Arithmetic of the Fabius function

TL;DR

This paper proves that certain scaled values of the Fabius function at dyadic points are integers and explores their arithmetical properties, unifying various notations and extending understanding of this classical function.

Contribution

The paper establishes the integrality of specific scaled Fabius function values at dyadic points and clarifies notation, advancing the arithmetic understanding of the function.

Findings

The numbers R_n are integers.

Identifies arithmetical properties of Fabius function at dyadic points.

Unifies notations related to the Fabius function.

Abstract

I solve here a question of Vladimir Reshetnikov in Mathoverflow (question 261649) about the values of Fabius function. Namely, I prove that the numbers $R_n:=2^{-\binom{n-1}{2}}(2n)! F(2^{-n})\prod_{m=1}^{\lfloor n/2\rfloor}(2^{2m}-1)$ are integers. We show also some other arithmetical properties of the values of Fabius function at dyadic points. The Fabius function was defined in 1935 by Jessen and Wintner and has been independently defined at least six times since. We attempt to unify notations related to the Fabius function.

Figures

Equations

$$F(x)=\sum _{n=0}^{\infty }(-1{)}^{w(n)}\mathrm{up}(x-2n-1),$$

$$\mathrm{up}(t)=\{\begin{array}{ll}F(t+1)& \text{for −1≤t≤0},\\ F(1-t)& \text{for 0≤t≤1},\\ 0& \text{for ∣t∣>1}.\end{array}$$

$$\mskip-200.0mu\operatorname{up}^{\prime}(x)=2\bigl{(}\operatorname{up}(2x+1)-\operatorname{up}(2x-1)\bigr{)},$$

$$F^{(k)}(x)=2^{\left(\genfrac{}{}{0ex}{}{k+1}{2}\right)}F(2^{k}x),$$

$${\mathrm{up}}^{(n)}(x)=2^{\left(\genfrac{}{}{0ex}{}{n+1}{2}\right)}F(2^n(x+1)),-1\le x\le 1.$$

$$\widehat{\mathrm{up}}(z)=\prod _{n=0}^{\infty }\frac{\sin (\pi z/2^n)}{\pi z/2^n}.$$

$$\mathrm{up}(x)={\int }_{\mathbb{R}}\widehat{\mathrm{up}}(t)e^{2\pi itx}dt.$$

$$\widehat{\mathrm{up}}(z)=\prod _{n=0}^{\infty }\frac{\sin (\pi z/2^n)}{\pi z/2^n}=\sum _{n=0}^{\infty }(-1{)}^n\frac{c_n}{(2n)!}(2\pi z{)}^{2n}.$$

$$c_n={\int }_{\mathbb{R}}{t}^{2n}\mathrm{up}(t)dt,$$

$$(2n+1)2^{2n}{c}_n=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{2n+1}{2k}\right)c_k.$$

$$c_0=1,(2n+1)(2^{2n}-1)c_n=\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{2n+1}{2k}\right)c_k.$$

$$c_n=\frac{F_n}{(2n+1)!!}\prod _{\nu =1}^n(2^{2\nu }-1{)}^{-1},$$

$$1,\frac{1}{9},\frac{19}{675},\frac{583}{59535},\frac{132809}{32531625},\frac{46840699}{24405225075},\frac{4068990560161}{4133856862760625},\dots$$

$$F_n:=c_n(2n+1)!!\prod _{\nu =1}^n(2^{2\nu }-1).$$

$$\frac{F_n}{(2n-1)!!}\prod _{\nu =1}^{n-1}(2^{2\nu }-1{)}^{-1}=\sum _{k=0}^{n-1}(\genfrac{}{}{0ex}{}{2n+1}{2k})\frac{F_k}{(2k+1)!!}\prod _{\nu =1}^k(2^{2\nu }-1{)}^{-1},$$

$$F_n=\sum _{k=0}^{n-1}{F}_k\left(\genfrac{}{}{0ex}{}{2n+1}{2k}\right)\frac{(2n-1)!!}{(2k+1)!!}\prod _{\nu =k+1}^{n-1}(2^{2\nu }-1),n\ge 1.$$

$$1,1,19,2915,2788989,14754820185,402830065455939,54259734183964303995,\dots$$

$$n{\int }_0^1{x}^{n-1}\mathrm{up}(x)dx=n!2^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}\mathrm{up}(1-2^{-n}).$$

$$\frac{c_n}{2}={\int }_0^1{x}^{2n}\mathrm{up}(x)dx=(2n)!2^{\left(\genfrac{}{}{0ex}{}{2n+1}{2}\right)}\mathrm{up}(1-2^{-2n-1}).$$

$$F(2^{-2n-1})=\mathrm{up}(1-2^{-2n-1})=\frac{2^{-\left(\genfrac{}{}{0ex}{}{2n+1}{2}\right)}}{2(2n)!}\frac{F_n}{(2n+1)(2n-1)\cdots 1}\prod _{k=1}^n(2^{2k}-1{)}^{-1}.$$

$$f(x)=\sum _{n=0}^{\infty }\frac{d_n}{n!}{x}^n$$

$$f(2x)=\frac{e^x-1}{x}f(x).$$

$$d_0=1,(n+1)(2^n-1)d_n=\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)d_k.$$

$$f(x):=1+x\int_{0}^{1}\operatorname{up}(t)e^{xt}\,dt=e^{\frac{x}{2}}\widehat{\operatorname{up}}\Bigl{(}\frac{ix}{4\pi}\Bigr{)}$$

$$\widehat{\mathrm{up}}(z)=\frac{\sin \pi z}{\pi z}\widehat{\mathrm{up}}(z/2).$$

$$f(2x)=e^{x}\widehat{\operatorname{up}}\Bigl{(}\frac{ix}{2\pi}\Bigr{)}=e^{x}\frac{\sin(ix/2)}{ix/2}\widehat{\operatorname{up}}\Bigl{(}\frac{ix}{4\pi}\Bigr{)}=e^{\frac{x}{2}}\frac{e^{-x/2}-e^{x/2}}{2i(ix/2)}f(x)=\frac{e^{x}-1}{x}f(x).$$

$$f(x)=1+x\sum _{n=0}^{\infty }\frac{x^n}{n!}{\int }_0^1{t}^n\mathrm{up}(t)dt.$$

$$d_n=n{\int }_0^1{t}^{n-1}\mathrm{up}(t)dt.$$

$$d_n=n!2^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}\mathrm{up}(1-2^{-n})=n!2^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}F(2^{-n}).$$

$$1,\frac{1}{2},\frac{5}{18},\frac{1}{6},\frac{143}{1350},\frac{19}{270},\frac{1153}{23814},\frac{583}{17010},\frac{1616353}{65063250},\frac{132809}{7229250},\frac{134926369}{9762090030},\dots$$

Full text

The Fabius function was defined in 1935 by Jessen and Wintner and has been independently defined at least six times since. We attempt to unify notations related to the Fabius function.

1. Introduction.

The Fabius function is a natural object. It is not surprising that it has been independently defined several times. As far as known it was defined independently in the following cases:

1. (1)

In 1935 by B. Jessen and A. Wintner [MR1501802]*Ex. 5, p. 62 (English). In five lines they showed that it is infinitely differentiable. 2. (2)

In 1966 by J. Fabius [MR0197656] (English); considered as the distribution function of a random variable. 3. (3)

In 1971 by V. A. Rvachëv [MR0296237] (Ukrainian); defined as a solution to the equation $y^{\prime}(x)=2(y(2x+1)-y(2x-1))$. 4. (4)

In 1981 by G. Kh. Kirov and G. A. Totkov [MR705073] (Bulgarian). 5. (5)

In 1982 by J. Arias de Reyna [MR666839] (Spanish); defined in the same way as V. A. Rvachëv. 6. (6)

In 1985 by R. Schnabl [MR1031820] (German).

Today there are many papers dealing with the Fabius function, due mainly to V. A. Rvachëv [MR0296237]–[MR1050928], (a summary of his result can be found in [MR1050928]). Recently [MR666839] has been translated to English and posted in arXiv:1702.05442. In this paper the function is defined and its main properties are proven.

The function $\operatorname{up}(x)$ takes rational values at dyadic points [MR0330388].111This Russian paper was not accessed directly. Rather, it came to notice through the works of V. A. Rvachëv [MR1050928]. In [MR666839] it was independently proved that the values of $\operatorname{up}(k/2^{n})$ are rational numbers. The object of this paper are the arithmetical properties of these rational values. Some of the results are also contained in [MR666839]. In particular, we solve here a question posed in MathOverflow by Vladimir Reshetnikov.

1.1. Some notations.

We consider two functions. Firstly, the Rvachëv function $\operatorname{up}(x)$ with support in $[-1,1]$ that satisfies on all ${\mathbb{R}}$ equation (2). V. A. Rvachëv denotes it by $\operatorname{up}(x)$; it is the same function that is called $\varphi(x)$ in the translation of [MR666839]. Secondly, the Fabius function $F(x)$, defined by Fabius, that satisfies $F^{\prime}(x)=2F(2x)$ for $x>0$. This function coincides with the function $\theta(x)$ in [MR666839], except that in [MR666839] it is defined on all of ${\mathbb{R}}$, putting $F(x)=0$ for $x<0$. Therefore $F^{\prime}(x)=2F(2x)$ is true for all $x\in{\mathbb{R}}$.

There is a connection between these two functions (see [MR666839]*Theorem 4). Specifically,

[TABLE]

where $w(n)$ is the sum of the digits of $n$ expressed in base $2$.

The Fabius function $F\colon[0,1]\to[0,1]$ is increasing with $F(0)=0$ and $F(1)=1$. Rvachëv’s function $\operatorname{up}\colon{\mathbb{R}}\to{\mathbb{R}}$ can be defined in terms of $F(t)$ by

[TABLE]

The function $\operatorname{up}(t)$ is an infinitely differentiable function of compact support $[-1,1]$. The relation with the Fabius function is only given for reference. The function $\operatorname{up}(t)$ has a nice independent definition, see Theorem 1 in [MR666839].

We use $w(n)$ to denote the sum of the digits of $n$ expressed in binary notation. We use $\nu_{2}(r)$ is the dyadic valuation of the rational number $r$, the exponent of $2$ in the factorization of $r$.

We will define two sequences of rational numbers. Firstly, $(c_{n})$ the moments of Rvachëv’s function $\operatorname{up}(x)$. Secondly, $(d_{n})$ the half moments of Rvachëv’s function. These are also related to the values of the Fabius function $F(x)$, see equation (21). There are two sequences of natural numbers associated to these rational numbers: $F_{n}$ associated to the moments $c_{n}$, and $G_{n}$ associated to the half moments.

A further two sequences are Reshetnikov’s numbers $R_{n}$, and the sequence $D_{n}$ of the common denominators of the values $F(a/2^{n})$ of the Fabius function at dyadic numbers with denominator $2^{n}$.

2. Properties of the Fabius function.

The Fabius function $\operatorname{up}(x)$ is a simple infinitely differentiable function of compact support. Visualising it as a bump function, it increases from [math] to a certain value and then decreases symmetrically to [math]. Its derivative is positive in the first interval and then negative in the next. Therefore it may be that the derivative is obtained from two homothetically deformed copies of the function.

In fact $\operatorname{up}(x)$ is infinitely differentiable of compact support equal to $[-1,1]$, and for all $x\in{\mathbb{R}}$ satisfies the equation

[TABLE]

with $\operatorname{up}(0)=1$.

The number $2$ is the only factor giving an interesting solution to the equation. The Fabius function (1) appears when one attempts to compute the derivatives $\operatorname{up}^{(n)}(x)$. We have $F^{\prime}(x)=2F(2x)$ for all $x\in{\mathbb{R}}$. Therefore ([MR666839]*eq. (22))

[TABLE]

and

[TABLE]

Its Fourier transform is an entire function

[TABLE]

The inversion formula yields the following expression

[TABLE]

3. The values of the Fabius function and two sequences of rational numbers.

3.1. The sequence of rational numbers $c_{n}$.

The Fourier transform $\widehat{\operatorname{up}}(z)$ is an entire function. In [MR666839]*eq. (4) and (6) it is shown that

[TABLE]

These coefficients $c_{n}$ are very important for the study of the Fabius function. They are given by the integrals [MR666839]*eq. (34)

[TABLE]

and satisfy the recurrence [MR666839]*eq. (7)

[TABLE]

The next Proposition is contained in [MR666839] but is a little cryptic, so it is expanded here.

Proposition 1**.**

The numbers $c_{n}$ satisfy the recurrence

[TABLE]

For any natural number $n$ we have

[TABLE]

where $F_{n}$ are natural numbers.

Proof.

The last term in the sum of (9) is equal to $(2n+1)c_{n}$. We obtain (10) from (9) by adding $-(2n+1)c_{n}$ to both sides. We have $c_{0}=\widehat{\operatorname{up}}(0)$ and this is equal to $1$. Then the recurrence can be used to compute the numbers:

[TABLE]

The recurrence shows that $c_{n}\geq 0$ for all $n$.

Equation (11) follows from the definition

[TABLE]

It remains to show that the numbers $F_{n}$ are natural. We understand that an empty product is equal to $1$, so that $F_{0}=1$. For $n\geq 1$ substituting (11) in the recurrence (10) yields, for $n\geq 1$,

[TABLE]

so that

[TABLE]

This expression for $F_{n}$ shows by induction that all the $F_{n}$ are natural numbers. The first $F_{n}$’s are

[TABLE]

∎

3.2. Relation to values of the Fabius function.

Notice that $F(2^{-n})=\operatorname{up}(1-2^{-n})$. In [MR666839]*Theorem 6 it is shown that

[TABLE]

Since $\operatorname{up}(x)$ is an even function, it follows that

[TABLE]

Therefore, as noticed in [MR666839] we have

[TABLE]

The values of $F(2^{-n})=\operatorname{up}(1-2^{-n})$ depend of another interesting sequence of rational numbers.

3.3. The sequence of rational numbers $d_{n}$.

The sequence of numbers $d_{n}$ is introduced in [MR666839]*eq.(38) and (39) as the coefficients of the power series

[TABLE]

satisfying

[TABLE]

So these numbers are defined also by the recurrence [MR666839]*eq. (40)

[TABLE]

The $d_{n}$ are connected with $\operatorname{up}(t)$ in [MR666839]*eq. (36). It is shown there that the function

[TABLE]

satisfies (18). We give the simple proof that was omitted in [MR666839].

Proposition 2**.**

Let $f$ be defined as in (20), then we have (18).

Proof.

The functional equation (2) of $\operatorname{up}(t)$ implies for the Fourier transform that

[TABLE]

Therefore

[TABLE]

∎

By (20) we have

[TABLE]

So, equating the coefficients of equal powers with the expansion (17) we obtain [MR666839]*eq. (42)

[TABLE]

Together with [MR666839]*eq. (33) this yields

[TABLE]

The values of the $d_{n}$ can be computed easily applying the recurrence (19). The first terms are

[TABLE]

Proposition 3**.**

The values of $d_{n}$ can be computed in terms of $c_{n}$:

[TABLE]

Proof.

For $n=0$ we have $c_{0}=d_{0}=1$ and (23) is trivial in this case. For $n\geq 1$ we have the following computation

[TABLE]

∎

As in the case of the $c_{n}$, there is a sequence of natural numbers hiding in $d_{n}$.

Proposition 4**.**

There is a sequence of natural numbers $(G_{n})$ such that

[TABLE]

Proof.

For $n\geq 0$ define $G_{n}$ by the equation (24). The recurrence (19) then implies

[TABLE]

By induction we obtain that all the $G_{n}$ are natural numbers. ∎

The first numbers $G_{n}$ starting from $n=0$ are

[TABLE]

By (22) we obtain the explicit value

[TABLE]

Equations (16) and (25) yields

[TABLE]

So that $(2n+1)F_{n}\mid G_{2n+1}$.

4. Reshetnikov question.

Recently Vladimir Reshetnikov posed a question (question 261649 in MathOverflow) about Fabius function. The results in [MR666839] makes it easy to answer. Note that [MR666839] has been translated to English and posted it in arXiv (English version arXiv:1702.05442). We refer to this paper for some of the results.

The question of Reshetnikov is the following:

Question 5** (V. Reshetnikov).**

For any natural number $n$ define the numbers

[TABLE]

It is true that all $R_{n}$ are natural numbers?

The first $R_{n}$ (starting from $n=1$) certainly are:

[TABLE]

Applying some of the results in [MR666839] we will show that in fact all $R_{n}$ are natural numbers.

In Section 6 we consider the problem of the computation of the exact values of the function $F(x)$ at dyadic points. This will allows us to state another conjecture related to this conjecture of Reshetnikov.

5. Answer to Reshetnikov’s question.

Reshetnikov’s numbers can be computed in terms of the $d_{n}$. This will be essential in the solution of Reshetnikov’s question.

Proposition 6**.**

For all $n\geq 1$ we have

[TABLE]

Proof.

In the definition (27) of the $R_{n}$ substitute the value of $F(2^{-n})$ obtained from (22). ∎

First we consider the odd case.

Theorem 7**.**

For all $n\geq 0$ the number $R_{2n+1}$ is an integer multiple of $F_{n}$.

Proof.

Since $\operatorname{up}$ is an even function, we have by (21)

[TABLE]

Therefore by (28)

[TABLE]

By (11)

[TABLE]

Simplifying

[TABLE]

So that $R_{2n+1}$ is a natural number and $F_{n}\mid R_{2n+1}$. ∎

Proposition 8**.**

For $n\geq 1$ we have

[TABLE]

so that the denominator of $R_{2n}$ can be only a power of $2$.

Proof.

By (23) and (28) we have

[TABLE]

so that by (11)

[TABLE]

Simplifying we obtain (30). Since the factor $\frac{(4n-1)!!}{(2k+1)!!}$ is an integer in (30) the assertion about the denominator follows. ∎

Theorem 9**.**

For all $n$ the number $R_{n}$ is a natural number.

Proof.

By Theorem 7 it remains only to prove that $R_{2n}$ is a natural number. For a rational number $r$ let $\nu_{2}(r)$ be the exponent of $2$ in the prime decomposition of $r$. By Proposition 8, $R_{2n}$ is a natural number if and only if $\nu_{2}(R_{2n})\geq 0$. By (28) we have $\nu_{2}(R_{n})=\nu_{2}(2d_{n})$. We will show by induction that $\nu_{2}(2d_{n})\geq 0$ and this will prove the Theorem.

Since the numbers $F_{n}$ are natural, (29) implies that $\nu_{2}(2d_{2n+1})=\nu_{2}(R_{2n+1})\geq 0$. We have $\nu_{2}(2d_{0})=\nu_{2}(2)=1\geq 0$. For any other even number we use induction. By (19) we have for $n\geq 1$

[TABLE]

Hence

[TABLE]

By induction hypothesis $\nu_{2}(2d_{k})\geq 0$ for $0\leq k\leq 2n-1$. It follows that $\nu_{2}(2d_{2n})\geq 0$ for all $n\geq 1$. ∎

6. Computation of $\operatorname{up}(m/2^{n})$.

At dyadic points $t=m/2^{n}$ the function takes rational values. This was proved in [MR666839]*Th. 9, where a procedure to compute these values was given. In [MR666839] it was desired to express $\operatorname{up}(q/2^{n})$ in terms of $\operatorname{up}(1-2^{-k-1})$. This was achieved by substituting $c_{k}$ in terms of $\operatorname{up}(1-2^{-k-1})$. This is correct, but for computations it is preferable not to make this last substitution. In this way the formula in [MR666839]*Thm. 7 can be written

[TABLE]

This is valid for $-2^{-n}\leq q\leq 2^{n}$. The sum in $h$ has few terms for $q$ near $-2^{n}$. Therefore since $\operatorname{up}(1-x)=\operatorname{up}(-1+x)$ we obtain, with $q=a-2^{n}$,

[TABLE]

The numbers $c_{k}$ can be computed with the recurrence (10). This equation, together with $\operatorname{up}(t)+\operatorname{up}(1-t)=1$ for $0\leq t\leq 1$, allows one to easily compute the values at dyadic points. Another way has been proposed by Haugland recently [H]. Notice also that our formula can be written as follows

[TABLE]

Equation (33) show that the values at dyadic points of $\operatorname{up}(x)$, and therefore also those of $F(x)$, are rational numbers. In principle it allow the explicit computation of $\operatorname{up}(a/2^{n})$ and therefore of $F(a/2^{n})$ for all integers $a$. Nevertheless there is a better way to compute the function. It is thought that this better method is due to V. A. Rvachëv, brought to attention by V. Reshetnikov. He posted a solution to a question of Pierrot Bolnez about how to compute $F(x)$. Reshetnikov’s answer consists of a program in Mathematica to compute the function. The program has been reverse engineered, and proved the relevant result. At other point Reshetnikov cites Rvachëv and says that he thinks it is not translated to English. So a proof of the formula on which the method is based can be useful.

We need a Lemma.

Lemma 1**.**

Let $x>0$ be a real number and let $k$ be the unique integer such that $2^{k}\leq x<2^{k+1}$. Then for any non negative integer $n$ we have

[TABLE]

Proof.

Putting $t=2^{k}(1+u)$ in the integral we obtain

[TABLE]

Letting $a=2^{-k}y$, we have $0\leq a<1$ and we obtain

[TABLE]

Differentiating $F^{(k)}(x)=2^{\binom{k+1}{2}}F(2^{k}t)$ repeatedly we obtain

[TABLE]

It follows that

[TABLE]

When $0<u<a$ we have $1<u+1<2$. For $0<x<2$ we have by definition $F(x)=\operatorname{up}(x-1)$. Therefore $F^{\prime}(x)=2\operatorname{up}(2x-1)-2\operatorname{up}(2x-3)$. This implies for $0<u<1$ that $F^{\prime}(u)=2\operatorname{up}(2u-1)$ and $F^{\prime}(u+1)=-2\operatorname{up}(2u-1)=-F^{\prime}(u)$ Hence

[TABLE]

It only remains to reverse our steps. We have

[TABLE]

∎

Proposition 10**.**

Let $x>0$ be a real number. Let $n$ be the unique integer such that $2^{-n}\leq x<2^{-n+1}$ and put $y=x-2^{-n}$. Then we have

[TABLE]

where the sum is [math] when $n<0$.

Proof.

We use Taylor’s Theorem with integral rest

[TABLE]

When $n<0$, the number $2^{-n}$ is an even natural number. For $k\geq 0$ the derivative $F^{(k)}(2^{-n})=0$. So that for any $N\geq 0$,

[TABLE]

By Lemma (1),

[TABLE]

Again, by Taylor’s Theorem, applied at the point [math] where all derivatives are also $=0$, this is equal to $-F(y)$ as we wanted to prove.

For $n\geq 0$ we apply again Taylor’s Theorem with $N=n$. In this case

[TABLE]

Lemma (1) gives us

[TABLE]

The last equality using Taylor’s Theorem at the point $t=0$ where all derivatives of $F$ are equal to [math].

To finish the proof notice that by (3) and (22) we have for $0\leq k\leq n$,

[TABLE]

∎

Remark 11*.*

Notice that Proposition 10 reduces the computation of $F(x)$ to that of $F(y)$. If $x$ is dyadic a repeated use ends in $F(0)$. The number of steps being merely the number of $1$ in the dyadic representation of $x$. If $x$ is not dyadic a finite number of steps reduce the computation to that of $F(y)$ for some $y$ small enough so that $F(y)$ is less than the absolute error requested. This procedure is implemented by Reshetnikov in his answer to Pierrot Bolnez question in Mathematica StackExchange.

7. The least common denominator of $\operatorname{up}(a/2^{n})$.

Reshetnikov’s conjecture tries to determine the denominator of the dyadic number $F(2^{-n})$. It is more natural to consider the common denominator $D_{n}$ of all numbers $F(a/2^{n})$ where $a$ is an odd number. This is a finite number because we only have to consider the case of $a=2k+1$ for $k=0$, $1$, …, $2^{n-1}-1$.

Therefore we define as follow.

Definition 12**.**

Let $D_{n}$ be le least common multiple of the denominators of the fractions $\operatorname{up}(\frac{2k+1}{2^{n}})$ with $2^{-n}<2k+1\leq 2^{n}$.

The first $D_{n}$ starting from $D_{0}=1$ are

[TABLE]

Theorem 13**.**

For any natural number $n\geq 1$ and any integer $-2^{n}<a<2^{n}$ the number

[TABLE]

is natural.

Equivalently

[TABLE]

Proof.

By (32) we know that

[TABLE]

is a linear combination with integer coefficients of the rational numbers $c_{k}$ with $0\leq k\leq n/2$. We have $c_{0}=1$ and, for $n\geq 1$, by (11) the numbers

[TABLE]

are natural. Therefore, if $m=\lfloor n/2\rfloor$ then all the numbers

[TABLE]

are natural. It follows that the number in (36) is an integer. Since it is also positive it is a natural number. ∎

Remark 14*.*

Theorem 13 applies to any $a$ and not only for $2^{n}-1$ as in Reshetnikov’s conjecture. The power of $2$ is better in Reshetnikov’s conjecture than in Theorem 13, but for other primes Theorem 13 is better than Reshetnikov’s conjecture. It is thought that the power of $2$ can be improved in Theorem 13.

Proposition 15**.**

For any natural number $n$ the number $D_{n}$ is a multiple of the denominator of $d_{n}/(2^{\binom{n}{2}}\,n!)$.

Proof.

This follows directly from equation (22) and the definition of $D_{n}$, since $D_{n}$ is a multiple of the denominator of $\operatorname{up}(1-2^{-n})$. ∎

The quotients $D_{n}/\operatorname{den}(d_{n}/(2^{\binom{n}{2}}\,n!))$ for $1\leq n\leq 17$ are

[TABLE]

Conjecture 16**.**

The following appear to be true:

• (a)

The quotient $A_{n}=2^{-\binom{n}{2}}D_{n}$ satisfies $A_{2n}=A_{2n+1}$ for $n\geq 1$.

• (b)

Let $K_{n}=A_{2n-1}$ for $n\geq 1$, so that by (a) we have $K_{1}=A_{1}$, and for $n\geq 2$

[TABLE]

Then $2\times(2n-1)!$ divides $K_{n}$.

• (c)

The quotients $H_{n}=\frac{K_{n}}{2(2n-1)!}$ are odd integers.

The first few $H_{n}$ starting from $H_{1}=1$ are

[TABLE]

If these conjectures are true we obtain for $n\geq 1$

[TABLE]

8. On the nature of the numbers $d_{n}$.

In this Section we show that the numerator and denominator of $2d_{n}$ are odd numbers.

We have noticed in the proof of Theorem 9 that $\nu_{2}(2d_{n})=\nu_{2}(R_{n})$. So that by equation (29) it follows that $\nu_{2}(2d_{2n+1})=\nu_{2}(F_{n})$. We will show now that the numbers $F_{n}$ are odd. This will prove our result for $2d_{n}$ when $n$ is an odd number. For this proof we will use Lucas’s Theorem [MR0023257].

Theorem 17** (Lucas).**

Let $p$ be a prime and $n$ and $k$ two positive integers that we write as

[TABLE]

in base $p$. Then we have the congruence

[TABLE]

From Lucas’s Theorem we get the following application:

Proposition 18**.**

(a) For $n\geq 1$ the cardinal of odd combinatory numbers $\binom{2n+1}{2k}$ with $0\leq k\leq n$ is exactly $2^{w(n)}$.

(b) For $n\geq 1$ the number of odd combinatory numbers $\binom{2n+1}{k}$ with $0\leq k\leq 2n+1$ is $2^{w(n)+1}$.

Proof.

Let the binary expansion of $n$ be $\varepsilon_{r}\varepsilon_{r-1}\dots\varepsilon_{0}$, that is,

[TABLE]

The expansion of $2n+1$ is $\varepsilon_{r}\varepsilon_{r-1}\dots\varepsilon_{0}1$. If $\binom{2n+1}{2k}$ is odd $k\leq n$, then $k\leq 2^{r+1}$. Hence its binary expansion is of the form $k=\delta_{r}\cdots\delta_{0}$. Therefore Lucas’s Theorem says that

[TABLE]

Here each factor $\binom{0}{1}=0$ and each factor $\binom{1}{\delta}=1$ for any value of $\delta$. So $\binom{2n+1}{2k}\equiv 1$ is equivalent to saying that $\delta_{j}=0$ when $\varepsilon_{j}=0$. Therefore the number of odd combinatory numbers is $2^{m}$, where $m$ is the number of digits equal to $1$ in the expansion of $n$. But this is to say $m=w(n)$.

The second assertion can be proved in the same way. ∎

Proposition 19**.**

All the numbers $F_{n}$ are odd.

Proof.

We have $F_{0}=F_{1}=1$. We proceed by induction assuming that we have proved that all $F_{k}$ are odd numbers for $0\leq k\leq n-1$. Equation (13) gives $F_{n}$ as a sum of terms. Each term is a product of an odd number for a combinatory number $\binom{2n+1}{2k}$, with $k$ running on $0\leq k\leq n-1$. The missing combinatory number (that is, for $k=n$ is $\binom{2n+1}{2n}=2n+1$) is an odd number. So, by Proposition 18, the term adding to $F_{n}$ are even except for $2^{w(n)}-1$ of them. Since $w(n)\geq 1$ this implies that $F_{n}$ is odd. ∎

Theorem 20**.**

For all $n\geq 1$ we have $\nu_{2}(2d_{n})=0$, that is, the rational number $2d_{n}$ is the quotient of two odd numbers.

Proof.

We have seen that $F_{n}$ is an odd number and that this implies $\nu_{2}(2d_{2n+1})=0$. So we only need to show this for $2d_{2n}$ and $n\geq 1$ (notice that $2d_{0}=2$, so that $\nu_{2}(2d_{0})=1$).

We have shown (Theorem 9) that $R_{n}$ are integers. By Proposition 6 there is an odd number

[TABLE]

such that $2d_{n}N_{n}=R_{n}$ is an integer. It is easy to see that for $k\leq n$ we have $N_{k}\mid N_{n}$. Therefore for all $1\leq k\leq n$ we have $d_{k}N_{n}$ is an integer.

We proceed by induction, assuming we have proved that $\nu(2d_{k})=0$ for $1\leq k<2n$ and we have to show that $\nu_{2}(2d_{2n})=0$. We multiply (31) for $n\geq 1$ by $N_{2n}$ and obtain

[TABLE]

By the induction hypothesis $2d_{k}N_{2n}$ is an odd natural number for $1\leq k\leq 2n-1$ and it is even for $k=0$. So we only need to show that there are an odd number of odd combinatory numbers $\binom{2n+1}{k}$ with $1\leq k\leq 2n-1$. This follows easily from Propostition 18. ∎

Theorem 21**.**

The number $R_{n}$ is odd for $n\geq 1$. Or equivalently

[TABLE]

Proof.

We have seen that $2d_{n}$ is a quotient of two odd numbers and $R_{n}$ is a natural number. Therefore equation (28) implies that $R_{n}$ is an odd number.

Equation (22) yields

[TABLE]

∎

9. Relation with Bernoulli numbers.

The numbers $c_{n}$ and $d_{n}$ are related to Bernoulli numbers. Perhaps this may be useful to solve Conjecture 16.

Proposition 22**.**

The numbers $d_{n}$ and $c_{n}$ are given by the recurrences

[TABLE]

and

[TABLE]

where $B_{k}$ are Bernoulli numbers $B_{0}=1$, $B_{1}=-\frac{1}{2}$, $B_{2}=\frac{1}{6}$, $B_{3}=0$, …

Proof.

By (7) and (18) the functions

[TABLE]

satisfy the equations

[TABLE]

Using the well known power series

[TABLE]

and equating coefficients of the same power of the variable we obtain our recurrences. ∎