Rainbow anti-Ramsey conjecture for unions of triangles

Let $n$ be sufficiently large and let $t in [0,n/3]$. Define $\Xi(n,t)$ by

$$\Xi(n,t) = \begin{cases} \binom{t}{2}+t(n-t)+\left\lceil\frac{n-t}{2}\right\rceil\left\lfloor\frac{n-t}{2}\right\rfloor & \text{if } t\in\left[0,\frac{2n-6}{9}\right],\\[1ex] \binom{2t+1}{2}+\left\lceil\frac{n}{2}\right\rceil\left\lfloor\frac{n}{2}\right\rfloor & \text{if } t\in\left[\frac{2n-6}{9},\frac{n-\sqrt{2n-3}-3}{4}\right],\\[1ex] \binom{2t+2}{2}+(2t+2)(n-2t-2) & \text{if } t\in\left[\frac{n-\sqrt{2n-3}-3}{4},\frac{5n+\sqrt{3n^2-2n+4}-20}{22}\right],\\[1ex] \binom{6t-n+6}{2}+(n-3t-3)(3t+3) & \text{if } t\in\left[\frac{5n+\sqrt{3n^2-2n+4}-20}{22},\frac{2n-\sqrt{16n-7}-3}{6}\right],\\[1ex] \binom{3t+5}{2}+(n-3t-6) & \text{if } t\in\left[\frac{2n-\sqrt{16n-7}-3}{6},\frac{n}{3}\right]. \end{cases}$$

Rainbow anti-Ramsey conjecture. The rainbow anti-Ramsey number for $(t+2)$ vertex-disjoint triangles satisfies

$$\mathrm{ar}(n,(t+2)K_3)=\Xi(n,t)+2.$$

This conjecture gives a piecewise formula for the anti-Ramsey threshold for unions of triangles across the full range $t\in[0,n/3]$. It is proposed as a step toward determining $\mathrm{ar}(n,(t+1)F)$ in general; the source does not state a resolution.