Put
$$ u_k = \frac{N_k}{\varphi(N_k)\log\log N_k}, \qquad N_k=\prod_{j\le k}p_j. $$
Nicolas proved that
$$ \mathrm{RH} \quad\Longleftrightarrow\quad u_k>e^\gamma $$
for every \(k\ge1\). I wanted to know what one step of this sequence is actually measuring.
For the next prime \(q=p_{k+1}\), write
$$ y=\vartheta(p_k)=\log N_k. $$
Then there is a unique number \(T(q)\) satisfying
$$ \boxed{ \log q = T(q) \left[ \exp\!\left(\frac{\log T(q)}{q-1}\right)-1 \right], } $$
and the entire increment has the exact sign law
$$ \boxed{ \operatorname{sgn}(\log u_{k+1}-\log u_k) = \operatorname{sgn}\bigl(\vartheta(p_k)-T(p_{k+1})\bigr). } $$
So each primorial step asks one very specific question: did the current Chebyshev theta value land above or below the threshold chosen by the next prime?
That is already cute. The part I keep staring at is what happens when the same local coordinate is pushed through the explicit formula for zeta zeros.
The increment is a positive displacement#
Define
$$ \Delta_q(y) = -\log(1-1/q) - \log\frac{\log(y+\log q)}{\log y}. $$
Its derivative is strictly positive:
$$ \Delta_q'(y) = \frac1{y\log y} - \frac1{(y+\log q)\log(y+\log q)} >0. $$
The threshold is exactly the unique zero of \(\Delta_q\), hence
$$ \boxed{ \Delta_q(y) = \int_{T(q)}^y \left[ \frac1{v\log v} - \frac1{(v+\log q)\log(v+\log q)} \right]\,dv. } $$
Nothing is hidden in an error term here. The sign of the ratio step is the sign of a displacement, integrated against a positive kernel.
For \(L=\log q\), the threshold begins
$$ \boxed{ T(q) = q-\frac{L(L+2)}{2(L+1)} +O\!\left(\frac{L^3}{q}\right). } $$
Equivalently, if \(h_k=p_{k+1}-p_k\) and \(S(x)=\vartheta(x)-x\),
$$ \vartheta(p_k)-T(p_{k+1}) = S(p_k)-h_k +\frac{\log p_{k+1}+1}{2} -\frac1{2(\log p_{k+1}+1)} +O\!\left(\frac{\log^2p_{k+1}}{p_{k+1}}\right). $$
The local coordinate balances the Chebyshev error against the next prime gap, with an explicit half-log correction.
A finite scout through \(5{,}000{,}000\) checked 348,511 consecutive transitions with no sign mismatch. Every transition in that range decreases, while the Nicolas margin remains positive and rather small:
$$ \log u_k-\gamma\approx5.81\times10^{-5} $$
at the end of the scout.
This does not turn monotonicity into a proof route. When the positive kernels are accumulated over primes, their continuum limit is exactly Nicolas's classical smoothed Chebyshev-error weight
$$ \left( \frac1{\log t}+\frac1{\log^2t} \right)\frac1{t^2}. $$
The discrete threshold is a clean local diagnostic, not a new global positive reserve.
One zero has an exact exponential-integral profile#
Let
$$ K(x) = \int_x^\infty (\vartheta(t)-t) \left( \frac1{\log t}+\frac1{\log^2t} \right)\frac{dt}{t^2}, $$
and normalize it by
$$ D(x)=-\sqrt x\log x\,K(x). $$
If \(y=\log x\), the contribution of one zeta zero \(\rho\) is exactly
$$ \boxed{ Z_\rho(y) = \frac{e^{(\rho-1/2)y}}{\rho} +y e^{y/2}E_1((1-\rho)y), } $$
where \(E_1\) is the exponential integral. Its leading term is
$$ Z_\rho(y) = \frac{e^{(\rho-1/2)y}}{\rho(1-\rho)} \left(1+O_\rho(1/y)\right). $$
Now suppose
$$ \rho=\frac12+\delta+i\gamma, \qquad \delta>0, $$
and include its full functional-equation quartet. If
$$ \frac1{\rho(1-\rho)}=a+ib, $$
the leading real profile is
$$ \boxed{ Q_{\delta,\gamma}(y) = 4\left[ a\cosh(\delta y)\cos(\gamma y) - b\sinh(\delta y)\sin(\gamma y) \right]. } $$
It has recurrent adverse intervals of fixed logarithmic width \(2\pi/(3\gamma)\). On those intervals the growing part is at most
$$ -|c|e^{\delta y}+2|c|e^{-\delta y}, \qquad c=\frac1{\rho(1-\rho)}. $$
The prime-square contribution tends only to \(2\). An off-axis quartet grows like \(e^{\delta y}/\gamma^2\), so eventually it wins.
The hiding scale is gamma to the two-over-delta#
Balancing
$$ \frac{e^{\delta y}}{\gamma^2}\asymp1 $$
gives
$$ y\asymp\frac{2\log\gamma}{\delta}, \qquad \boxed{x=e^y\asymp\gamma^{2/\delta}.} $$
That exponent is wonderfully rude.
For the verifier's isolated-quartet models:
| \(\delta\) | \(\gamma\) | approximate \(\log_{10}x\) |
|---|---|---|
| \(0.1\) | \(14\) | \(32.1\) |
| \(0.05\) | \(1000\) | \(138.1\) |
| \(0.01\) | \(14\) | \(319.8\) |
A zero only one hundredth to the right of the critical line at height \(14\) can therefore hide from this primorial criterion until a prime scale around \(10^{320}\), up to constants exponential in \(1/\delta\). The associated primorial is exponentially larger again.
This is not a universal finite verification bound. The propagation theorem needs the rightmost off-axis real part to be attained and separated by a positive gap from all lower zero families. Under that hypothesis, the right-edge contribution is a nonzero mean-zero uniformly almost-periodic wave. It has relatively dense negative intervals, and the factor \(e^{(\beta_*-1/2)y}\) eventually amplifies them past the fixed reserve.
If the spectral edge is unattained, or approached by zeros with no positive real-part gap, the argument stops. Nicolas already gives qualitative sign changes when RH is false; this calculation says how violently dispersed the first visible failure may be for an isolated quartet, and names the exact global zero-family case that a stronger argument would still have to handle.
So the local arithmetic is surprisingly crisp:
$$ \text{one next-prime threshold} \quad+\quad \text{one positive displacement kernel}. $$
The spectral consequence is not crisp at all. A false zero can be perfectly real and still wait hundreds of decimal orders before the Nicolas sequence has any practical chance of noticing it.
Notebook references: K-0160, K-0161