The three-halves-unit shifted triangular-hole tiling enumeration conjecture

Let $a,b,c,m$ be nonnegative integers, with $a$ of parity different from the parity of both $b$ and $c$. Consider a hexagon with side lengths $a,b+m,c,a+m,b,c+m$, from which an equilateral triangle of side length $m$ is removed in the three-halves-unit off-center position described in the source. Let $\operatorname{H}$ denote the hyperfactorial function, and define $P_2(a,b,c,m)$ by

$$P_2(a,b,c,m)=\begin{cases} ((a+b)^2-1)((a+c)^2-1)+4am(a^2+2ab+b^2+2ac+3bc+c^2)+2am+3bm+3cm+2m^2-1,&a\text{ even},\\ ((a+b)^2-1)((a+c)^2-1)+4(a+b+c+m)m(a^2+bc-1),&a\text{ odd.} \end{cases}$$

Three-halves-unit shifted-hole tiling conjecture. The number of lozenge tilings of this hexagon with the triangular hole equals

$$\begin{aligned} &\frac{1}{16}\frac{\operatorname{H}(a+m)\operatorname{H}(b+m)\operatorname{H}(c+m)\operatorname{H}(a+b+c+m)}{\operatorname{H}(a+b+m)\operatorname{H}(a+c+m)\operatorname{H}(b+c+m)}\\ &\quad\times\frac{\operatorname{H}\left(\frac m2\right)^2\prod_{x\in\{a,b,c\}}\operatorname{H}\left(\left\lceil\frac x2\right\rceil\right)\operatorname{H}\left(\left\lfloor\frac x2\right\rfloor\right)}{\prod_{x\in\{a,b,c\}}\operatorname{H}\left(\frac m2+\left\lceil\frac x2\right\rceil\right)\operatorname{H}\left(\frac m2+\left\lfloor\frac x2\right\rfloor\right)}\\ &\quad\times\frac{\operatorname{H}\left(\left\lceil\frac{a+b}{2}\right\rceil+\frac m2\right)\operatorname{H}\left(\left\lfloor\frac{a+b}{2}\right\rfloor+\frac m2\right)\operatorname{H}\left(\left\lfloor\frac{a+c}{2}\right\rfloor+\frac m2\right)\operatorname{H}\left(\left\lceil\frac{a+c}{2}\right\rceil+\frac m2\right)\operatorname{H}\left(\frac{b+c}{2}+\frac m2\right)^2}{\operatorname{H}\left(\frac m2+\left\lceil\frac{a+b+c}{2}\right\rceil\right)\operatorname{H}\left(\frac m2+\left\lfloor\frac{a+b+c}{2}\right\rfloor\right)\operatorname{H}\left(\left\lfloor\frac{a+b}{2}\right\rfloor-1\right)\operatorname{H}\left(\left\lceil\frac{a+c}{2}\right\rceil+1\right)\operatorname{H}\left(\frac{b+c}{2}\right)}\\ &\quad\times\frac{\operatorname{H}\left(m+\left\lceil\frac{a+b+c}{2}\right\rceil\right)\operatorname{H}\left(m+\left\lfloor\frac{a+b+c}{2}\right\rfloor\right)}{\operatorname{H}\left(\left\lfloor\frac{a+c}{2}\right\rfloor+m-1\right)\operatorname{H}\left(\frac{b+c}{2}+m\right)\operatorname{H}\left(\left\lceil\frac{a+b}{2}\right\rceil+m+1\right)}P_2(a,b,c,m). \end{aligned}$$

The statement is introduced as something that “seems to be true”; no resolution is supplied in the source.