Forrester's hard-edge conditioned gap constant conjecture

Let $E_\beta^{\rm hard}(n;(0,s);\beta a/2)$ and $E_\beta^{\rm hard}(0;(0,s);\beta a/2)$ be the hard-edge probabilities of exactly $n$ and zero eigenvalues, respectively. For $\beta n\in\mathbb Z_{\geq0}$, define

$$\tau_{\beta a/2,\beta}^{\rm hard}(n)=\frac{2^{-(a+n)\beta n}}{n!}\left(\frac\beta2\right)^{n(a+n-1)\beta/2}\prod_{j=1}^{\beta n}\frac{\Gamma(a+2j/\beta)}{\sqrt{2\pi}}\frac{\prod_{j=0}^{n-1}\Gamma(1+(j+1)\beta/2)}{\prod_{j=n}^{2n-1}\Gamma(1+(j+a)\beta/2)}.$$

Forrester's conjecture. As $s\to\infty$,

$$\frac{E_\beta^{\rm hard}(n;(0,s);\beta a/2)}{E_\beta^{\rm hard}(0;(0,s);\beta a/2)}=\tau_{\beta a/2,\beta}^{\rm hard}(n)\exp\left(-\beta\left\{-n\sqrt{s}+\left(\frac{n^2}{2}+\frac{na}{2}\right)\log s^{1/2}\right\}\right)\left(1+O(s^{-1/2})\right).$$

This extends the conditioned hard-edge asymptotic by predicting its constant term for the stated integrality condition.