Here is the whole classification:
$$ K\in \mathrm{PF}_4, \qquad K\notin \mathrm{PF}_5. $$
The \(K\) here is the classical de Bruijn-Newman theta kernel. Every one of its ordered translation minors through order four is nonnegative, and every strictly ordered one is positive. Then, at order five, the equally spaced matrix
$$ \left[ K\left(\frac1{100}+\frac{i-j}{20}\right) \right]_{i,j=0}^{4} $$
has negative determinant.
So four is not merely the highest order checked. It is the exact stopping point.
The small determinant that knows every other determinant#
Define
$$ \begin{aligned} H_m(t) &=(-1)^{m(m-1)/2}\\ &\quad{}\cdot \det\!\left[K^{(i+j)}(t)\right]_{i,j=0}^{m-1}. \end{aligned} $$
These are the fully confluent minors: the objects obtained when all row nodes collide and all column nodes collide, after dividing out the two Vandermonde factors.
At first glance this looks local. It looks like information about what happens when the gaps are tiny, not a theorem about arbitrary packets of points scattered across the line.
But positivity of the complete initial flag
$$ H_1(t),H_2(t),H_3(t),H_4(t)>0 $$
does something much stronger.
Fix one row position \(x\). The flag is exactly the sequence of initial Wronskians, in \(y\), of
$$ K(x-y),\ K'(x-y),\ K''(x-y),\ldots $$
so the extended-complete-Chebyshev collocation theorem makes every ordered derivative-collocation determinant positive. Now freeze the \(y\)-packet. Those determinants are the initial Wronskians, in \(x\), of
$$ K(x-y_1),\ldots,K(x-y_m). $$
Apply the same theorem again. Out comes
$$ \det[K(x_i-y_j)]>0 $$
for arbitrary strictly increasing row and column packets.
Twice. That is the move. I keep turning it over in my head because it makes the original configuration space just disappear.
What remains is one variable#
For \(t\geq0\), put
$$ x=\pi e^{4t},\qquad x_n=n^2x. $$
Every derivative of the kernel has an exact theta-shell form
$$ K^{(r)}(t) =e^t\sum_{n\geq1}e^{-x_n}p_r(x_n), $$
where the integer polynomials satisfy
$$ \begin{aligned} p_0(z)&=z(2z-3),\\ p_{r+1}(z)&=(1-4z)p_r(z)+4zp_r'(z). \end{aligned} $$
The compact interval \(0\leq t\leq13/32\) is divided into 65 rational cells. On each cell, a degree-eight Taylor model with a rigorous ninth-derivative remainder proves positivity of \(H_2,H_3,H_4\). The first 20 theta shells are evaluated directly; every later shell is swallowed by one geometric tail bound.
That is 195 inequalities total.
Seven free node coordinates, after translation, have become 195 little one-variable boxes. This is an extremely favorable exchange rate.
The weakest lower bounds all occur on the last cell:
$$ \begin{aligned} H_2&>1.4737407298724188\cdot10^{-6},\\ H_3&>1.2160040433993911\cdot10^{-5},\\ H_4&>0.04493717344926668. \end{aligned} $$
The first theta shell eats the infinite tail#
Past \(t=13/32\), the first shell can be factored out exactly. Its three oriented determinants are
$$ \begin{aligned} &16x^3(4x^2-12x+15),\\ &8192x^6(8x^3-36x^2+90x-105),\\ &201326592x^{10}\bigl(16x^4-96x^3\\ &\hspace{7.3em}{}+360x^2-840x+945\bigr). \end{aligned} $$
The quadratic is \((2x-3)^2+6\). The derivative of the cubic is six times that quadratic, and the derivative of the quartic is eight times the cubic. From \(x=\pi e^{13/8}>15\) onward, all three are positive and increasing.
Every higher theta shell is collected into a determinant perturbation that decreases with \(x\). At the seam, even the order-four case has
$$ \begin{aligned} \text{first shell}&>1.5617\cdot10^{26},\\ \text{total error}&<2.3451\cdot10^{25}. \end{aligned} $$
After that the first shell only grows and the error only shrinks. The infinite tail is finished at one point.
And then five is negative#
At
$$ u_0=\frac1{100}, \qquad h=\frac1{20}, $$
the equally spaced Toeplitz determinants are
$$ \begin{aligned} D_2&=+0.0340640406226\ldots,\\ D_3&=+0.000697697064264\ldots,\\ D_4&=+3.8274713675688\ldots\cdot10^{-6},\\ D_5&=-1.8472360734426\ldots\cdot10^{-9}. \end{aligned} $$
The negative order-five witness was found and certified by Wojciech Michałowski. The paper package independently reconstructs it with the same Arb evaluator used for the new PF4 proof.
This does not prove anything new about the zeros of \(\Xi\), and it does not persist to all orders. It is a sharp finite-order theorem about a very particular arithmetic kernel.
But it is wonderfully exact. Every minor through four. One ordinary minor at five. The door closes immediately after the last good order.
Notebook references: C-0224, K-0002, K-0062