"Choix de Bruxelles": A New Operation on Positive Integers

TL;DR

The paper introduces the 'Choix de Bruxelles' operation on positive integers, exploring its properties and iterative behavior, where numbers are transformed by halving or doubling substrings of their decimal representations.

Contribution

It defines a novel operation on integers and analyzes its properties and iterative dynamics, which has not been studied before.

Findings

Characterization of possible transformations of integers

Analysis of the iterative behavior of the operation

Identification of fixed points and cycles

Abstract

The "Choix de Bruxelles" operation replaces a positive integer n by any of the numbers that can be obtained by halving or doubling a substring of the decimal representation of n. For example, 16 can become any of 16, 26, 13, 112, 8, or 32. We investigate the properties of this interesting operation and its iterates.

Tables

Table 1. Table 1: Numbers arising when “Choix de Bruxelles” is applied to the numbers 1 to 16.

Table 2. Table 2: τ ​ ( n ) = 𝜏 𝑛 absent \tau(n)= number of steps to reach n 𝑛 n from 1 1 1 , or − 1 1 -1 if n 𝑛 n cannot be reached.

Table 3. Table 3: Values of r ​ ( n ) 𝑟 𝑛 r(n) .

Equations

$$10218,40218,20118,20418,20228,20214,\text{andleavevmodenobreak }202116(!).$$

$$10218\text{leavevmode}\&\text{leavevmode}40218\text{leavevmodenobreak again},20428,2029(!),20236,$$

$$20218;10118,40418,20109,20436;40428;10109\text{leavevmodenobreak andleavevmodenobreak }40436.$$

$$\begin{array}[]{rrrrrrrrrrrrrrrrrrrr}n:&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&\cdots\\ M_{0}(n):&1&1&3&2&5&3&7&4&9&5&11&6&13&7&15&8&\cdots\\ M_{1}(n):&2&4&6&8&10&12&14&16&18&20&22&24&26&28&110&112&\cdots\end{array}$$

$$n\text{leavevmode}=\text{leavevmode}{d}_1{d}_2{d}_3\dots d_k$$

$$t_q\text{leavevmode}=\text{leavevmode}{d}_1\dots d_{p-1}\parallel 2(d_p\dots d_q)\parallel d_{q+1}\dots d_k,$$

$$\frac{n}{10}\text{leavevmode}<\text{leavevmode}m\text{leavevmode}<\text{leavevmode}10n,$$

$$M_1(n)=A10^{i+1}+10^i+x10^{i-1}+2B$$

$$1\text{leftrightline}2\text{leftrightline}4\text{leftrightline}8\text{leftrightline}16\text{leftrightline}112\text{leftrightline}56\text{leftrightline}28\text{leftrightline}14\text{leftrightline}12\text{leftrightline}6\text{leftrightline}3$$

$$\begin{array}[]{rrrrrrrrr}n:&1&2&3&99&369&999&1999&9879\\ \tau(n):&0&1&11&12&13&14&15&16\end{array}$$

$$\begin{array}[]{rrrrrrrrrrrrrr}5m:&5&10&15&20&25&30&35&40&45&50&55&60&\cdots\\ \text{steps}:&0&1&11&2&11&12&8&3&10&12&9&11&\cdots\\ \end{array}$$

$$\begin{array}[]{rrrrrrrrrr}5m:&5&10&15&30&100&200&400&9875&19995\\ \text{steps}:&0&1&11&12&13&14&15&16&17\\ \end{array}$$

$$\displaystyle 10+4i\leavevmode\nobreak\$$

$$\displaystyle 11+4i\leavevmode\nobreak\$$

$$\displaystyle 12+4i\leavevmode\nobreak\$$

$$\displaystyle 13+4i\leavevmode\nobreak\$$

$$8.112\text{leavevmode}10^{n-6}\text{leavevmode}<\text{leavevmode}R(n)\text{leavevmode}\le \text{leavevmode}8.113\text{leavevmode}10^{n-6}.$$

$${\log }_{10}n+5\text{leavevmode}<\text{leavevmode}\tau (n)\text{leavevmode}\le \text{leavevmode}12{\log }_{10}n.$$

Full text

1 Introduction

Let $n$ be a positive integer with decimal expansion $d_{1}d_{2}d_{3}\ldots d_{k}$. The “Choix de Bruxelles” operation replaces $n$ by any of the numbers that can be formed by taking a number $s$ represented by a substring $d_{p}d_{p+1}\ldots d_{q}$, with $1\leq p\leq q\leq k$, where the initial digit $d_{p}$ is not zero, and replacing the substring in situ by the decimal expansion of $2s$ or, if $s$ is even, by the decimal expansion of $s/2$. One may also leave $n$ unchanged (corresponding to choosing the empty substring).

Since the definition may be confusing at first glance, we give a detailed example. Suppose $n=20218$. By choosing substrings of length $1$, we can obtain any of

[TABLE]

(For the last of these, we replaced $8$ by $16$ in situ.) Using substrings of length $2$ we can obtain

[TABLE]

and using substrings of lengths $0,3,4,5$ we obtain

[TABLE]

It is the possibility of increasing or decreasing the number of digits by one that makes this operation interesting. Changing $16$ to $112$, for example, increases the number by a factor of $7$. In fact, as we see in Theorem 2.3, $n$ may change by a factor ranging from $\frac{1}{10}$ to $10$. Note that the operation is symmetric: if $n$ can be changed to $m$, then $m$ can be changed to $n$.

The name “Choix de Bruxelles” arose from a combination of several ideas: this a kind of mathematical game, like “sprouts” [1], sprouts are “choux de Bruxelles”, the operation was proposed by the first author, who lives in Brussels, and involves making choices.

Our goal is to investigate the properties of this operation, and to study what happens when it is iterated.

Section 2 studies the range of numbers that can be reached by applying the operation once. Theorems 2.1 and 2.2 give the largest and smallest numbers that can be obtained from $n$, and Theorem 2.3 describes their range.

Section 3 shows that by repeatedly applying the operation, any number not ending with [math] or $5$ can be transformed into any other such number, and any number ending in [math] or $5$ can be be transformed into any other number of that form, but the two classes always remain separate. The final section studies how many steps are need to reach $n$ from $1$, assuming that $n$ does not end in [math] or $5$.

One could also consider this operator in other bases. However, the binary version (see sequence A323465222Six-digit numbers prefixed by A refer to entries in the On-Line Encyclopedia of Integer Sequences. in [2]) is not very interesting—the operation preserves the binary weight—so we have not pursued this.

Notation. For clarity we will sometimes use the symbol $\parallel$ to indicate that the digits of two numbers are to be concatenated. For example, if $x=7$ and $y=8$, $(2x)\parallel y$ indicates the number $148$.

2 Numbers that can be reached in one step.

Table 1 shows the numbers that can be reached from $n$ in one step, for $1\leq n\leq 16$. The rows of this table form sequence A323460 in the OEIS [2].

The smallest ($M_{0}(n)$, A323462) and largest ($M_{1}(n)$, A323288) numbers in each row are as follows:

[TABLE]

Theorem 2.1**.**

The largest number $M_{1}(n)$ that can be obtained from

[TABLE]

by the Choix de Bruxelles operation is either $2n$, if all $d_{i}<5$, and otherwise is obtained by doubling the substring $d_{p}\ldots d_{k}$ where $d_{p}$ is the right-most digit $\geq 5$.

Proof.

If all $d_{i}<5$, the number of digits will not change during the operation, and so the best we can do is to double every digit, getting $M_{1}(n)=2n$.

If there are digits $d_{i}\geq 5$, then doubling any substring beginning with such a $d_{i}$ will increase the number of digits by one. The doubled substring will begin with $1$ instead of $d_{i}$, and so should be as far to the right as possible. (E.g., if we double a substring starting at $d_{3}=7$, say, we get a number $d_{1}d_{2}1\ldots$, whereas if we double a substring starting at $d_{5}=6$ we get $d_{1}d_{2}d_{3}d_{4}1\ldots$, which is a larger number since $d_{3}\geq 5$.)

Finally, if $d_{p}$ is the right-most digit $\geq 5$, let $t_{q}$ (where $p\leq q\leq k$) denote the result of doubling just the substring from $d_{p}$ through $d_{q}$. Then

[TABLE]

and this number is maximized by taking $q=k$. ∎

Theorem 2.2**.**

The smallest number $M_{0}(n)$ that can be obtained from (2.2) by the Choix de Bruxelles operation is as follows: If there is a substring $d_{r}...d_{s}$ of (2.2) which starts with $d_{r}=1$ and ends with an even digit $d_{s}$, take the string of that form which starts with the left-most $1$ and ends with the right-most even digit, and halve it. Otherwise, if there is any even digit, take the substring from $d_{1}$ to the right-most even digit and halve it. In the remaining case, all $d_{i}$ are odd, and $M_{0}(n)=n$.

The proof uses similar arguments to the proof of Theorem 2.1, and is omitted.

We can use Theorems 2.1 and 2.2 to get precise bounds on the range of numbers produced by the operation.

Theorem 2.3**.**

The numbers $m$ obtained by applying the Choix de Bruxelles operation to $n$ lie in the range

[TABLE]

and there are values of $n$ for which $m$ is arbitrarily close to either bound.

Proof.

For the upper bound, by Theorem 2.1 we can assume there is a digit of $n$ that is $\geq 5$. Let $d$ be the right-most such digit, and write $n=A\parallel d\parallel B=A\,10^{i}+d\,10^{i-1}+B$. Let $2d=10+x$, $x\in\{0,2,4,6,8\}$. Then

[TABLE]

and $M_{1}(n)<10n$ follows immediately. Numbers of the form $n=99\ldots 9$ with $d=9$, $i=1$, $B=0$ have $M_{1}(n)=99\ldots 918=10n-72$, so $M_{1}(n)/n=10-72/n$, which comes arbitrarily close to the bound.

For the lower bound, from Theorem 2.1 the only nontrivial case is when there is a substring $B=1\ldots e$, $e$ even, and then $n=A\,10^{i+j}+B\,10^{j}+C$, $M_{0}(n)=A\,10^{i+j-1}+\frac{B}{2}10^{j}+C$, which implies $n<10\,M_{0}(n)$. Numbers of the form $n=10^{t}+10$, $t\geq 2$, with $M_{0}(n)=10^{t-1}+5$, $M_{0}(n)/n=\frac{1}{10}(1+4/10^{t-1}+\cdots)$ come arbitrarily close to this bound. ∎

3 Which numbers can be reached from 1?

If we start with $1$ and repeatedly apply the Choix de Bruxelles operation, which numbers can we reach? From333We write $\leftrightline$ rather than $\rightarrow$ in describing these transformations, since the operation is symmetrical.

[TABLE]

we can reach $3$ in $11$ steps444Found by Lorenzo Angelini., as well as $2,4,8$ and $6$, and $\cdots\leftrightline 28\leftrightline 14\leftrightline 7$ and $\cdots\leftrightline 28\leftrightline 14\leftrightline 18\leftrightline 9$ give us all the numbers less than $10$ except $5$. Further experimenting suggests that $5$ and $10$ may be impossible to reach. This is true, as we now show.

Let us define an undirected graph $G$ with vertices labeled by the positive integers, where vertices $n$ and $m$ are joined by an edge if Choix de Bruxelles takes $n$ to $m$.

Theorem 3.1**.**

The graph $G$ has two connected components, one containing all numbers whose decimal expansion does not end in [math] or $5$, the other containing all numbers which do end in [math] or $5$.

Proof.

(i) All numbers not ending in [math] or $5$ are connected to $1$. If not, let $n$ be the smallest number, not ending in [math] or $5$, that cannot be reached from 1. Any number $m$ that is reachable from $n$ must be larger than $n$, or else by symmetry $m$ would be a smaller counter-example. This means that all digits of $n$ are odd, and cannot be $3$, $7$, or $9$ (since by (3.1) etc. they could be reduced to $1$). If $n$ has an internal digit $5$, we can use $51\leftrightline 102\leftrightline 52\leftrightline 26$ to get a smaller number. So any $5$s must appear in a string at the end of $n$, which contradicts our assumptions. The only remaining possibility is that $n=11\ldots 11$. But even these numbers can be reduced by using $11\leftrightline 12\leftrightline 6$.

(ii) Any number ending in [math] or $5$ is connected to $5$. This follows from (i), using $n0\leftrightline 10\leftrightline 5$ and $n5\leftrightline n\parallel 10\leftrightline 10\leftrightline 5$.

(iii) The Choix de Bruxelles operations never change a final [math] or $5$ into any digit other than [math] or $5$. ∎

4 How many steps to reach $n$?

Let $\tau(n)$ be the number of steps to reach $n$ from $1$ using the Choix de Bruxelles operation, or $-1$ if $n$ cannot be reached from $1$. We found the values of $\tau(n)$ for $n\leq 10000$ by computer (A323454). The initial values are shown in Table 2.

We will derive upper and lower bounds on $\tau(n)$ for large $n$. The record high values of $\tau(n)$ that are presently known, and the values of $n$ where they occur, are as follows (see A323463):

[TABLE]

In particular, the entry $\tau(n)=12$ for $n=99$ means that every number less than $100$ and not ending in [math] or $5$ can be reached from $1$ in at most $12$ steps.

We also need analogous data for the other class of numbers. The numbers of steps to reach $5m$ from $5$ (A323484) are:

[TABLE]

and the record high values that are presently known, and the values of $5m$ where they occur, are as follows (see A323464):

[TABLE]

In particular, the entry $13$ for $n=100$ means that every multiple of $5$ less than $100$ can be reached from $5$ in at most $12$ steps, and so all numbers less than $100$ can be reached from either $1$ or $5$ in at most $12$ steps.

To get an upper bound on $\tau(n)$ for larger $n$, consider a $k$-digit number $n$ given by (2.2). We can repeatedly replace the two leading digits by a single digit (by $1$ or $5$, in fact) at the cost of at most $12$ steps. So we can reach $1$ or $5$ in at most $12(k-1)$ steps. Since $k=\lfloor 1+\log_{10}n\rfloor$, this takes at most $12\log_{10}n$ steps. In particular, $\tau(n)\leq 12\log_{10}n$.

To get a lower bound on $\tau(n)$, we note that Theorem 2.3 implies that at least $\log_{10}n$ steps will be needed to reach $n$ from $1$. We can make this more precise by finding $R(n)$, the largest number that can be reached from $1$ in $n$ steps. The values of $R(n)$ can almost be found by a greedy algorithm. Set $r(0)=1$, and, for $n>0$, let $r(n)$ be the largest number that can be obtained by applying Choix de Bruxelles to $r(n-1)$ (of course $R(n)\geq r(n)$). Theorem 2.1 tells us how to calculate $r(n)$, and the first $20$ values are shown in Table 3. We find that $r(n+4)=8112\parallel r(n)$ for $n\geq 10$.

Theorem 4.1**.**

$R(n)$, the largest number that can be reached from $1$ in $n$ steps, is equal to $r(n)$ for $n\neq 7$, and $R(7)=512$.

Proof.

By computer calculation, the result is true for $n\leq 16$. Suppose $n\geq 17$. The candidates for $R(n)$ are all the numbers that can be obtained by applying Choix de Bruxelles to the $(n-1)$st generation numbers—numbers that can be reached from $1$ in $n-1$ steps. In view of Theorem 2.3, we can discard any $(n-1)$st generation numbers that are less than $\frac{1}{10}r(n)$. The sets of remaining candidates form a repeating pattern of period four, which, starting in generation $14$, are as follows (here $i=1,2,3,\cdots$ and $P$ denote $i$ copies of the string $8112$):

[TABLE]

At each new generation, the largest number is obtained by expanding the final number in each row (using Theorem 2.3), and the resulting numbers are the $r(n)$, as claimed. ∎

In summary, we have:

Theorem 4.2**.**

For $n\geq 14$,

[TABLE]

The upper bound in Theorem 4.2 states that, starting at $1$, we cannot reach a number greater than $c\,10^{n-6}$ in $n$ steps, where $c=8.113$. By solving $c\,10^{n-6}=m$, we see that to reach $m$ from $1$ we require at least $n=6+\log_{10}(m/c)=\log_{10}m+5.09\ldots$ steps.

By combining this with our earlier result, we have our final theorem.

Theorem 4.3**.**

The number of steps needed to reach $n$ from $1$ by the Choix de Bruxelles operation, assuming $n$ does not end in [math] or $5$, is bounded by

[TABLE]