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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 $$F'(x) = 2 \,F(2 x) \quad \text{for} \,\, 0<x<1/2.\tag2$$

Fabius graph

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?

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    $\begingroup$ My initial thought was to prove that your "fudge factor" used when going from $F$ to $a$ would be more than enough to make each term in the defining sum integral, but it's more delicate than that (as you probably knew). For example the $2^{-n(n+1)/2}/n!$ term when multiplied by the fudge factor is not in general an integer; there are powers of 2 in the denominator. So one has to think harder. $\endgroup$ – Kevin Buzzard Feb 7 '17 at 23:39
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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.

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  • $\begingroup$ Can you say something about the values of this function, which is somehow similar? mathoverflow.net/questions/94038/… $\endgroup$ – Pietro Majer Feb 23 '17 at 23:24
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    $\begingroup$ @Pietro Majer Not for the moment. I am now trying to put in order my second arXiv paper arXiv:1702.06487. Adding some new material to it. Your question appear to be interesting. I will try. $\endgroup$ – juan Feb 24 '17 at 10:43
  • $\begingroup$ A newer version of the paper: math.colgate.edu/~integers/s51/s51.pdf $\endgroup$ – Vladimir Reshetnikov Jun 13 '18 at 23:07

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