A conjecture about certain values of the Fabius function 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) = \frac1{n! \, 2^{\binom{n+1}2}} \, \sum_{m\ge0}\binom n {2m} \, c_m,\tag3$$
where $c_m$ are defined by the recurrence
$$c_0 = 1, \quad c_n = \frac1{(4^n - 1)(2n+1)} \, \sum_{m\ge1} \binom{2n+1}{2m+1} \, c_{n-m}.\tag4$$
Note that only finite number of terms in each sum are non-zero.
The values $F\left(2^{-n}\right) $ appear as A272755 / A272757 in the OEIS.
Let
$$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$$
(see more terms here)
I conjecture that all terms of this sequence are integers. How can we prove (disprove) this conjecture?
 A: I have posted  in arXiv:1702.05442 the English translation of a paper about Fabius 
function that I published in Spanish in 1982 (we will refer to it as (A)). 
With the Theorems in this paper
the question can be answered without much difficulty. I have posted  also in 
arXiv:1702.06487
a new paper with the complete answer to this question.  Here we only show the main points of this proof.
I will call $R_n$ your numbers $a_n$.
First it is shown that 
$$R_{2n+1}=F_n\frac{(4n+1)!!}{(2n-1)!!}.$$
where $F_n$ are natural numbers introduced in my old paper (A).
This shows the conjecture for $R_n$  in case $n$ is an odd number.
For the other case, first prove that the numbers $R_n$ are related by the equation
$$R_n=2d_n (2n-1)!! \prod_{k=1}^{\lfloor n/2\rfloor}(2^{2k}-1)$$
with some   rational numbers $d_n$ introduced in (A).
These numbers $d_n$ are related to the $F_n$ so that at the end
we get 
$$R_{2n}=\sum_{k=0}^n \frac{2 F_k}{2^{2n}}\binom{2n}{2k}\frac{(4n-1)!!}{(2k+1)!!}
\prod_{\ell=k+1}^n(2^{2\ell}-1)$$
This shows that the denominator of $R_{2n}$ is at most a power of $2$. 
The relation between $R_n$ and $d_n$ implies that the same power will appear in 
the denominator of $2d_n$. An easy induction shows that this power is $0$.  
We have other results about the values of Fabius function in  arXiv:1702.06487 
including a relation with Bernoulli numbers. This paper is a preliminary 
version.
