The Fabius function is the unique function ${\bf F}:\mathbb R\to[-1, 1]$ satisfying the following conditions:

- a functional–integral equation$\require{action} \require{enclose}{^{\texttip{\dagger}{a poet or philosopher could say "it knows and replays its own future"}}}$ $\displaystyle{\bf F}(x)={\small\int_0^{2{\tiny\text{ }}x}}{\bf F}\left(t\right)dt,\,$ and
- a normalization condition ${\bf F}(1)=1.$

It is noteworthy that despite being smooth (class $C^\infty$)${^{\texttip{\dagger}{i.e. continuous and having continuous derivatives of any order}}}$ everywhere, ${\bf F}(x)$ is not real-analytic at any point $x\ge0$ — its Taylor series there is either divergent, or a polynomial with a finite number of terms.${^{\texttip{\dagger}{the latter happens at dyadic rational points}}}$

Unlike some other functions that appear specifically constructed for that purpose, the Fabius function naturally occurs in research of several seemingly unrelated problems — perhaps, this is why it has been discovered and re-discovered independently many times by several mathematicians. It also has some practical applications in computational mathematics.

The Fabius function ${\bf F}(x)$ is constant $0$ for all $x\le0,$ but some authors prefer to define it only on $[0,\infty)$ or only on $[0,1].$ Some authors study a very closely related Rvachev${^{\texttip{\dagger}{also variously spelled as Rvachëv or Rvachyov, a.k.a. "atomic function"}}}$ function $\operatorname{up}(x)$ defined as $\operatorname{up}(x) = {\bf F}(x+1)$ for $-1\le x\le1$ and $\operatorname{up}(x) = 0$ otherwise.

It is known that the Fabius function assumes rational values at dyadic rational arguments — efficient algorithms to compute those values have been published or posted online, and have been discussed on this site.$^{[1]}$$\!^{[2]}$$\!^{[3]}$$\!^{[4]}$$\!^{[5]}$$\!^{[6]}$$\!^{[7]}$$\!^{[8]}$$\!^{[9]}$

I have been looking for a non-recursive, self-contained formula for the Fabius function for quite a long time. After lots of experimentation and looking for patterns in its values I came up with a conjectured empirical formula. Let${^{\texttip{\dagger}{the superscript is just an index here, not to be confused with a power}}}$
$$\mathscr F^m_n = \frac1{2^{n^2}\left(\frac12;{\tiny\text{ }}\frac12\right)_n}\,\sum _{k=0}^n\frac{\binom n k_{1/2}}{2^{{\tiny\text{ }}k{\tiny\text{ }}(k-1)}(n+k)!}\,\sum _{\ell=0}^{2^k{\tiny\text{ }}m-1}\,(-1)^{s_2\left(\ell\right)}\,\left(\ell-2^km+\tfrac12\right)^{n+k}{\small,}\tag{$\small\spadesuit$}$$
where $k,\,\ell,\,m,\,n$ are non-negative integers, $\displaystyle\small\left(a;{\tiny\text{ }}q\right)_n=\prod_{k=0}^{n-1} (1-a{\tiny\text{ }}q^k)$ is the *q*-Pochhammer symbol${^{\texttip{\dagger}{it assumes only rational values in this formula}}}$, ${\binom n k}_q=\displaystyle\small\frac{\left(q;{\tiny\text{ }}q\right)_n}{\left(q;{\tiny\text{ }}q\right)_k\left(q;{\tiny\text{ }}q\right)_{n-k}}\vphantom{\Huge|}$ is the *q*-binomial coefficient${^{\texttip{\dagger}{it assumes only rational values in this formula}}}$, and $s_2(n)\vphantom{\Huge|}$ is the sum of binary digits${^{\texttip{\dagger}{i.e. the number of 1's in the base-2 representation}}}$ of $n$ (note that $(-1)^{s_2\left(n\right)} = t_n\vphantom{\Huge|}$ is just the signed version of the Thue–Morse sequence, satisfying recurrence $t_0 = 1,\,t_n = (-1)^n \, t_{\lfloor n/2\rfloor};\vphantom{\Huge|}$ see also${^{[10]}}$${\!^{[11]}}$${\!^{[12]}}$).

I conjecture that for all non-negative integers $m,\,n,$ the following identity holds: $${\bf F}\!\left({\small\frac m{2^n}}\right)=\mathscr F^m_n.\tag{$\small\diamondsuit$}$$ Note that there is no requirement for $\frac m{2^n}$ to be a proper irreducible fraction — it might well be that $m$ is even, or $m>2^n$. I have not been able to rigorously prove it yet, but it produces exact rational values that agree with those computed using known correct algorithms for all dyadic rational arguments I have tried. Of course, this formula did not just appear out of the blue — its structure was inspired by known algorithms, and it went a long way to this relatively concise form (at some point it had about $5$ levels of nested summations/products).

- Can we prove that $\small(\diamondsuit)$ is indeed a correct representation of the Fabius function at dyadic rationals?
- If the conjecture is difficult to tackle at once, we can try to at least prove that the function defined by $\small(\spadesuit)$ shares some known properties with the Fabius function, e.g.:
- $0\le\mathscr F^m_n\le1,$
- $\mathscr F^m_n = \mathscr F^{2{\tiny\text{ }}m}_{n+1}\vphantom{\Large|}$ — the result is the same across all representations of a rational number $\frac m{2^n}$,
- $\mathscr F^m_n = 1-\mathscr F^{(2^n-m)}_n\vphantom{\Large|}$ for all $\small0\le m\le2^{n}$ — rotation symmetry,
- $\mathscr F^m_n = \mathscr F^{(2^{n+1}-m)}_n\vphantom{\Large|}$ for all $\small0\le m\le2^{n+1}$ — reflection symmetry,
- for a fixed $n,$ and $\small0\le m\le2^n$ the values of $\mathscr F^m_n\vphantom{\Large|}$ strictly increase,
- all points $\small\left(\frac m{2^n},\,\mathscr F^m_n\right)\vphantom{\Large|}$ lie on a continuous curve — for any convergent sequence of dyadic rationals $\frac m{2^n}\vphantom{\Large|}$, the sequence of corresponding values $\mathscr F^m_n\vphantom{\Large|}$ also converges.

If the conjecture $\small(\diamondsuit)$ it true, then for all $0\le x\le1$ (not necessary rational) $${\bf F}\!\left(x\right)=\lim_{n\to\infty}\,\frac1{2^{n^2}\left(\frac12;{\tiny\text{ }}\frac12\right)_n}\,\sum _{k=0}^n\frac{\binom n k_{1/2}}{2^{{\tiny\text{ }}k{\tiny\text{ }}(k-1)}(n+k)!}\,\sum _{\ell=0}^{\left\lfloor2^{n+k}{\tiny\text{ }}x-1\right\rfloor}\,(-1)^{s_2\left(\ell\right)}\,\left(\ell-2^{n+k}{\tiny\text{ }}x+\tfrac12\right)^{n+k}.\tag{$\small\heartsuit$}$$ For dyadic rational $x$ the limit is trivial, because the sequence under the limit becomes constant for large enough $n$. Also, there is a known series (by Rvachev) that expresses values of the Fabius function for all $0\le x\le1$ via its values at negative powers of $2$: $${\bf F}\!\left(x\right)=\sum _{n=1}^\infty\frac{(-1)^{\left\lfloor 2^n{\tiny\text{ }}x\right\rfloor }-1}{2}\,(-1)^{s_2\left(\lfloor2^n{\tiny\text{ }}x\rfloor\right)}\;\sum_{k=0}^n\frac{2^{\frac{k{\tiny\text{ }}(k+1)}2}}{k!}\,\left(x-2^{-n}\left\lfloor 2^n{\tiny\text{ }}x\right\rfloor\right)^k\;{\bf F}\!\left(2^{{\tiny\text{ }}k-n}\right).\tag{$\small\clubsuit$}$$ Again, for dyadic rational $x$ this series terminates after a finite number of terms, producing an exact rational value.