# 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$$(see more terms here) I conjecture that all terms of this sequence are integers. How can we prove (disprove) this conjecture? • My initial thought was to prove that your "fudge factor" used when going from F to a would be more than enough to make each term in the defining sum integral, but it's more delicate than that (as you probably knew). For example the 2^{-n(n+1)/2}/n! term when multiplied by the fudge factor is not in general an integer; there are powers of 2 in the denominator. So one has to think harder. – Kevin Buzzard Feb 7 '17 at 23:39 ## 1 Answer I have posted in arXiv:1702.05442 the English translation of a paper about Fabius function that I published in Spanish in 1982 (we will refer to it as (A)). With the Theorems in this paper the question can be answered without much difficulty. I have posted also in arXiv:1702.06487 a new paper with the complete answer to this question. Here we only show the main points of this proof. I will call R_n your numbers a_n. First it is shown that$$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.

• Can you say something about the values of this function, which is somehow similar? mathoverflow.net/questions/94038/… – Pietro Majer Feb 23 '17 at 23:24
• @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. – juan Feb 24 '17 at 10:43
• A newer version of the paper: math.colgate.edu/~integers/s51/s51.pdf – Vladimir Reshetnikov Jun 13 '18 at 23:07