The Fabius function is a smooth monotone function $F:[0,1]\to[0,1]$, satisfying functional equations $$F(0)=0, \quad F(1-x)=1-F(x)\tag1$$ and $$F'(x) = 2 \,F(2 x) \quad \text{for} \,\, 0<x<1/2.\tag2$$

The function $F$ assumes rational values at dyadic rational arguments. In particular, it is known $\!^{[1]}$$\!^{[2]}$ that $$F\left(2^{-n}\right) = 2^{-\frac{n(n+1)}{2}}\left[\frac1{n!}+\sum_{m=1}^{\lfloor n/2\rfloor}\frac{(-1)^m \, c_m}{(n - 2 m)!}\right],\tag3$$ where $$c_m = \frac1{4^m - 1}\left[\frac{(-1)^m}{(2m+1)!}+\sum_{k=1}^{m-1} \frac{(-1)^k \, c_{m-k}}{(2 k + 1)!}\right],\tag4$$ and empty sums are assumed to be zero as usual. The values $F\left(2^{-n}\right) $ appear as A272755/A272757 in the OEIS.

Let $$a_n = F\left(2^{-n}\right) \, 2^{\binom {n-1}2} \, (2n)! \, \prod_{m=1}^{\lfloor n/2\rfloor}\left(4^m - 1\right).\tag5$$ This sequence begins $$1, \, 5, \, 15, \, 1001, \, 5985, \, 2853675, \, 26261235, \, 72808620885, \, 915304354965 \, ...\tag6$$ (see more terms here)

I conjecture that all terms of this sequence are integers. How can we prove (disprove) this conjecture?