6 Main theorem

Combining the primary estimate and the residual bound yields the main asymptotic expansion.

Theorem 26 Main asymptotic count

✓

There exist \(C_0,c{\gt}0\) such that for all sufficiently large \(n\) and \(t\ge C_0 n^3\),

\[ \frac{N_{n,4t}}{A_{n,4t}} =1-\frac{\binom {n}{3}}{8t} +O\! \left(\frac{n^2}{t}+\frac{n^{5/2}}{t^{3/2}} +\frac{n^6}{t^2}+e^{-cn^2}\right). \]

Proof ▶

By the three-way decomposition, \(\mathbb {P}(S_{4t}=0) =K_n\operatorname {Re}\! \int _{\mathcal{D}_\mathrm {core}}\psi ^{4t}\, d\lambda +\operatorname {Re}E\), where \(E\) collects off-core and residual contributions. Proposition 22 gives the primary estimate. Proposition 25 gives \(|E|=O(e^{-cn^2})\hat A_{n,4t}\), absorbed by the existing \(O(e^{-cn^2})\) term. Since \(N_{n,4t}=2^{4nt}\mathbb {P}(S_{4t}=0)\) and \(A_{n,4t}=2^{4nt}\hat A_{n,4t}\), dividing by \(\hat A_{n,4t}\) yields the claim.

Corollary 27 Uniform counting

✓

For every \(\varepsilon {\gt}0\) there exists \(K{\gt}0\) such that

\[ \left|\frac{N_{n,4t}}{A_{n,4t}}-1\right|{\lt}\varepsilon \qquad \text{for all } n\ge 2 \text{ and } t\ge Kn^3. \]

Proof ▶

For \(n\ge N_0\) (sufficiently large), the expansion in Theorem 26 bounds the ratio by \(\varepsilon \) when \(K\) is large enough. For each \(n{\lt}N_0\), the fixed-\(n\) asymptotics (Lemma 7) provide a threshold \(T_n\) beyond which the ratio is within \(\varepsilon \); enlarging \(K\) so that \(Kn^3\ge T_n\) for all small \(n\) completes the proof.