The factorisation conjecture for the overconvergent $U$ operator

Let $p\in\{2,3,5\}$, let $r$ range over some open interval containing $\frac{1}{2}$, and let $\mathcal{S}_k(r)$ be the space on which the overconvergent $U$ operator acts. A factorisation of the form

$$U^{(r)}=A^{(r)}DB^{(r)}$$

uses matrices $A^{(r)}$ and $B^{(r)}$ with entries in $\mathcal{O}_{\mathbb{C}_p}$, each congruent to the identity modulo $p$. The factorisation conjecture. For all such $p$ and $r$, this factorisation exists, with diagonal entries of $D$ satisfying

$$\begin{array}{r|c|c|} p & D_{ii} & \nu_p D_{ii}\\ \hline 2 & \dfrac{2^{4i + 1} (3i)!^2 i!^2}{3\cdot (2i)!^4} & 1 + 2 \nu_2\left(\frac{(3i)!}{i!}\right)\\ 3 & \dfrac{3^{3i} (6i)!(2i)!i!}{2\cdot(3i)!^3} & 2i + 2 \nu_3 \left( \frac{(2i)!}{i!}\right)\\ 5 & \dfrac{5^{2i} (10i)!(3i)!^2i!}{3\cdot(5i)!^3(2i)!} & i + 2\nu_5\left(\frac{(3i)!}{i!}\right)\\ \end{array}$$

The factorisation would make the slopes of $U$ accessible from the diagonal entries of $D$; the claim is open for $p=3$ and $p=5$, although computations of the matrix entries suggest it holds for $r\in(\frac{1}{3},\frac{2}{3})$ in both cases.