Edit: updated to show all the previously unsolved numbers, and those I have solved.
The bountied best answer to the previous question left these 26 numbers unsolved:
$499 \ 501 \ 549 \ 607 \ 652 \ 653 \ 787 \ 795 \ 802 \ 803 \ 805 \ 806 \ 807 $
$ 821 \ 829 \ 853 \ 857 \ 859 \ 869 \ 877 \ 878 \ 879 \ 883 \ 884 \ 891 \ 892 $
Assuming the Gamma(n) function introduced is (n-1)! I updated my generator, and use $\Gamma$ to indicate the Gamma factorial so that for example $ \Gamma(4) = 3 \times 2 \times 1 = 6 $.
I solved 12 more numbers.
$ 501 = \Gamma(11!) - (0! + 10) + (100!!)!! $
$ 549 = (\frac{0! + 0!}{.1} + .1) \times (\Gamma(11!) + 0! + 0!) $
$ 607 = (0! + 0! + 0!)! + 0! + (11!)! - \Gamma(11!) $
$ 652 = \frac{(100!!)!! + 0! + 0!}{.1} - \Gamma(11!) $
$ 787 = ((0! + 0! + 0!)! + .1) \times (\Gamma(11!) + 0!) + .1 $
$ 795 = ((0! + 0! + 0!)! + .101) \times \Gamma(11!) $
$ 829 = \Gamma((0! + 0! + 0!)!) - \sqrt{(\Gamma(11!) + 0!)} + (11!)! $
$ 853 = (\Gamma(11!) + 0! + 0!) \times (11! + 0!) - 0! $
$ 879 = \frac{\Gamma(11!)}{.11} - 0! + ((0! + 0! + 0!)!)! $
$ 884 = ((\frac{0! + 0!}{.1})!! + .1) \times ((11! + 0!)!! - 0!) $
$ 891 = \frac{(100!!)!! + 0!}{.1} + \Gamma(11!) + 0! $
$ 892 = \frac{(100!!)!! + 0! + 0!}{.1} + \Gamma(11!) $
Can anyone get the other 14 numbers? I expect they can, because my generator did not previously find all those in the linked best answer from @isaacg.
Update:
This answer only extended answers to the previous question by implementing the solution of integer values passed to the Gamma function, which is a simple factorial. But the Gamma function also applies to real numbers.
I am no mathematician and this is new to me, but I found this YouTube video The Gamma Function for Half Integer Values which discusses the Gamma function for positive and negative fractional half-numbers.
For $\frac{1}{2}$ he gives (and this is the starting point)
$$ \Gamma(\frac{1}{2}) = \sqrt{\pi} $$
The speaker then derives the expansions for the other positive fractions. I can follow most of it, but I think it goes wrong in one place
For $k = 1$ is $\Gamma(\frac{1}{2} + 1) = \Gamma(\frac{3}{2}) = \frac{\sqrt{\pi}}{2^1} \cdot 1 $
For $k = 2$ is $\Gamma(\frac{1}{2} + 2) = \Gamma(\frac{5}{2}) = \frac{\sqrt{\pi}}{2^2} \cdot 3 \cdot 1 $
For $k = 3$ is $\Gamma(\frac{1}{2} + 3) = \Gamma(\frac{7}{2}) = \frac{\sqrt{\pi}}{2^3} \cdot 5 \cdot 3 \cdot 1 $
For $k = 4$ is $\Gamma(\frac{1}{2} + 4) = \Gamma(\frac{9}{2}) = \frac{\sqrt{\pi}}{2^4} \cdot 7 \cdot 5 \cdot 3 \cdot 1 $
He then forms a general expression, but when he comes to the last term says he does not understand a double factorial, fumbles around, and writes something like
$$\Gamma(\frac{1}{2} + k) = \Gamma(\frac{2k + 1}{2}) = \frac{\sqrt{\pi}}{2^k} \cdot (2k + 1)!!$$
But that looks wrong to me, IMO the general case ($k > 0$) should be
$$\Gamma(\frac{1}{2} + k) = \frac{\sqrt{\pi}}{2^k} \cdot (2k - 1)!!$$
For negative fractions I make the general expression ($k > 0$) to be
$$\Gamma(\frac{1}{2} - k) = \frac{(-1)^k \cdot 2^k \cdot \sqrt{\pi}}{(2k - 1)!!}$$
although I used $k$ a bit differently from his working. Obviously the $\sqrt{\pi}$ isn't useful when looking for integer results, but it will cancel out when by dividing one Gamma function by another.
Anyway, I implemented the Gamma function for half integer values, but my program took forever with all the new values to permute. When I restricted the source values to 3 bits, it completed but did not find any more results.
Perhaps there are mistakes and someone else can follow it up, and of course, there are the other real values that are not simple half-fractions.