2}.}\text{\hspace{1em}(VH8)}$$

The elementary correction remains the positive original $r_A$. The original complete coefficients, actual real parts, both signs, all heights and multiplicities, and the already paid cut and pole compensation return through (VE10). The smaller head pays no new signed main upper bound.

The original source is still the conditional least integer attaining the global Robin-ratio maximum under a violation, with $A=\log N>(7/2)10^{46}$. That standing lower clock alone is not a certificate that this source exceeds the unspecified eventual threshold of (VH7)–(VH8). The earlier all-clock (VG) interface remains available. A proof still must control the same head and corrections against the original $\mathcal{E}(L)$ and retain the strict core, including any uncovered finite clock range. No Robin or RH conclusion is supplied.

One positive heat time pays a further finite signed Robin window

The smaller negative-time head above still needs its signed upper bound. A separate finite payment is available from the already retained positive-time resolvent (HR1) and exact real-parameter explicit formula. This payment uses the actual original zeros throughout. It does not require the eventual eligibility threshold of (VH7), transfer original zero density to another function, or assume RH above the verified height.

Reuse the complete coefficient remainder (H1), the original positive elementary correction after (G9), and the strict core (G6)–(G7) in the Nicolas note. The existing finite payment (GL3)–(GL4) already excludes the same selected source through $A\le (7/2)10^{46}$. The new interval is

$$\frac{7}{2}10^{46}<A\le 4\cdot 10^{46},L=\log A,H=3\cdot 10^{12},U=\frac{7}{2\cdot 10^{24}}.$$

The source is the conditional least integer $N>5040$ attaining the global Robin-ratio maximum under a violation, with $A=\log N$. It is not the least counterexample. The earlier verified-height and complete reciprocal-square inputs are unchanged:

$$c_0:=\sum _{\rho }\frac{m_{\rho }}{\rho (1-\rho )}<0.05,S_2(H):=\sum _{\gamma >H}\frac{m_{\rho }}{{\gamma }^2}<1.48\cdot 10^{-12}.\text{\hspace{1em}(VI1)}$$

Every zero with $|\gamma |\le H$ is on the critical line under the same cited finite-verification premise. All sums below retain actual real parts and multiplicities, and include both ordinate signs. This is a paper-level application of existing analytic inputs, with no new source theorem, zero computation, mathematical-priority claim or Lean certification.

Split the whole rational response at one common time

Put $d_{\rho }=\rho (1-\rho )$ and write the normalized original response as

$$\sqrt{A}L{Z}_{\mathrm{orig}}(A)=(1+1/L)\sum _{\rho }{m}_{\rho }\frac{A^{\rho -1/2}}{d_{\rho }}+R_A.$$

This is the full (H1) coefficient, not a leading-term replacement:

$$|R_A|\le \frac{2(L+2)}{L^2}\sum _{\rho }{m}_{\rho }\frac{A^{\beta -1/2}}{|\rho ||1-\rho {|}^2}.\text{\hspace{1em}(VI2)}$$

The exact existing heat-resolvent identity gives

$$\sum _{\rho }{m}_{\rho }\frac{A^{\rho -1/2}}{d_{\rho }}=\sum _{\rho }{m}_{\rho }\frac{A^{\rho -1/2}{e}^{-U{d}_{\rho }}}{d_{\rho }}+A^{-1/2}{\int }_0^U\mathcal{H}(u,u-L)du,\mathcal{H}(u,v)=\sum _{\rho }{m}_{\rho }{e}^{u{\rho }^2-v\rho }.\text{\hspace{1em}(VI3)}$$

Here $\Re d_{\rho }={\gamma }^2+\beta (1-\beta )>0$. The absolute integral per occurrence is at most $\sqrt{A}/{\gamma }^2$, and the full reciprocal-square sum is finite. Thus the whole-zero sum and heat integral interchange absolutely. There is no height truncation, exceptional zero or value inserted at $u=0$. Reflection and conjugation make these complete expressions real.