Put
$$ a_j^{(\kappa)} = \frac{1}{4j}W_0\left(\frac{j}{\kappa}\right), \qquad j\ge 1, $$
on the off-diagonal of an infinite zero-diagonal Jacobi matrix \(\mathcal A_\kappa\).
The coefficients tend to zero, so this is a compact selfadjoint operator. Count its positive eigenvalues above \(1/T\):
$$ N_\kappa(T) = \#\left\{ \lambda\in\operatorname{spec}(\mathcal A_\kappa): \lambda>T^{-1} \right\}. $$
Then
$$ \boxed{ N_\kappa(T) = \frac{T}{2\pi} \left( \log\frac{T}{\kappa}-1 \right) +o(T). } $$
So there is exactly one value of \(\kappa\) for which the first two extensive terms match the smooth Riemann--von Mangoldt count:
$$ \boxed{\kappa=2\pi.} $$
I like this much more than I expected to. The leading \(T\log T\) term only sees the rough scale \((\log j)/j\). The next \(T\)-term reaches into the Lambert \(W\), finds the hidden normalization constant, and points at \(2\pi\).
Constant paths, then multiplicative blocks#
A constant zero-diagonal path of length \(m\) with edge weight \(a\) has eigenvalues
$$ 2a\cos\frac{r\pi}{m+1}, \qquad 1\le r\le m. $$
The proportion above a positive threshold \(\lambda\) is therefore
$$ F\left(\frac{\lambda}{2a}\right), \qquad F(s)=\frac1\pi\arccos(s)\mathbf 1_{\{0\le s\le1\}}. $$
The Lambert coefficients vary slowly. Partition their active range into multiplicative blocks, delete the edges between blocks, and compare each piece with a constant path. The deleted edges cost only the number of block boundaries, while the within-block coefficient error contributes \(o(T)\). This leaves the phase sum
$$ N_\kappa(T) = \sum_{j\ge1} F\left(\frac{1}{2Ta_j^{(\kappa)}}\right) +o(T). $$
Now make the Lambert substitution
$$ j=\kappa we^w, \qquad w=W_0(j/\kappa). $$
It turns the coefficient into the almost offensively simple expression
$$ a_j^{(\kappa)} = \frac{1}{4\kappa e^w}. $$
If \(L_T=\log(T/(2\kappa))\), the phase argument becomes \(e^{w-L_T}\). After one more substitution, the whole count is governed by two universal integrals:
$$ \int_0^\infty e^{-u}\arccos(e^{-u})\,du=1, $$
$$ \int_0^\infty e^{-u}(1-u)\arccos(e^{-u})\,du = \log 2-1. $$
Those two numbers produce the two terms above. The \(\log 2\) combines with the turning point \(T/(2\kappa)\), and the surviving constant is precisely \(-\log\kappa-1\). Matching zeta's count chooses \(\kappa=2\pi\).
The resulting coefficient law is
$$ \boxed{ a_j^* = \frac{1}{4j}W_0\left(\frac{j}{2\pi}\right). } $$
Not merely \(C\log j/j\). The additive Lambert normalization is the part the second counting term can actually see.
The count can be run backwards#
There is now a partial converse, and this is the bit that made me sit up.
Suppose \(a_j\) is eventually decreasing, put \(s_j=(2a_j)^{-1}\), and let
$$ M(T)=\#\{j:s_j\le T\}. $$
The constant-path phase sum is not just an approximation with a convenient kernel. It is an exact Abel convolution:
$$ \frac{\Phi(T)}T = \frac1\pi\int_0^1 \frac{M(Tu)/T}{\sqrt{1-u^2}}\,du. $$
On logarithmic scale, the kernel has Fourier transform
$$ \widehat K(\xi) = \frac{\Gamma(1+i\xi/2)} {2\sqrt\pi\,\Gamma(3/2+i\xi/2)}. $$
Gamma has no zeros, so this symbol never vanishes on the real axis. Wiener inversion can therefore recover the unsmoothed coefficient count, provided its detrended logarithmic profile is bounded and asymptotically uniformly continuous. Under those regularity hypotheses, the two-term phase count forces
$$ M(T) = \frac T2\log\frac{T}{4\pi}+o(T). $$
Quantile inversion is sharp enough to recover the additive law itself:
$$ \boxed{ 4ja_j-W_0\left(\frac j{2\pi}\right)\longrightarrow0. } $$
So the Lambert profile is not merely one explicit way to produce the count. Among sufficiently regular ordered profiles, the count chooses it back. Rapid log-scale oscillations are the exact loophole: the arcsine convolution can hide them unless the slow-oscillation hypothesis is supplied.
The same profile has a forbidden region#
There is a second reason this model is useful.
Take a positive eigenvalue \(\lambda\) of a finite zero-diagonal Jacobi matrix and look into a trailing region where the future edge weights are at most \(A\), with \(\lambda>2A\). Each coordinate ratio is a boundary coupling times a diagonal suffix resolvent.
Expand that resolvent in return walks. Every length-\(2r\) endpoint return is a Dyck walk, so there are at most the Catalan number
$$ C_r=\frac{1}{r+1}\binom{2r}{r} $$
of them. Summing the Catalan generating function gives
$$ \left\langle e_0,(\lambda I-H)^{-1}e_0 \right\rangle \le \frac{2}{ \lambda+\sqrt{\lambda^2-4A^2} }. $$
Thus the terminal coordinate ratios decay with an exact discrete WKB action. For the macroscopic Lambert profile \(a(t)=A/t\), that action becomes
$$ \boxed{ \mathcal I^\sharp(\tau) = \operatorname{arcosh}\frac1\tau -\sqrt{1-\tau^2}. } $$
This controls not just the last coordinate of an eigenvector but an entire terminal packet. A separate compression identity turns that packet mass into the error in a trailing spectral enclosure through partial Christoffel ratios.
So the coefficient profile selected by the two-term zero count is also the profile with a computable forbidden-tail action. Spectral counting chooses the model; Catalan walks explain how its eigenvectors tunnel. That pairing is the part I keep coming back to.
The tunneling action is also an equilibrium cost#
The same function appears again in a continuum obstacle problem for reciprocal zeta zeros. On the natural lifted scale
$$ z=\frac{2\pi n}{\log n}\gamma^{-1}, $$
the limiting positive lattice capacity is \(dz/z^2\). The unit-mass logarithmic equilibrium for the squared Vandermonde interaction \(\log|z^2-w^2|\), constrained by that capacity, is explicit. With \(b=\pi/2\),
$$
\rho_*(z)
=
\begin{cases}
\displaystyle
\frac{1-\sqrt{1-(z/b)^2}}{z^2},
&0 The tail is saturated, with mass \(2/\pi\), and the density deficit opens as a
square root at \(b\). Removing a particle from \(z>b\) and placing it in the
band costs $$
\mathcal A(z)
=
4\left[
\operatorname{arcosh}\left(\frac zb\right)
-\sqrt{1-\frac{b^2}{z^2}}
\right].
$$ After the change \(y=4(z/b)^2\), this is exactly the terminal-packet WKB action
above. The \(3/2\) soft-edge opening is the same too. I did not expect the
Catalan resolvent bound and a constrained reciprocal-zero equilibrium to hand
back literally the same action. The compact-model counting theorem is unconditional. Under RH, reciprocal positive zeta-zero ordinates have the same counting
function, so this model becomes the forced normalization for any sufficiently
regular reciprocal-zero square-root Jacobi profile. The reverse theorem also
requires eventual coefficient ordering, an \(o(T)\) Jacobi phase-space
approximation, and bounded slow oscillation of the detrended coefficient
count. Those hypotheses are not yet proved for the completed-theta finite
sections, which have also not been proved to obey the additive Lambert law $$
4j a_j-W_0\left(\frac{j}{2\pi}\right)\longrightarrow0.
$$ They also still need positivity, control through the opening turning layer,
and the required homotopy estimates. The reciprocal-zero equilibrium is exact once its continuum lattice scaling
is granted. Convergence of the endogenous finite Gaussian lattice or hole
ensemble to that equilibrium, and the theta-observable local asymptotics, are
still open. What is proved is narrower and, to me, prettier: a completely explicit
compact tridiagonal operator reproduces both large terms of the classical
zero count; a zero-free Mellin symbol can force the same coefficient law back
from the count; and the forbidden-tail action reappears as an equilibrium
hole cost. Notebook references: K-0116, K-0117, K-0118, K-0119, K-0122, K-0123The exact boundary#