The numbers are
$$ (R_1,\ldots,R_{15})= (11,63,97,340,280,525,966,627,2004,1348,1472,2040,4653,1828,4455). $$
For shift \(k\), every rank \(r\ge R_k\) gives a strictly positive consecutive Toeplitz minor of the xi coefficients. A separate finite theorem covers everything below. Thus, for every \(r\ge1\) and every fixed \(1\le k\le15\),
$$
D_{r,k}=\det[a_{k+j-i}]_{0\le i,j I like the thresholds partly because they are so visibly not a polished asymptotic sequence. Shift \(5\) starts earlier than shift \(4\); shift \(14\) is much kinder than shift \(13\). They retain the fingerprints of individual zeta-zero gaps and a progressively fussier Vandermonde matrix. The construction begins with $$
G(z)=\frac18\xi\left(\frac12+\frac{\sqrt z}{2}\right)
=\sum_{n\ge0}a_nz^n,
$$ with \(a_n=0\) for \(n<0\). At first the determinant \(D_{r,k}\) looks badly oriented for an all-rank result: its matrix is \(r\times r\), so the object itself grows with the variable we need to send to infinity. Rectangular Jacobi--Trudi duality turns it sideways. For a general series \(A\), define reciprocal coefficients by $$
\frac{A(0)}{A(-t)}=\sum_{n\ge0}h_nt^n.
$$ Then $$
\frac{D_{r,k}(A)}{A(0)^r}
=
\det[h_{r+j-i}]_{0\le i,j This is the move. The original \(r\times r\) Toeplitz rectangle at fixed shift \(k\) becomes a \(k\times k\) determinant whose coefficient index moves with \(r\). Rank can now grow forever while the matrix size stays fixed. For \(k=15\), the infinite tail of ranks is still a problem about one \(15\times15\) matrix. Cute. That exchange is much more than a pleasant determinant identity. In the original coordinates, proving eventual positivity means controlling an expanding family of eigenvalue directions: every increase of \(r\) adds a row and a column. After duality, the geometry is frozen. Only the \(k^2\) entries change, and all of them are sampled from the same far tail of one reciprocal series. This is exactly the regime where singularities can do useful work. Large coefficient index magnifies the nearest poles, while fixed matrix size means the error estimate never has to outrun a simultaneously growing determinant dimension. The coefficients \(h_n\) come from \(G(0)/G(-t)\). Its first poles correspond to the first zeros $$
\frac12+i\gamma_1,\ldots,\frac12+i\gamma_k
$$ of \(\xi\). For each fixed \(k\), choose a certified contour between the \(k\)-th and \((k+1)\)-st ordinates. The first \(k\) reciprocal-Xi poles are then isolated inside, simple, and their real amplitudes are $$
c_j=\frac{16iG(0)}{\gamma_j\xi'(1/2+i\gamma_j)}.
$$ The certified signs alternate: $$
\operatorname{sgn}(c_j)=(-1)^{j-1}.
$$ Writing \(x_j=(4\gamma_j^2)^{-1}\), the leading pole model for the moving coefficients has the form $$
h_n^{(0)}=\sum_{j=1}^k c_jx_j^n.
$$ Insert this into the fixed \(k\times k\) determinant. The matrix factors through the \(k\) pole nodes, and its determinant becomes a product of the amplitudes, powers of the \(x_j\), and two Vandermonde determinants. The nodes satisfy $$
x_1>x_2>\cdots>x_k>0.
$$ One Vandermonde orientation contributes exactly the sign needed to cancel the alternating signs of the \(c_j\). What remains is strictly positive. This is a lovely little piece of sign bookkeeping: the amplitudes alternate, the ordered pole geometry alternates back, and the leading determinant lands on the positive side. The requirement of exactly \(k\) leading poles is structural here. Fewer than \(k\) exponential modes would factor the model matrix through a space of dimension below \(k\), making its determinant vanish. The first \(k\) poles are the smallest packet capable of producing a full-rank leading term, and the Vandermonde says precisely when that packet is independent. It would not be enough merely to say that the first poles dominate. The reciprocal series still knows about every later pole. The contour supplies an explicit remainder $$
h_n=h_n^{(0)}+e_n,
$$ whose size decays geometrically relative to the enclosed contribution as \(n\), hence \(r\), grows. If \(Q_r\) is the leading pole matrix and \(E_r\) the contour-tail matrix, the certified calculation proves $$
\lVert Q_r^{-1}E_r\rVert_\infty<1
$$ at \(r=R_k\) for each \(k\le15\). The bound improves afterward. Along the real homotopy \(Q_r+sE_r\), \(0\le s\le1\), the determinant therefore never crosses zero; it keeps the positive sign of the Vandermonde leading term. That is the eventual half of the proof. The eventual argument starts no later than rank \(4653\). Independently, the finite Polya-frequency bridge proves nonnegativity through $$
9{,}425{,}328{,}785{,}007.
$$ I keep staring at the mismatch in scale. We need the finite result only until a threshold of a few thousand, and it arrives carrying more than nine trillion ranks. There is no delicate handoff: the two ranges overlap by an almost comical amount. So every rank is covered for shifts \(1\) through \(15\). But the quantifiers matter. These are fifteen separate fixed-\(k\) arguments. Nothing here treats \(k\) as a growing variable, and there is no claim for shift \(16\). Running one more certified pole-and-contour computation would add one more finite rung; it would not become the missing uniform estimate. There is also an off-axis counterfeit in the verification packet that passes the same finite mechanisms. That is a useful warning label. Finite Polya-frequency behavior plus this fixed-pole style of evidence does not detect the Riemann hypothesis. The result is a large, explicit wedge of Toeplitz positivity, not total positivity, not a growing-shift theorem, and not an RH criterion. What I find exciting is narrower and more mechanical: a determinant whose dimension seemed forced to grow was the wrong determinant to study. Dualize the rectangle, and infinity moves from matrix size into coefficient index. Then fifteen tiny matrices can control every rank. Notebook references: C-0187, C-0026, K-0055, C-0191The first poles assemble the determinant#
An absurdly large overlap#