Bounds for sets of remainders

TL;DR

This paper studies the sequence counting distinct remainders for numbers modulo all smaller divisors, providing asymptotic bounds, analyzing differences between terms, and exploring related iterated remainder sets.

Contribution

It establishes explicit asymptotic formulas for the sequence, bounds on term differences, and analyzes properties of iterated remainder sets related to Pierce expansions.

Findings

s(n) = c * n + O(n/(\log n \log \log n))

Differences between s(n) and s(n+1) are at most one, with arbitrarily large decreases

Bounds are provided for the size of iterated remainder sets

Abstract

Let $s(n)$ be the number of different remainders $n \bmod k$, where $1 \leq k \leq \lfloor n/2 \rfloor$. This rather natural sequence is sequence A283190 in the OEIS and while some basic facts are known, it seems that surprisingly it has barely been studied. First, we prove that $s(n) = c \cdot n + O(n/(\log n \log \log n))$, where $c$ is an explicit constant. Then we focus on differences between consecutive terms $s(n)$ and $s(n+1)$. It turns out that the value can always increase by at most one, but there exist arbitrarily large decreases. We show that the differences are bounded by $O(\log \log n)$. Finally, we consider ''iterated remainder sets''. These are related to a problem arising from Pierce expansions, and we prove bounds for the size of these sets as well.

Tables

Table 1. Table 1 . First eighteen values of s ​ ( n ) s(n) .

$n$	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18
$s\left(n\right)$	0	1	1	1	2	1	2	2	2	3	4	2	3	3	3	4	5	3

Equations

$$S(n):=\{n\mathrm{mod}k:k\in \{1,2,\dots ,\lfloor n/2\rfloor \}\}.$$

$$s(n):=|S(n)|.$$

$$\underset{n\to \infty }{\lim }\frac{s(n)}{n}=\sum _{p\in \mathbb{P}}\frac{1}{p(p+1)}\cdot \prod _{\begin{array}{c}q\in \mathbb{P},\\ q<p\end{array}}\left(1-\frac{1}{q}\right)\approx \mathrm{0.2296.}$$

$$c:=\sum _p\frac{1}{p(p+1)}\cdot \prod _{\begin{array}{c}{p}^{\prime }<p\end{array}}\left(1-\frac{1}{p^{\prime }}\right)\approx \mathrm{0.2296.}$$

$$s(n)=c\cdot n+O\left(\frac{n}{\log n\log \log n}\right).$$

$$\underset{n\to \infty }{\lim }\frac{s(n)}{n}=c.$$

$$s(n)-O(\log \log n)\le s(n+1)\le s(n)+1.$$

$$\underset{n\to \infty }{\mathrm{liminf}}s(n+1)-s(n)=-\infty .$$

$$S_0(n)$$

$$S_j(n)$$

$$s_j(n):=|S_j(n)|.$$

$$\frac{1}{(j+2)!}\le \underset{n\to \infty }{\mathrm{liminf}}\frac{s_j(n)}{n}\le \underset{n\to \infty }{\mathrm{limsup}}\frac{s_j(n)}{n}\le \frac{1}{j+2}.$$

$$n=k\cdot q+r$$

$$n-r=k\cdot q$$

$$D_p:=\{m\in \mathbb{Z}:p\mid m,\text{ and }{p}^{\prime }\nmid m\text{ for every prime }{p}^{\prime }<p\}.$$

$$|D_p\cap [a+1,a+t]|=t\cdot \frac{1}{p}\cdot \prod _{\begin{array}{c}{p}^{\prime }<p\end{array}}\left(1-\frac{1}{p^{\prime }}\right)+E(a,t,p),$$

$$|E(a,t,p)|\le 2^{\pi (p-1)}.$$

$$|D_p\cap [a+1,a+t]|=$$

$$+\sum _{p_1<p_2<p}\left(\lfloor \frac{a+t}{p{p}_1{p}_2}\rfloor -\lfloor \frac{a}{p{p}_1{p}_2}\rfloor \right)-+\dots$$

$$\pm \left(\lfloor \frac{a+t}{p{\prod }_{p^{\prime }<p}{p}^{\prime }}\rfloor -\lfloor \frac{a}{p{\prod }_{p^{\prime }<p}{p}^{\prime }}\rfloor \right).$$

$$|D_p\cap [a+1,a+t]|=$$

$$=t\cdot \frac{1}{p}\cdot \prod _{\begin{array}{c}{p}^{\prime }<p\end{array}}\left(1-\frac{1}{p^{\prime }}\right)+E(a,t,p),$$

$$\pi (n)\le 1.3\cdot \frac{n}{\log n}.$$

$$\sum _{p>n}\frac{1}{p^2}=O\left(\frac{1}{n\log n}\right).$$

$$\sum _{p>n}\frac{1}{p^2}$$

$$\le C_2\cdot \frac{1}{(\log (n/\log n){)}^2}\cdot {\int }_{n/\log n}^{\infty }\frac{1}{t^2}dt$$

$$\le C_3\cdot \frac{1}{(\log n{)}^2}\cdot \frac{1}{n/\log n}=C_3\cdot \frac{1}{n\log n}.$$

$$R_p(n):=\left\{r:1\le r\le \frac{n-p}{p+1},p\mid n-r,\text{ and }{p}^{\prime }\nmid n-r\text{ for every prime }{p}^{\prime }<p\right\}.$$

$$S(n)\setminus \{0\}=\bigcup _{p<n}{R}_p(n).$$

$$s(n)-1=|S(n)\setminus \{0\}|=\sum _{p<n}|R_p(n)|.$$

$$M_p(n):=\left\{m\in \left[n-\frac{n-p}{p+1},n-1\right]:p\mid m,\text{ and }{p}^{\prime }\nmid m\text{ for every prime }{p}^{\prime }<p\right\},$$

$$s(n)=1+\sum _{p<n}|M_p(n)|=\sum _{p<X}|M_p(n)|+\underset{=:E_1(n)}{\underbrace{1+\sum _{X\le p<n}|M_p(n)|}}.$$

$$|E_1(n)|=1+\sum _{X\le p<n}|M_p(n)|\le 1+\sum _{p\ge X}\frac{n}{(p+1)p}\le 1+n\cdot \sum _{p\ge X}\frac{1}{p^2}=O\left(\frac{n}{X\log X}\right),$$

Full text

Research of the second author supported by NSERC grant 2024-03725.

1.  Introduction

Let us fix an integer $n$. Then a natural question is the following: Which are the remainders $r=n\bmod k$? The sum of such remainders with $1\leq k\leq n$ was already considered by Édouard Lucas [5, p. 373] and more recently in [8, 10, 3]. In the present paper, we are simply interested in the number of distinct remainders. Surprisingly, it seems that this has barely been studied.

If $k>n$, then we get $r=n$ for all $k$.

If $\lfloor n/2\rfloor+1\leq k\leq n$, then $r=n-k$, so in this range we simply get all the integers between [math] and $\lceil n/2\rceil-1$.

Therefore, the interesting cases are where $1\leq k\leq\lfloor n/2\rfloor$, and we define the set

[TABLE]

Note that clearly $S(n)\subseteq\{0,1,\ldots,\lfloor n/2\rfloor-1\}$. Now let us define

[TABLE]

In other words, $s(n)$ is the number of different values $n\bmod k$ for $1\leq k\leq\lfloor n/2\rfloor$. This is precisely sequence A283190 in the OEIS (On-Line Encyclopedia of Integer Sequences) [6]. The first few numbers of the sequence are presented in Table 1.

Although it seems like a very natural sequence, it was entered into the OEIS only in 2017 by Thomas Kerscher. Robert Israel then observed that $s(n)/n$ seems to converge to approximately $0.2296$. He asked about the actual value of the constant on the StackExchange website [4]. This was answered by the user Empy2 (and added as a comment to the OEIS entry by Michael Peake):

[TABLE]

The answer on the StackExchange website also contains an explanation for why this is true. While the argument is relatively simple, we haven’t been able to find a rigorous proof in print. In this paper, in Section 3, we give a proof of (1), including a bound for the error term.

Then we investigate the differences between consecutive terms of $s(n)$. Looking at the first few values of the sequence $s(n)$, one sees that the values of $s(n+1)$ compared to $s(n)$ usually either stay the same or increase or decrease by $1$. They never seem to increase by more than $1$, but sometimes they decrease by $2$ or, as it turns out, even more. In Section 4 we show that these differences can get arbitrarily large, while being double logarithmically bounded in terms of $n$.

In Section 5, we generalize the set $S(n)=:S_{1}(n)$ to “iterated remainder sets” via $S_{j+1}(n):=\{n\bmod{k}\colon k\in S_{j}(n)\setminus\{0\}\}$ for $j\geq 1$. These sets are related to an older problem arising from Pierce expansions. This problem was considered in [9, 2, 1], and the main question is still open: If we fix a positive integer $n$, choose another integer $1\leq a\leq n$, and repeatedly set $a:=n\bmod a$, what is the largest number of steps performed before reaching $a=0$? More on this and the relation to our iterated remainder sets in Section 5. In Section 6 we prove some bounds for these sets.

Finally, in Section 7 we pose some open problems.

But first of all, we present all main results in the next section.

2.  Main results and some lemmas

Let us define the constant from (1), namely

[TABLE]

In Section 3 we will prove the following asymptotics for $s(n)$.

Theorem 1**.**

We have

[TABLE]

In particular,

[TABLE]

We do not believe that our bound for the error term is sharp. Numerical experiments suggest that perhaps $O(n^{1/3})$ might be closer to the truth; see Section 7. In any case, we know that overall $s(n)$ grows linearly, but due to the error term this does not give us much information on the differences $s(n+1)-s(n)$.

As mentioned in the introduction, looking at the first few values of $s(n)$, it seems that the values of $s(n+1)$ compared to $s(n)$ usually either stay the same or increase or decrease by $1$. It turns out that they never increase by more than $1$, but sometimes they do decrease by $2$ or even more. For example, $s(131)=33$ and $s(132)=30$. The first time that the value decreases by $4$ happens at $n=17291$, where $s(17291)=3975$ and $s(17292)=3971$. We have searched up to $n=10^{7}$ and have not found a decrease by more than $4$ in this range. However, it turns out that there do exist arbitrarily large decreases, and we will give a construction for such $n$. On the other hand, we can prove bounds on the decreases in terms of $n$. In Section 4 we will, in particular, prove the following results.

Theorem 2**.**

For $n\geq 1$ we have

[TABLE]

Moreover,

[TABLE]

Now let us define the sets of iterated remainders of $n$ inductively by

[TABLE]

Note that, in particular, $S(n)=S_{1}(n)$. Moreover, analogously to $s(n)$, we define

[TABLE]

In Section 6, we will prove the following bounds.

Theorem 3**.**

For $j\geq 0$ we have

[TABLE]

While the bounds are clearly not sharp (for example, set $j=1$ and compare to Theorem 1), it seems that for $j\geq 2$ there is indeed a gap between the limit inferior and limit superior; see Section 7.

Before moving on, we state two simple lemmas. All results on $s(n)$ will be based on the following equivalence.

Lemma 1**.**

Let $r\in\{0,1,\ldots,\lfloor n/2\rfloor-1\}$. Then $r\in S(n)$ if and only if $n-r$ has a proper divisor $k\geq r+1$.

Proof.

First, assume $r\in S(n)$. This means that there exist integers $k\in\{1,2,\ldots,\lfloor n/2\rfloor\}$ and $q$ such that

[TABLE]

and $r\leq k-1$. In other words, $k$ is a divisor of $n-r$ with $k\geq r+1$. Moreover, $k\leq\lfloor n/2\rfloor$ implies $q\geq 2$, and therefore $k$ is indeed a proper divisor of $n-r$.

Now assume conversely that $n-r$ has a proper divisor $k\geq r+1$. Then we have

[TABLE]

for some $q\geq 2$. This means that $r=n\bmod k$. Moreover, $q\geq 2$ implies $k=(n-r)/q\leq\lfloor n/2\rfloor$, and so by definition $r\in S(n)$. ∎

Finally, let us give a slightly more precise statement than $S(n)\subseteq\{0,1,\ldots\lfloor n/2\rfloor-1\}$.

Lemma 2**.**

We have $S(n)\subseteq\{0,1,\ldots,\lfloor(n-2)/3\rfloor\}$.

Proof.

Let $r\in S(n)$. Then by Lemma 1, the number $n-r$ has a proper divisor $k\geq r+1$. In other words, $n-r=k\cdot q$ with $q\geq 2$. This implies $n-r\geq(r+1)\cdot 2$, and rewriting the inequality yields $n\geq 3r+2$. ∎

Note that throughout the paper, $p$ will denote a prime. In particular, if we sum or take the union over an index $p$, the numbers $p$ are implied to be primes.

3.  Asymptotics for $s(n)$ (proof of Theorem 1)

In this section we want to prove that $s(n)$ asymptotically behaves like $c\cdot n$ and compute a bound for the error term. We will do this with a straightforward sieving argument. In preparation for this, let us define the set of integers that are divisible by $p$ but not by any smaller prime $p^{\prime}<p$:

[TABLE]

As usual, $\pi(n)$ will denote the number of primes $p\leq n$.

The next lemma is standard; we provide a proof for completeness.

Lemma 3**.**

Let $a,t$ be positive integers and $p$ a prime. Then

[TABLE]

and the error term $E(a,t,p)$ is bounded by

[TABLE]

Proof.

This follows from a simple inclusion–exclusion argument. Indeed, we have

[TABLE]

Each difference of the shape $\lfloor(a+t)/q\rfloor-\lfloor a/q\rfloor$ is equal to $t/q+\delta$, with some error term $\delta$ with $\lvert\delta\rvert<1$. There are exactly $2^{\pi(p-1)}$ such differences. Therefore,

[TABLE]

with $\lvert E(a,t,p)\rvert<2^{\pi(p-1)}$, as desired. ∎

We will also use the next lemma, which is a classical upper bound for the prime counting function [7, Corollary 1].

Lemma 4**.**

For every $n\geq 2$ we have

[TABLE]

Finally, we will use the following simple estimate.

Lemma 5**.**

We have

[TABLE]

Proof.

By the prime number theorem, the $k$-th $p_{k}$ prime is asymptotically of the size $p_{k}\sim k\log k$. The lemma follows from a density argument. To be rigorous, we can do the following estimates for sufficiently large $n$, where $C_{1},C_{2},C_{3}$ are some positive constants:

[TABLE]

∎

Now we are ready to prove the asymptotics for $s(n)$.

Proof of Theorem 1.

We want to use Lemma 1 to count the remainders $r\in S(n)\setminus\{0\}$. Note that $n-r$ having a proper divisor $k\geq r+1$ is equivalent to $n-r$ having a prime factor $p\leq(n-r)/(r+1)$. Moreover, note that the last inequality is equivalent to $r\leq(n-p)/(p+1)$. We want to count the remainders $r$ systematically by going through the prime numbers $p$. Therefore, we define for every $n$ and every prime $p<n$ the set

[TABLE]

Then we have

[TABLE]

Since all sets in the union are disjoint by construction, we have in particular

[TABLE]

Instead of counting the numbers $r\in R_{p}(n)$, we can equivalently count the numbers $m=n-r$. In other words, if we define

[TABLE]

then we have $\lvert R_{p}(n)\rvert=\lvert M_{p}(n)\rvert$. Let $X=X(n)$ be some threshold function with $1<X(n)<n$ that we will fix later. Our strategy is to compute

[TABLE]

We do this because on the one hand, if $p$ is small compared to $n$, we can compute $\lvert M_{p}(n)\rvert$ relatively precisely. On the other hand, if $p$ is sufficiently large, $\lvert M_{p}(n)\rvert$ is small, and we can estimate the error term $E_{1}(n)$ trivially. Let us do the latter first. We have

[TABLE]

where for the last estimate we used Lemma 5.

In order to compute the main term in (4), for each $p<X$ we apply Lemma 3 with $a=n-\frac{n-p}{p+1}-1$ and $t=\frac{n-p}{p+1}$. This gives us

[TABLE]

with

[TABLE]

Now we can write

[TABLE]

with

[TABLE]

where in the error term we estimated the first sum very crudely by $X$, and for the second sum we used the definition of $\pi(x)$.

Finally, recall that

[TABLE]

Thus, we can write

[TABLE]

with

[TABLE]

where for the last estimate we used Lemma 5.

Overall, we have from (4) and (7) that

[TABLE]

Using the bounds for the error terms from (5), (6), and (8), and setting $X=X(n)=\log n$, we get

[TABLE]

The theorem follows upon noting that with Lemma 4 for sufficiently large $n$ we get

[TABLE]

∎

Remark 1**.**

At first sight, the last estimate in the proof above seems very rough. However, increasing $X(n)$ from $\log n$ even to just $(\log n)^{1+\varepsilon}$ does not work using the same arguments. In order to improve the bound, one would have to find stronger error bounds.

4.  Differences between consecutive terms of $s(n)$ (proof of Theorem 2)

In this section we try to better understand how $s(n)$ changes as $n$ increases by 1.

First, observe the following relation between elements in $S(n+1)$ and $S(n)$.

Lemma 6**.**

For every $n\geq 1$ we have

[TABLE]

Proof.

Let $r\in S(n+1)\setminus\{0\}$. Then by Lemma 1 the integer $n+1-r$ has a proper divisor $k\geq r+1$. Now since $n-(r-1)=n+1-r$ has a proper divisor $k\geq r+1\geq r$, we have $r-1\in S(n)$. ∎

From this, we immediately see that the value of $s(n)$ can increase by at most one at a time:

Proposition 1**.**

For every $n\geq 1$ we have

[TABLE]

Proof.

We have

[TABLE]

where we applied Lemma 6 for the last inequality. ∎

Note that $0=n\bmod 1$ is always in $S(n+1)$ and recall Lemma 6 and it’s proof. We can interpret it in the following way: The set $S(n+1)$ consists precisely of the element [math] and all the elements $r+1$ that were “transferred” from $S(n)$ to $S(n+1)$ by being increased by $1$. Thus, in order to understand the difference $s(n+1)-s(n)$ we need to understand how many elements $r$ were not transferred. We denote the set of “not transferred elements” by

[TABLE]

Then we have

[TABLE]

Next, we characterize the set $T(n,n+1)$.

Lemma 7**.**

We have $r\in T(n,n+1)$ if and only if $r+1$ is the largest proper divisor of $n-r$.

Proof.

This follows directly from Lemma 1: $r\in S(n)$ is equivalent to $n-r$ having a proper divisor $\geq r+1$, and $r+1\notin S(n+1)$ is equivalent to $(n+1)-(r+1)=n-r$ not having a proper divisor $\geq r+2$. ∎

In order to count such occurrences, we will use the next simple lemma.

Lemma 8**.**

Let $n,d$ be positive integers. Then $d$ is the largest proper divisor of $n$ if and only if $n/d=p$ for some prime $p$ and every prime factor of $d$ is $\geq p$.

Now for even $n$, the situation is rather simple:

Proposition 2**.**

Let $n\geq 2$ be even. Then

[TABLE]

In particular,

[TABLE]

Proof.

By Lemma 7 we have $r\in T(n,n+1)$ if and only if $r+1$ is the largest proper divisor of $n-r$. By Lemma 8 this is the case if and only if $(n-r)/(r+1)=p$ for some prime $p$ and every prime factor of $r+1$ is $\geq p$. Let us find all $r$ where this is the case. Assume first that $r$ is odd. Then $(n-r)/(r+1)=\text{odd}/\text{even}=p$, which is impossible. Now assume that $r$ is even, and we have $(n-r)/(r+1)=\text{even}/\text{odd}=p$. This means that $p=2$ and $n=3r+2$. So this happens if and only if $n\equiv 2\pmod{3}$ and $r=(n-2)/3$.

The second part of the lemma now follows immediately from formula (9). ∎

For odd $n$, we will prove that $\lvert T(n,n+1)\rvert$ is at most double logarithmic in $n$.

Let us first use Lemmas 7 and 8 to characterize the sets $T(n,n+1)$ more precisely.

Lemma 9**.**

Assume that $n+1=p_{1}^{x_{1}}\cdots p_{\ell}^{x_{\ell}}$ is the prime factorization of $n+1$ with $p_{1}<\dots<p_{\ell}$. Let $1\leq r<(n-2)/3$. Then $r\in T(n,n+1)$ if and only if all of the following conditions are satisfied:

1. (i)

$r+1=p_{I}^{x_{I}}\cdots p_{\ell}^{x_{\ell}}$ for some index $2\leq I\leq\ell$; 2. (ii)

$(n+1)/(r+1)-1=p_{1}^{x_{1}}\cdots p_{I-1}^{x_{I-1}}-1=p$ for some prime $p$; 3. (iii)

$p\leq p_{I}$.

Moreover, $r=0\in T(n,n+1)$ if and only if $n=p_{1}^{x_{1}}\cdots p_{\ell}^{x_{\ell}}-1$ is prime. Finally, if $n\equiv 2\pmod{3}$, then $r=(n-2)/3\in T(n,n+1)$.

Proof.

By Lemma 7 we have $r\in T(n,n+1)$ if and only if $r+1$ is the largest proper divisor of $n-r$. By Lemma 8 this is equivalent to

[TABLE]

Let $1\leq r<(n-2)/3$ and assume that $n+1=p_{1}^{x_{1}}\cdots p_{\ell}^{x_{\ell}}$. We check that (10) and (11) are equivalent to the three conditions in the statement of the lemma.

Assume first that (10) and (11) hold. Then since $r+1\mid n+1$, we have

[TABLE]

Let $I$ be the index such that $p_{1}<\dots<p_{I-1}<p\leq p_{I}<\dots<p_{\ell}$ (where $I=1$ is allowed and means that $p\leq p_{1}$). Then (11) implies that $y_{1}=\dots=y_{I-1}=0$. Moreover, for every $I\leq i\leq\ell$ we can write

[TABLE]

with some integer $A\geq 1$. Then $p_{I}\geq p$ implies $x_{i}-y_{i}=0$ for all $i=I,\ldots,\ell$, except if $p_{i}=3$, $x_{i}-y_{i}=1$, $A=1$ and $p=2$. The exceptional situation is $(n+1)/(r+1)=3$, which is equivalent to $r=(n-2)/3$ and was excluded. In the other situations, we have $y_{I}=x_{I},\ldots,y_{\ell}=x_{\ell}$, and the three conditions in the lemma are clearly satisfied (the condition $I\geq 2$ follows from $(n+1)/(r+1)=p+1>1$).

Conversely, it is clear that the three conditions in the lemma imply (10) and (11).

The statement for $r=0$ holds because always $0\in S(n)$ and by Lemma 1 we have $1\in S(n+1)$ if and only if $n=(n+1)-1$ is composite.

Finally, assume $r=(n-2)/3$ is an integer. Then $n\bmod(n+1)/3=r$ and so $r\in S(n)$. On the other hand, $r+1>(n-2)/3=\lfloor(n+1-2)/3\rfloor$ and so by Lemma 2 we have $r+1\notin S(n+1)$. ∎

Proposition 3**.**

For every odd $n\geq 1$ we have

[TABLE]

Proof.

In view of formula (9) it suffices to prove that

[TABLE]

In other words, we want to show that there are at most $O(\log\log n)$ numbers $r$ satisfying the three conditions in Lemma 9. Fix an odd $n\geq 1$ an let

[TABLE]

be the prime factorization of $n$ with $p_{1}<\dots<p_{\ell}$. Then by Lemma 9, $r+1$ must be of the shape $r+1=p_{I}^{x_{I}}\cdots p_{\ell}^{x_{\ell}}$ for some index $2\leq I\leq\ell$. Moreover, for these indices $I$, we have

[TABLE]

Assume that there are $t$ such numbers $r$, corresponding to the indices $I_{1}<\dots<I_{t}$. Then we can weaken the above inequality to

[TABLE]

Using this inequality inductively, we get

[TABLE]

This implies

[TABLE]

and so $t=O(\log\log n)$, as desired. ∎

Proposition 4**.**

We have

[TABLE]

Proof.

Let us set

[TABLE]

In view of formula (9), our strategy is to use Lemma 9 to recursively construct numbers $n^{(1)},n^{(2)},\ldots$ with the properties

• •

$n^{(j)}$ is prime and

• •

$t(n^{(j)})\geq j$.

For good intuition, let us set the first three values straight away:

[TABLE]

This works because of Lemma 9 and the following facts: $2^{2}-1$ is prime and $2^{2}-1\geq 3$, and $2^{2}\cdot 3-1$ is prime and $2^{2}\cdot 3-1\geq 11$, and also $2^{2}\cdot 3\cdot 11-1$ is prime.

Now assume we have already constructed $n^{(j)}$ for some $j\geq 3$. Set

[TABLE]

For reasons that will become apparent in (12), we want to choose an integer $y$ with the following properties:

1. (a)

$\gcd(xP+y,\,P)=1$ for all integers $x$. This is equivalent to $y\not\equiv 0\pmod{p}$ for all $p\mid P$. 2. (b)

$\gcd(y(n^{(j)}+1)-1,\,P)=1$. This is equivalent to $y(n^{(j)}+1)\not\equiv 1\pmod{p}$ for all $p\mid P$ with $p\nmid n^{(j)}+1$.

Overall, we want $y$ to satisfy

[TABLE]

For each $p\geq 3$ this is clearly possible, since at most two residue classes need to be avoided. For the case $p=2$, note that since $n^{(j)}$ is a prime, $n^{(j)}+1$ is even, so $p\mid n^{(j)}+1$ and we only need to avoid $y\equiv 0\pmod{2}$.

Therefore, by the Chinese remainder theorem, there exists an integer $1\leq y\leq P$ with properties (a) and (b).

We fix such an integer $y$ and consider the arithmetic progression

[TABLE]

Now our property (b) implies that

[TABLE]

Thus, by Dirichlet’s prime number theorem, there exists an integer $x\geq 1$ such that $a_{x}$ is a prime. We set

[TABLE]

By construction, $n^{(j+1)}$ is prime, and we finally only need to check that indeed $t(n^{(j+1)})\geq t(n^{(j)})+1\geq j+1$.

It is easy to see that by construction $n^{(j+1)}\equiv 2\pmod{3}$ for all $j\geq 2$ and $n^{(j+1)}$ is prime, so the special cases from Lemma 9, namely $r=0$ and $r=(n^{(j+1)}-2)/3$, are always in $T(n^{(j+1)},n^{(j+1)}+1)$.

Moreover, let us write $n^{(j)}=p_{1}^{x_{1}}\cdots p_{\ell}^{x_{\ell}}-1$ with $p_{1}<\dots<p_{\ell}$. Then we have $n^{(j+1)}=p_{1}^{x_{1}}\cdots p_{\ell}^{x_{\ell}}\cdot Q-1$, where all prime factors in $Q=xP+y$ are larger than $p_{\ell}$ by property (a) and the definition of $P$. Thus, if $r$ satisfied the three conditions in Lemma 9 for $n^{(j)}$, then $r^{\prime}=(r-1)\cdot Q+1$ satisfies the three conditions in Lemma 9 for $n^{(j+1)}$. The increase comes from the fact that now $r^{\prime}=Q-1$ satisfies the three conditions as well. ∎

Proof of Theorem 2.

Combine Propositions 1, 2, 3, and 4. ∎

5.  Iterated remainder sets and their relation to the “n mod a problem”

Recall that we defined the sets of iterated remainders of $n$ inductively by

[TABLE]

These sets are related to an older open problem about the length of Pierce expansion. This problem was first studied by Shallit [9], and can be phrased in the following way.

Fix a positive integer $n$ and choose another integer $1\leq a\leq n$. Set $a_{0}:=a$ and $a_{j+1}:=n\bmod a_{j}$ for $j\geq 0$ and as long as $a_{j}>0$. For example, for $(n,a)=(35,22)$, we get $a_{0}=22,a_{1}=13,a_{2}=9,a_{3}=8,a_{4}=3,a_{5}=2,a_{6}=1,a_{7}=0$. Now let us define $P(n,a)$ to be the integer $t$ such that $a_{t}=0$. So, for example, $P(35,22)=7$. Finally, let us set

[TABLE]

The problem is to obtain upper and lower bounds for $P(n)$ in terms of $n$. In particular, experiments suggest that the upper bound should be sublinear; but this seems to be hard to prove. The best known bounds are due to Chase and Pandey [1], who slightly improved the bounds by Erdős and Shallit [2]: We have

[TABLE]

for sufficiently large $n$.

Now this problem directly relates to our sets $S_{j}(n)$ via the next simple lemmas.

Lemma 10**.**

Let $j\geq 1$ and $n\geq 1$. Then $r\in S_{j}(n)$ if and only if there exists an integer $\lfloor n/2\rfloor+1\leq a\leq n$ such that in the above notation $a_{j+1}=r$.

Proof.

This follows directly from the definitions. The index shift comes from the fact that we are assuming $\lfloor n/2\rfloor+1\leq a\leq n$ and so $a_{1}$ can be exactly every element from $S_{0}(n)$. ∎

Lemma 11**.**

The following statements are equivalent:

1. (1)

$P(n)=t$; 2. (2)

$S_{t+1}(n)=\{0\}$; 3. (3)

$\lvert S_{t+1}(n)\rvert=1$ and $\lvert S_{t+1+j}(n)\rvert=0$ for all $j\geq 1$.

Proof.

When computing $P(n)=\max_{1\leq a\leq n}P(n,a)$, we may restrict ourselves to $\lfloor n/2\rfloor+1\leq a\leq n$, since if $a<\lfloor{n/2}+1\rfloor$, the starting value $n-a$ gives $P(n,n-a)=P(n,a)+1$.

Now the equivalences follow from Lemma 10 and the fact that $S_{j}(n)=\emptyset$ if and only if $S_{j-1}=\emptyset$ or $S_{j-1}=\{0\}$. ∎

Remark 2**.**

The known bounds (13) imply that for sufficiently large $n$, we have $\lvert S_{j}(n)\rvert=0$ for all $j\geq n^{1/3}$. On the other hand, there exists a constant $c$ such that for all sufficiently large $n$ we have $\lvert S_{j}(n)\rvert\geq 1$ for $j\leq c\log n/\log\log n$.

6.  Bounds for iterated remainders

Recall that we defined

[TABLE]

In this section we prove upper and lower bounds for $s_{j}(n)$.

We start with a simple upper bound.

Lemma 12**.**

For all $j\geq 0$ and $n\geq 1$ we have

[TABLE]

Proof.

The first inequality is clear since $\min S_{j}(n)=0$. We show the second inequality by induction. For $j=0$ this is clearly satisfied by definition. Now assume that $\max S_{j}(n)\leq n/(j+2)$ for some $j\geq 0$. Then for every $r\in S_{j+1}(n)$ there exists a $k\in S_{j}(n)$ with $k>r$ such that $n=qk+r$. By induction hypothesis we have $k\leq n/(j+2)$; hence $q\geq j+2$. Now

[TABLE]

implies

[TABLE]

as desired. ∎

Remark 3**.**

The arguments from [2] for the upper bound on $P(n)$ in fact give stronger upper bounds when $j$ is relatively large compared to $n$. For example, since $\max S_{j}(n)$ strictly decreases as $j$ increases, one can show that $\max S_{j}(n)$ is roughly bounded by $2\sqrt{n}-j$ for $j>\sqrt{n}$. One can do even better (see [2, proof of Theorem 2]), using the fact that the number of divisors of $m$ is $O(m^{\varepsilon})$. It turns out that for $j>n^{1/3+\varepsilon}$ we get roughly the upper bound $2n^{2/3+\varepsilon}-j\cdot n^{1/3}$.

Finally, we prove a lower bound. In particular, we show that the sequence $s_{j}(n)$ grows linearly (even if $\lim_{n\to\infty}s_{j}(n)/n$ might not exist).

Lemma 13**.**

For every $j\geq 0$ and $n\geq N(j)$ there exists an integer $x_{j}=x_{j}(n)$ such that

[TABLE]

In particular,

[TABLE]

Proof.

Fix some $J\geq 0$ and let $n\geq N(J)$. We want to prove (14) for $j=0,1,\ldots,J$ with finite induction. For $j=0$ the inclusion (14) is clearly true with $x_{0}=0$, since $S_{0}(n)=\{0,1,\ldots,\lfloor n/2\rfloor\}$. Now assume that (14) holds for some $0\leq j\leq J-1$. Our goal is to show

[TABLE]

i.e., we set $x_{j+1}=n-(j+2)x_{j}$. Assume $r$ is in the set on the right hand side. Then the condition on the residue class implies that we can write $r=n-(j+2)x_{j}-q(j+2)!$ with some integer $q$. This implies

[TABLE]

where we set $k:=x_{j}+q(j+1)!$. Thus, in order to prove $r\in S_{j+1}(n)$ it suffices to show that $k\in S_{j}(n)$ and $k\geq r+1$.

Since $n-r=(j+2)k$, the condition $k\geq r+1$ is equivalent to $n-r\geq(j+2)(r+1)$. This is equivalent to $r\leq(n-j-2)/(j+3)$, which is satisfied by assumption.

Clearly, $k\equiv x_{j}\bmod{(j+1)!}$, so in order to show $k\in S_{j}(n)$ we only need to check $j\leq k\leq(n-j-1)/(j+2)$. First, $k\leq(n-j-1)/(j+2)$ is equivalent to $n-r\leq(j+2)(n-j-1)/(j+2)=n-j-1$ so this indeed holds for $r\geq j+1$. Finally, $k\geq j$ is equivalent to $n-r\geq(j+2)j$, which holds since $r\leq(n-j-2)/(j+3)\leq n-(j+2)j$ for sufficiently large $n$.

The bound (15) follows immediately from (14), since for large $n$ the interval length is $(n-j-1)/(j+2)-j+1\sim n/(j+2)$ and every $(j+1)!$-th number is included. ∎

Proof of Theorem 3.

Combine Lemmas 12 and 13. ∎

7.  Numerical experiments and open problems

Recall that in Theorem 1 we proved

[TABLE]

The bound for the error term seems very large. We have computed the values for $s(n)$ for $n\leq 10^{7}$ and determined the points $(n,s(n)-c\cdot n)$ where $s(n)-c\cdot n$ reaches a new maximum or minimum. These points are plotted in Figure 1, together with the graph of $2n^{1/3}$. This suggests that perhaps the correct bound for the error term might be $O(n^{1/3})$. In any case, we propose the following problem.

Problem 1**.**

Improve the bound for the error term for $s(n)$ in Theorem 1.

For the iterated remainder sets, recall that Theorem 3 says that

[TABLE]

In particular, there is a gap between our lower and our upper bound. Indeed, numerical experiments strongly suggest that $\lim_{n\to\infty}s_{j}(n)/n$ does not exist for $j\geq 2$. The three plots in Figure 2 show the values of $s_{1}(n),s_{2}(n),s_{3}(n)$, respectively, for $n\leq 10^{4}$. For $s_{2}(n)$ and $s_{3}(n)$ we see some “bands” of values. The blue points correspond to $n\equiv 0\bmod 6$, the green points correspond to $n\equiv 2,4\bmod 6$, the yellow points to $n\equiv 3\bmod 6$, and the red points to $n\equiv 1,5\bmod 6$. It seems that really the precise divisibility properties of $n$ determine the value of $s_{2}(n),s_{3}(n)$. To support this further, in Figure 3 we have plotted the values of $s_{2}(n)$ only for $n\equiv 1\bmod 6$ in the range $[6\cdot 10^{4},6\cdot 10^{4}+10^{3}]$. In particular, we only consider numbers $n$ not divisible by $2$ and $3$. Indeed, the numbers $n$ that are divisible by $5$ (points coloured red) yield the smallest relative values. In any case, the easiest problem in this context might be Problem 2.

Problem 2**.**

Prove that for $j\geq 2$ the limit $\lim_{n\to\infty}s_{j}(n)/n$ does not exist.

Acknowledgements

We thank Jeffrey Shallit for his encouragement, for sharing his numbers up to $s(10^{7})$ with us, and for suggesting to consider iterated sets.