There are two pairs:
$$ (10.295195,\ 61.272145) $$
and
$$ (9.93082,\ 48.07157). $$
The second pair is smaller in both coordinates. It is very tempting to read that as "the first optimization, but better." It is not. These constants live in two genuinely different theorems, and the difference is exactly where the proof is allowed to begin.
Write
$$ V(t)=(\log t)^{2/3}(\log\log t)^{1/3}. $$
The global statement says that for every \(q\ge 3\), every Dirichlet character \(\chi\pmod q\), and every real \(|t|\ge 10\),
$$ L(\sigma+it,\chi)\ne 0 \quad\text{if}\quad \sigma\ge 1-\frac{1}{10.295195\log q+61.272145V(|t|)}. $$
The large-height statement says that there is an absolute, effectively computable \(Y\) such that, for every real \(|t|\ge Y\),
$$ \zeta(\sigma+it)\ne0 \quad\text{if}\quad \sigma\ge1-\frac{1}{48.07157V(|t|)}, $$
and, for the same \(Y\), every \(q\ge3\) and every \(\chi\pmod q\) satisfy
$$ L(\sigma+it,\chi)\ne0 \quad\text{if}\quad \sigma\ge 1-\frac{1}{9.93082\log q+48.07157V(|t|)}. $$
That \(Y\) is not numerically evaluated. This is not decorative fine print. It is the hinge.
There is a small direction-of-inequality trap here too. A smaller denominator gives a wider zero-free region: subtracting its reciprocal moves the boundary farther left. So \((9.93082,48.07157)\) really is stronger wherever it applies. But "wherever it applies" is part of the theorem, not metadata that can be peeled off afterward. Without a numerical upper bound for \(Y\), it cannot certify a particular finite height, replace a finite zero table, or be dropped into an argument whose variables begin at a printed cutoff.
The global pair has to survive the ground#
To start at \(|t|=10\), the proof cannot simply optimize the limiting conductor and height ratios. It has to bridge an actual finite range.
At the fixed height \(\log T_0=1943.408\), the primitive conductor \(d\) splits into three branches. The principal branch \(d=1\) is covered by a global zeta region. Conductors \(3\le d\le400000\) use Kadiri's \(5.60\) region. Conductors \(d\ge400001\) use McCurley's region, whose possible exceptional zero is real and therefore does not obstruct this nonzero-height argument.
The rude little number is the McCurley margin at the corner \(d=400001\), \(\log|t|=1943.408\):
$$ 0.0004039446269509\ldots>0. $$
That finite-height bridge then has to meet a fixed-threshold propagation envelope. At this architecture, the two active constraints cross at
$$ Q_*=10.2951944189592868\ldots,\qquad C_*=61.2721437036028819\ldots. $$
The displayed rational pair \((10.295195,61.272145)\) sits strictly beyond both constraints. It improves Khale's published global pair \((10.5,61.5)\), but the optimization claim is only for this fixed \(T_0\), conductor split, and Kadiri-McCurley-Khale propagation architecture. It is not an unrestricted best-possible theorem.
This pair is therefore an endpoint compromise. Push the \(\log q\) coefficient down and the large-conductor bridge asks for more height coefficient; push the height coefficient down and the propagation envelope pushes back. The crossing identifies the local Pareto frontier of this proof machine. That is a different question from asking what ratios survive after every term smaller than \(\log q+V(t)\) has faded away.
The large-height pair comes from a different object#
The sharper pair begins with a degree-\(46\) trigonometric polynomial
$$ P(x)=1+\sum_{k=1}^{46}b_k\cos(kx). $$
Its certificate is wonderfully concrete. Start with a \(47\times47\) integer lower-triangular matrix \(\mathcal L\), form
$$ \mathcal Q=\mathcal L\mathcal L^{\mathsf T}, $$
and set
$$ P(x)= \frac{\|\mathcal L^{\mathsf T}(1,e^{ix},\ldots,e^{46ix})^{\mathsf T}\|_2^2} {\operatorname{tr}(\mathcal Q)}. $$
So \(P(x)\ge0\) for every real \(x\) by an exact sum-of-squares identity, not a numerical eigenvalue guess. Exact integer correlations produce all \(46\) cosine coefficients as rational numbers and prove every \(b_k>0\). The limiting ratios then evaluate, with directed arithmetic, to
$$ \frac{b}{2\bar\alpha} =9.93081138131449\ldots<9.93082 $$
and
$$ \frac{B_*^{2/3}H_P(E_P)}{\bar\alpha} =48.07155656004025\ldots<48.07157. $$
Here the strict decimal reserves are allowed to absorb the fixed-degree lower-order conductor, zero-count, transform, and outer-line terms once the height is sufficiently large. That is what creates the effective \(Y\). The theorem proves that such a computable threshold exists; it does not print the threshold.
The Dirichlet step also needs one common seed interval for every character modulo \(q\), because the detector simultaneously uses powers \(\chi^j\). An effective log-free family estimate supplies such an interval at sufficiently large absolute height, and the new zeta coefficient is needed to exclude the principal-character branch inside the same argument. Primitive conductors divide \(q\); extra Euler factors from imprimitive characters have their zeros on \(\Re s=0\). After those reductions, the fixed polynomial propagates the common seed upward.
So the two optimizations are not competitors on one domain. The global pair pays for contact with the ground. The large-height pair waits until the scaffolding becomes negligible. I keep wanting to draw them as two curves that only pretend to be two rows of the same table.
The Goldbach specialization has two normalization traps#
Feeding the global pair into the explicit Goldbach summatory framework lowers its formal logarithmic threshold to
$$ 64923900841612.7437791589\ldots, $$
with least sufficient integer ceiling \(64923900841613\). It also decreases the zero-region input and widens the associated at-most-one-zero region.
But two printed uniform coefficients, \(5.805\) and \(7.246\), do not follow from that specialization. Positive auxiliary terms grow with \(q\); evaluating them at \(q=400001\) does not create a uniform upper bound. The audit finds source values \(11.0323617\ldots>7.246\) at \(q=10^7,\log x=10^8\), and \(766.06109\ldots>5.805\) at \(q=10^{35}\) after the new substitution. This identifies a gap in those numerical deductions, not a disproof of the Goldbach summatory conclusions.
There is also a clean factor-of-two ambush in the optional Lambert-\(W\) inversion. If
$$ A_q= \frac{\sqrt q\log^2q/100-10.295195\log q}{61.272145}, $$
then the required strict sufficient condition is
$$ x> \exp\left( \exp\left(\frac12W(2A_q^3)\right) \right). $$
The \(2\) is forced because, with \(z=2\log\log x\),
$$ V(x)^3=\frac12ze^z. $$
Using \(W(A_q^3)\) reaches only \(V(x)^3=A_q^3/2\). Tiny normalization error, enormous downstream argument. Classic.
Neither region removes a possible real exceptional Dirichlet zero at height zero. Neither implies RH or GRH. What they do give is more interestingly precise: one certified region that really starts at \(10\), and one sharper region whose exact algebra is ready long before its numerical starting height is.
Notebook references: C-0129, C-0121, C-0125, C-0117