Conjectured maximum purity of absolutely PPT qutrit-qudit states

Let $\rho\in M_3\otimes M_n$ be an absolutely PPT state with $n\geq 2$, and let $\lambda_1,\ldots,\lambda_{3n}$ be its eigenvalues. Maximum-purity conjecture. The maximum purity of absolutely PPT states is

$$\begin{cases} \frac{4}{9n} &\text{if } n\equiv 0\pmod 4,\\ \frac{36n+8}{(9n+1)^2} &\text{if } n\equiv 1\pmod 4,\\ \frac{36n+16}{(9n+2)^2} &\text{if } n\equiv 2\pmod 4,\\ \frac{36n-8}{(9n-1)^2} &\text{if } n\equiv 3\pmod 4. \end{cases}$$

Moreover, the maximum is attained when

$$\begin{cases} \lambda_1=\cdots=\lambda_{\frac{3n}{4}}=\frac2{3n},\quad \lambda_{\frac{3n}{4}+1}=\cdots=\lambda_{3n}=\frac2{9n} &\text{if } n\equiv 0\pmod 4,\\ \lambda_1=\cdots=\lambda_{\frac{3n+1}{4}}=\frac6{9n+1},\quad \lambda_{\frac{3n+5}{4}}=\cdots=\lambda_{3n}=\frac2{9n+1} &\text{if } n\equiv 1\pmod 4,\\ \lambda_1=\cdots=\lambda_{\frac{3n+2}{4}}=\frac6{9n+2},\quad \lambda_{\frac{3n+6}{4}}=\cdots=\lambda_{3n}=\frac2{9n+2} &\text{if } n\equiv 2\pmod 4,\\ \lambda_1=\cdots=\lambda_{\frac{3n-1}{4}}=\frac6{9n-1},\quad \lambda_{\frac{3n+3}{4}}=\cdots=\lambda_{3n}=\frac2{9n-1} &\text{if } n\equiv 3\pmod 4. \end{cases}$$

The formulas are motivated by numerical estimates, which become smaller than the conjectured values as the dimension increases; the conjecture remains open.