On the discriminator of Lucas sequences

Abstract

We consider the family of Lucas sequences uniquely determined by $U_{n+2}(k)=(4k+2)U_{n+1}(k) -U_n(k),$ with initial values $U_0(k)=0$ and $U_1(k)=1$ and $k\ge 1$ an arbitrary integer. For any integer $n\ge 1$ the discriminator function $\mathcal {D}_k(n)$ of $U_n(k)$ is defined as the smallest integer m such that $U_0(k),U_1(k),\ldots ,U_{n-1}(k)$ are pairwise incongruent modulo m. Numerical work of Shallit on $\mathcal {D}_k(n)$ suggests that it has a relatively simple characterization. In this paper we will prove that this is indeed the case by showing that for every $k\ge 1$ there is a constant $n_k$ such that ${\mathcal D}_{k}(n)$ has a simple characterization for every $n\ge n_k$. The case $k=1$ turns out to be fundamentally different from the case $k>1$.

FRENCH ABSTRACT

Pour un entier arbitraire $k\ge 1$, on considère la famille de suites de Lucas déterminée de manière unique par la relation de récurrence $U_{n+2}(k)=(4k+2)U_{n+1}(k) -U_n(k),$ et les valeurs initiales $U_0(k)=0$ et $U_1(k)=1$. Pour tout entier $n\ge 1$, la fonction discriminante $\mathcal {D}_k(n)$ de $U_n(k)$ est définie comme le plus petit m tel que $U_0(k),U_1(k),\ldots ,U_{n-1}(k)$ soient deux à deux non congruents modulo m. Des travaux numériques de Shallit sur $\mathcal {D}_k(n)$ suggère qu’il en existe une caractérisation relativement simple. Dans cet article, on démontre que c’est en effet le cas en établissant que pour tout $k\ge 1$, il existe une constante $n_k$ telle que ${\mathcal D}_{k}(n)$ possède une caractérisation simple pour tout $n\ge n_k$. Le cas $k=1$ se révèle être de nature totalement différente du cas $k>1$.

1 Introduction

The discriminator of a sequence $\mathbf{a}=\{a_n\}_{n\ge 1}$ of distinct integers is the sequence given by

$$\begin{aligned} {\mathcal D}_\mathbf{a}(n)=\min \{m: a_0,\ldots ,a_{n-1}~{\text { are pairwise distinct modulo}}~m\}. \end{aligned}$$

In other words, ${\mathcal D}_\mathbf{a}(n)$ is the smallest integer m that allows one to discriminate (tell apart) the integers $a_0,\ldots , a_{n-1}$ on reducing modulo m.

Note that since $a_0,\ldots ,a_{n-1}$ are n distinct residue classes modulo ${\mathcal D}_\mathbf{a}(n)$ it follows that ${\mathcal D}_\mathbf{a}(n)\ge n$. On the other hand obviously

$$\begin{aligned} {\mathcal D}_\mathbf{a}(n)\le \max \{a_0,\ldots ,a_{n-1}\} -\min \{a_0,\ldots ,a_{n-1}\}. \end{aligned}$$

Put

$$\begin{aligned} {\mathcal D}_\mathbf{a}=\{{\mathcal D}_\mathbf{a}(n): n\ge 1\}. \end{aligned}$$

The main problem is to give an easy description or characterization of the discriminator (in many cases such a characterization does not seem to exist). The discriminator was named and introduced by Arnold, Benkoski and McCabe in [1]. They considered the sequence $\mathbf u$ with terms $u_j = j^2$. Meanwhile the case where $u_j=f(j)$ with f a polynomial has been well-studied, see, for example, [3, 9, 10, 16]. The most general result in this direction is due to Zieve [16], who improved on an earlier result by Moree [9].

In this paper we study the discriminator problem for Lucas sequences (for a basic account of Lucas sequences see, for example, Ribenboim [13, 2.IV]). Our main results are Theorem 1 ($k=1$) and Theorem 3 ($k>2$). Taken together with Theorem 2 ($k=2$) they evaluate the discriminator ${\mathcal D}_k(n)$ for the infinite family of second-order recurrences (1), with for each k at most finitely many n that are not covered.

All members in the family (1) have a characteristic equation that is irreducible over the rationals. Very recently, Ciolan and Moree [6] determined the discriminator for another infinite family, this time with all members having a reducible characteristic equation. For every prime $q\ge 7$ they computed the discriminator of the sequence

$$\begin{aligned} u_q(j)=\frac{3^j-q(-1)^{j+(q-1)/2}}{4},\quad j=1,2,3,\ldots \end{aligned}$$

that was first considered in this context by Jerzy Browkin. The case $q=5$ was earlier dealt with by Moree and Zumalacárregui [11], who showed that, for this value of q,  the smallest positive integer m discriminating $u_q(1),\ldots ,u_q(n)$ modulo m equals $\min \{2^e,5^f\},$ where e is the smallest integer such that $2^e\ge n$ and f is the smallest integer such that $5^f\ge 5n/4$.

Despite structural similarities between the present paper and [6] (for example the index of appearance z in the present paper plays the same role as the period $\rho$ in [6]), there are also many differences. For example, Ciolan and Moree have to work much harder to exclude small prime numbers as discriminator values. This is related to the sequence of good discriminator candidate values in that case being much sparser, namely being $O(\log x)$ for the values $\le x$, versus $\gg \log ^2 x$. In our case one has to work with elements and ideals in quadratic number fields, whereas in [6] in the proof of the main result the realm of the rationals is never left.

Let $k\ge 1$. For $n\ge 0$ consider the sequence $\{U_n(k)\}_{n\ge 0}$ uniquely determined by

$$\begin{aligned} U_{n+2}(k) =(4k+2)U_{n+1}(k)-U_n(k),\quad U_0(k)=0,\quad U_1(k)=1. \end{aligned}$$

(1)

For $k=1$, the sequence $\{U_n(1)\}_{n\ge 0}$ is

$$\begin{aligned} 0, 1, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214,\ldots . \end{aligned}$$

This is A001109 in OEIS. On noting that

$$\begin{aligned} U_{n+2}(k)-U_{n+1}(k)=4kU_{n+1}(k) +U_{n+1}(k)-U_n(k)\ge 1, \end{aligned}$$

one sees that the sequence $U_n(k)$ consists of strictly increasing non-negative numbers. Therefore we can consider ${\mathcal D}_{U(k)}$, which for notational convenience we denote by ${\mathcal D}_{k}$.

In May 2016, Jeffrey Shallit, who was the first to consider ${\mathcal D}_k$, wrote to the third author that numerical evidence suggests that ${\mathcal D}_{1}(n)$ is the smallest number of the form $250\cdot 2^i$ or $2^i$ greater than or equal to n, but that he was reluctant to conjecture such an unpredictable thing. More extensive numerical experiments show that if we compute ${\mathcal D}_1(n)$ for all $n\le 2^{10}$, then they are powers of 2 except for $n\in [129,150]$, and other similar instances such as

$$\begin{aligned} n\in [2^a+1,2^{a-6}\cdot 75],\text { for which }{\mathcal D}_1(n)=2^{a-6}\cdot 125\text { and }a\in \{7,8,9\}. \end{aligned}$$

Thus the situation is more subtle than Shallit expected and this is confirmed by Theorem 1.

The fractional part of a real number x is denoted by $\{x\}$ and its floor by $\lfloor x\rfloor$.

Theorem 1

Let $v_n$ be the smallest power of two such that $v_n\ge n$. Let $w_n$ be the smallest integer of the form $2^a5^b$ satisfying $2^a5^b\ge 5n/3$ with $a,b\ge 1$. Then

$$\begin{aligned} {\mathcal D}_{1}(n)=\min \{v_n,w_n\}. \end{aligned}$$

Let

$$\begin{aligned} {\mathcal M}=\left\{ m\ge 1: \left\{ m \frac{\log 5}{\log 2}\right\} \ge 1-\frac{\log (6/5)}{\log 2}\right\} =\{3,6,9,12,15,18,21,\ldots \}. \end{aligned}$$

We have

$$\begin{aligned} \{{\mathcal D}_1(2),{\mathcal D}_1(3),{\mathcal D}_1(4), \ldots \}=\{2^a5^b:a\ge 1,~b\in {\mathcal M}\cup \{0\}\}. \end{aligned}$$

A straightforward application of Weyl’s criterion (cf. the proof of [11, Proposition 1] or [6, Proposition 1]) gives

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{\#\{m\in {\mathcal M}:m\le x\}}{x}=\frac{\log (6/5)}{\log 2} =0.263034\ldots . \end{aligned}$$

In contrast to the case $k=1$, the case $k=2$ turns out to be especially easy.

Theorem 2

Let $e\ge 0$ and $f\ge 1$ be the smallest integers such that $2^e\ge n$, respectively $3\cdot 2^f\ge n$. Then ${\mathcal D}_2(n)=\min \{2^e,3\cdot 2^f\}$.

Our second main result shows that the behavior of the discriminator ${\mathcal D}_k$ with $k>2$ is very different from that of ${\mathcal D}_1$.

Theorem 3

Put

and

Let $k>2$. We have

$$\begin{aligned} {\mathcal D}_{k}(n)\le \min \{m\ge n: m\in {\mathcal A}_{k}\cup {\mathcal B}_{k}\}, \end{aligned}$$

(2)

with equality if the interval [n, 3n / 2) contains an integer $m\in {\mathcal A}_{k}\cup {\mathcal B}_{k}$. There are at most finitely many n for which in (2) strict inequality holds. Furthermore, we have

$$\begin{aligned} {\mathcal D}_{k}(n)=n\iff n\in {\mathcal A}_{k}\cup {\mathcal B}_{k}. \end{aligned}$$

(3)

The condition on the interval [n, 3n / 2) is sufficient, but not always necessary. The proof also works for $k=2$ in which case the interval becomes [n, 5n / 3). However, we prefer to give a short proof from scratch of Theorem 2 (in Sect. 6.1).

Theorems 2 and 3 taken together have the following corollary.

Corollary 1

For $k>1$ there is a finite set ${\mathcal F}_k$ such that

$$\begin{aligned} {\mathcal D}_{k}= {\mathcal A}_k\cup {\mathcal B}_k\cup {\mathcal F}_k. \end{aligned}$$

(4)

Note that ${\mathcal A}_1=\{1\},~ {\mathcal B}_1=\{2^e:e\ge 1\}$ and that by Theorem 1 identity (4) holds true with ${\mathcal F}_1=\{2^a\cdot 5^m: a\ge 1 \text { and } m\in {\mathcal M}\}.$ In particular, ${\mathcal F}_1$ is not finite. In contrast to this, Theorem 2 says that ${\mathcal F}_2$ is empty and Theorem 3 says that ${\mathcal F}_k$ is finite for $k>1$. In part II [5] the problem of explicitly computing ${\mathcal F}_k$ is considered.

Despite the progress made in this paper, for most second order recurrences (and the Fibonacci numbers belong to this class), the discriminator remains quite mysterious, even conjecturally. Thus in this paper we only reveal the tip of an iceberg.

2 Preliminaries

We start with some considerations about U(k) valid for any $k\ge 1$. The characteristic equation of this recurrence is

$$\begin{aligned} x^2-(4k+2)x+1=0. \end{aligned}$$

Its roots are $(\alpha (k),\alpha (k)^{-1})$, where

$$\begin{aligned} \alpha (k)=2k+1+2{\sqrt{k(k+1)}}. \end{aligned}$$

Its discriminant is

$$\begin{aligned} \Delta (k)=\Big (\alpha (k)-\frac{1}{\alpha (k)}\Big )^2=16k(k+1). \end{aligned}$$

We have $\alpha (k)=\beta (k)^2$, where $\beta (k)={\sqrt{k+1}}+{\sqrt{k}}$. Thus,

$$\begin{aligned} U_n(k)=\frac{\alpha (k)^{n}-\alpha (k)^{-n}}{\alpha (k)-\alpha (k)^{-1}}=\frac{\beta (k)^{2n}-\beta (k)^{-2n}}{\beta (k)^2-\beta (k)^{-2}} \end{aligned}$$

is both the Lucas sequence having roots $(\alpha (k),\alpha (k)^{-1})$, as well as the sequence of even indexed members of the Lehmer sequence having roots $(\beta (k),\beta (k)^{-1})$ (cf. Bilu et al. [2] or Ribenboim [13, pp. 69–74]).

First we study the congruence in case m is an arbitrary integer. By the Chinese Remainder Theorem, it suffices to study this congruence only in the case where m is a prime power. In this section we will only deal with the easiest case where m is a power of two.

Lemma 1

If , then .

Proof

This is clear for $a=0$. When $a=1$, we have $U_0(k)=0,~U_1(k)=1$ and . Thus, . This shows that for all $n\ge 0$. Therefore implies that , which is what we wanted. We now proceed by induction on a. Assume that $a>1$ and that the lemma has been proved for $a-1$. Let $i\le j$ be such that . In particular, and so . It is easy to check that putting $V_n(k)$ for the sequence given by $V_0(k)=2,~V_1(k)=4k+2$, we have

$$\begin{aligned} U_j(k)-U_i(k)=U_{(j-i)/2}(k)V_{(j+i)/2}(k). \end{aligned}$$

The sequence $\{V_n(k)\}_{n\ge 0}$ satisfies the same recurrence as $\{U_n(k)\}_{n\ge 0}$, namely

$$\begin{aligned} V_{n+2}(k)=(4k+2)V_{n+1}(k)-V_n(k). \end{aligned}$$

Note that $V_n(k)=\alpha (k)^n+\alpha (k)^{-n}$. Further, by induction on n using the fact that $2\Vert V_0(k)$ and $2\Vert V_1(k)$ and the recurrence for V(k), we conclude that if $2\Vert V_n(k)$ and $2\Vert V_{n+1}(k)$, then

so $2\Vert V_{n+2}(k)$. Hence, since $2^{a}\mid U_i(k)-U_j(k)=U_{(i-j)/2}(k)V_{(i+j)/2}(k)$, and $2\Vert V_{(i+j)/2}(k)$, we get that $2^{a-1}\mid U_{(i-j)/2}(k)$. Thus, and by the induction hypothesis we get that . Thus, and the induction is complete. $\square$

Corollary 2

We have ${\mathcal D}_{k}(n)\le \min \{2^e:2^e\ge n\}$.

3 Index of appearance

We now need to study the congruence for odd primes p and integers $b\ge 1$. We start with the easy case when $j=0$. Given m, the smallest $n\ge 1$ such that exists, cf. [2], and is called the index of appearance of m in U(k) and is denoted by z(m). (For notational convenience we suppress the dependence of z(m) on k.) The following result is well-known, cf. Bilu and Hanrot [2]. We write $\nu _p(m)$ for the exponent of the prime p in the factorization of the positive integer m. For an odd prime p we write $(\frac{\bullet }{p})$ for the Legendre symbol with respect to p.

Lemma 2

The index of appearance z of the sequence U(k) has the following properties.

• (i) If $p\mid \Delta (k)$, then $z(p)=p$.

• (ii) If $p\not \mid \Delta (k)$, then $z(p)\mid p-e$, where $e=(\frac{\Delta (k)}{p})$.

• (iii) Let $c=\nu _p(U_{z(p)}(k))$. Then $z(p^{b})=p^{\max \{b-c,0\}} z(p).$

• (iv) If $p|U_m(k)$, then z(p)|m.

• (v) If $n=m_1\ldots m_s$ with $m_1,\ldots ,m_s$ pairwise coprime, then

  $$\begin{aligned} z(m_1\dots m_s)={\text { lcm}}[z(m_1),\ldots ,z(m_s)]. \end{aligned}$$

Part i says that $z(p^b)=p^b$ in case $p\mid \Delta (k)$ and $b\ge 1$. The next result describes what happens for arbitrary b and $p>2$.

Lemma 3

Assume that $p>2$ is such that $p\mid \Delta (k)$. Let $z(p^b)$ be the index of appearance of $p^b$ in the sequence U(k).

• (i) If $p>3$, then $\nu _p(U_p)=1$. In particular, $z(p^b)=p^b$ holds for all $b\ge 1$.

• (ii) If $p=3$, then

  $$\begin{aligned} U_3=16k(k+1)+3. \end{aligned}$$

  In particular, $\nu _3(U_3)=c>1$ exactly when . In these cases, $z(p^b)=p^{\max \{b-c,0\}}$. Hence, $z(p^b)\mid p^{b-1}$ for all $b\ge 2$.

Proof

Recall that $\Delta (k)=16k(k+1)$. Part i is known. As for ii, we compute

$$\begin{aligned} U_3=\frac{\alpha ^3-\alpha ^{-3}}{\alpha -\alpha ^{-1}}=\alpha ^2+1+\alpha ^{-2}=16k(k+1)+3. \end{aligned}$$

Since by assumption $3\mid 16k(k+1)$, it follows that either $3\mid k$ or $3\mid (k+1)$. In the first case, $k=3k_0$ and

$$\begin{aligned} U_3=3(16k_0(3k_0+1)+1). \end{aligned}$$

The number within the parentheses is congruent to , which is a multiple of 3 exactly when ; hence, . In the second case, $k+1=3k_1$, so

$$\begin{aligned} U_3=3(16k_1(3k_1-1)+1) \end{aligned}$$

and the number in parenthesis is congruent to which is a multiple of 3 exactly when , so . $\square$

3.1 Index of appearance in case $k=1$

For notational convenience we ignore where appropriate the index $k=1$ in $U(k),~\alpha (k),~\beta (k)$ and so we only write $U,~\alpha ,~\beta$. We have $\Delta (1)=8$ and the relevant quadratic field is ${\mathbb K}={\mathbb Q}[{\sqrt{2}}]$, which has ${\mathbb Z}[{\sqrt{2}}]$ as its ring of integers. If $\gamma ,\delta \in \mathbb Z[\sqrt{2}]$, then we write if and only if $(\gamma -\delta )/p\in \mathbb Z[\sqrt{2}]$. If $\rho =a+b\sqrt{2}\in {\mathbb K}$ with a and b rational numbers, then the norm $N_{\mathbb K}(\rho )=\rho \cdot {\overline{\rho }}=a^2-2b^2,$ where ${\overline{\rho }}$ is the conjugate of $\rho$ obtained by sending $\sqrt{2}$ to $-\sqrt{2}$.

For odd p, z(p) is a divisor of either $p-1$ or $p+1$ by Lemma 2 ii. The next lemma shows that even more is true. The result is an easy consequence of the fundamental work [4] by Carmichael. For the benefit of the reader we will also give a self-contained proof.

Lemma 4

Let $k=1$ and p be an odd prime. Then

$$\begin{aligned} z(p^b)\mid p^{b-1}\left( p-\left( \frac{2}{p}\right) \right) /2. \end{aligned}$$

Proof

Put

$$\begin{aligned} W_n=\frac{(1+\sqrt{2})^n-(1-\sqrt{2})^n}{4\sqrt{2}}. \end{aligned}$$

By [4, Theorem XIII], cf. Williams [15, p. 85], it follows that $p^b$ divides $W_{p^{b-1}(p-\frac{2}{p})}$. The result now follows on noting that $U_n(1)=W_{2n}/2$ and that $p^{b-1}(p-(\frac{2}{p}))$ is even. $\square$

Proof

(More self-contained)

(i) The case $e=(\frac{2}{p})=1$.

Then . We have

Here we used Euler’s theorem that . Since $\beta$ is a unit, we infer from . Thus,

The same congruence holds for $\alpha$ replaced by $\alpha ^{-1}$. Hence, subtracting the two congruences we get that . Thus, p divides the difference $\alpha ^{(p-1)/2}-\alpha ^{-(p-1)/2}$. On noting that

$$\begin{aligned} N_{\mathbb K}(\alpha ^{(p-1)/2}-\alpha ^{-(p-1)/2})=32U_{(p-1)/2}^2, \end{aligned}$$

we infer that $p\mid U_{(p-1)/2}$. Thus, $z(p)\mid (p-1)/2$, therefore $z(p^b)$ divides $p^{b-1}(p-1)/2$ by Lemma 2 iii.

(ii) The case $e=(\frac{2}{p})=-1$.

Then . Now we have

Thus, . In particular,

(5)

The same congruence holds for $\alpha$ replaced by $\alpha ^{-1}$. Subtracting the two congruences, we get that . Noting that

$$\begin{aligned} N_{\mathbb K}(\alpha ^{(p+1)/2}-\alpha ^{-(p+1)/2})=32U_{(p+1)/2}^2, \end{aligned}$$

we obtain that $p\mid U_{(p+1)/2}$. We have, in particular, $z(p)\mid (p+1)/2$ and hence $z(p^b)\mid p^{b-1}(p+1)/2$ by Lemma 2 iii. $\square$

Let us recall the following well-known result.

Lemma 5

Let p be odd such that $e=(\frac{2}{p})=-1$ and let $b\ge 1$ be an integer. Then $z(p^b)$ is the minimal $m\ge 1$ such that .

Proof

Assume that $m\ge 1$ is such that . Then . Subtracting both congruences we get that $p^b$ divides $\alpha ^m-\alpha ^{-m}$. Computing norms we see that $p^{2b}\mid N_{\mathbb K}(\alpha ^m-\alpha ^{-m})$, and so $p^{2b}\mid 32U_m^2,$ and therefore $p^b|U_m$, showing that $z(p^b)|m$. Next assume that $p^b\mid U_m$ for some $m\ge 1$. Then , so . Thus, $p^b\mid (\alpha ^m-1)(\alpha ^m+1)$. The assumption on e implies that p is inert in ${\mathbb Z}[{\sqrt{2}}]$. Since, moreover, p cannot divide both $\alpha ^m-1$ and $\alpha ^m+1$, it follows that $p^b$ must divide either $\alpha ^m-1$ or $\alpha ^m+1$. $\square$

4 Structure of the discriminator ${\mathcal D}_1$

Now we are ready to restrict the number of values the discriminator can assume.

Lemma 6

Let $m={\mathcal D}_1(n)$ for some $n>1$. Then

• (i) m has at most one odd prime divisor.

• (ii) If m is divisible by exactly one odd prime p, then $e=(\frac{2}{p})=-1$ and $z(p)=(p+1)/2$.

• (iii) If m is not a power of 2, then m can be written as $2^ap^{b}$ with $a,b\ge 1$ and .

Proof

Assume that ${\mathcal D}_1(n)=m$ and write it as

$$\begin{aligned} m=2^a p_1^{b_1}\dots p_r^{b_r}, \end{aligned}$$

where the $p_i$ are distinct odd primes. Assume first that $r\ge 2$. Then $n\le z(m)$ (otherwise if $z(m)<n$, it follows that , a contradiction). On recalling (Lemma 2 v) that $z(m)={\text { lcm}}[z(2^a),z(p_1^{b_1}),\ldots ,z(p_r^{b_r})],$ we obtain the inequality

$$\begin{aligned} z(m)\le 2^ap_1^{b_1-1}\dots p_r^{b_k-1} \left( \frac{p_1+1}{2}\right) \cdots \left( \frac{p_r+1}{2}\right) <\frac{m}{2}, \end{aligned}$$

(6)

where the last inequality needs proof. Indeed, it is equivalent with the inequality

$$\begin{aligned} \prod _{i=1}^r \left( \frac{p_i+1}{2}\right) <p_1\dots p_r. \end{aligned}$$

It suffices to justify that

$$\begin{aligned} \left( \frac{p_1+1}{2}\right) \left( \frac{p_2+1}{2}\right)<\frac{p_1p_2}{2}\quad {\text { and}}\quad \frac{p_i+1}{2}<p_i\quad {\text { for}}\quad i=3,\ldots ,r. \end{aligned}$$

The second inequality is clear. The first is equivalent to $p_1p_2>p_1+p_2+1$. Assuming $3\le p_1<p_2$, this inequality is implied by $p_2(p_1-2)>1$, which is obviously true.

Since $z(m)<m/2$ by (6), it follows that the interval [z(m), 2z(m)) contains a power of 2, say $2^b< 2z(m)<m$. But then since $2^b\ge z(m)\ge n$, it follows that $U_0,\ldots ,U_{n-1}$ are already distinct modulo $2^b$ and $2^b<m$, which contradicts the definition of the discriminator. Thus, the only possibility is that $r\in \{0,1\}$. If $r=1$ and $e_1=(\frac{2}{p_1})=1$, then

$$\begin{aligned} z(m)=z(2^ap_1^{b_1})\le 2^ap_1^{b_1-1}(p_1-1)/2<m/2, \end{aligned}$$

and so the same contradiction applies. Assume now that $e_1=-1$ and that $z(p_1)$ is a proper divisor of $(p+1)/2$. Then

$$\begin{aligned} z(m)\le 2^a p_1^{b_1-1}z(p_1)\le 2^ap_1^{b_1-1} (p_1+1)/4<m/2, \end{aligned}$$

and again the same contradiction applies.

It remains to prove part iii. We write $m=2^ap_1^{b_1}$. We know that $a\ge 1$ and $e=-1$. Thus, . If , then

$$\begin{aligned} z(m)={\text { lcm}}[z(2^a), z(p^{b})]\mid 2^{a} p^{b-1}(p+1)/4. \end{aligned}$$

In particular, $z(m)<m/2$, and we get again a contradiction. Thus, . $\square$

Lemma 7

If $n>1$, then ${\mathcal D}_{1}(n)$ is even.

Proof

Assume that ${\mathcal D}_1(n)=m$ is odd. By the previous lemma, it follows that $m=p_1^{b_1}$, where $({2\over p_1})=-1$ and $z(p_1)=(p_1+1)/2$. Further, in this situation (5) applies and we have

$$\begin{aligned} \alpha ^{(p_1+1)/2}=-1+p_1\gamma \end{aligned}$$

for some algebraic integer $\gamma \in {\mathbb Z}[{\sqrt{2}}]$. By induction on $m\ge 0$ one establishes that

Let

$$\begin{aligned} i=\left\lfloor \frac{p_1^{b_1-1}(p_1+1)}{4}\right\rfloor -1\quad {\text { and}}\quad j=\frac{p_1^{b_1-1}(p_1+1)}{2}-\left( \left\lfloor \frac{p_1^{b_1-1}(p_1+1)}{4}\right\rfloor -1\right) . \end{aligned}$$

Since $b_1\ge 1$ and $p_1\ge 3$, we have that $i\ge 0$. Further,

$$\begin{aligned} j\ge \frac{p_1^{b_1-1}(p_1+1)}{2}-\frac{p_1^{b_1-1}(p_1+1)}{4}+1=\frac{p_1^{b_1-1}(p_1+1)}{4}+1\ge i+2, \end{aligned}$$

and

$$\begin{aligned} j \le \frac{p_1^{b_1-1}(p_1+1)}{2}-\left( \frac{p_1^{b_1-1}(p_1+1)}{4}-\frac{3}{4}\right) +1= \frac{p_1^{b_1-1}(p_1+1)}{4}+\frac{7}{4}. \end{aligned}$$

(7)

Since $i+j=p_1^{b_1-1}(p_1+1)/2$, we have Thus,

and also

Taking the difference of the latter two congruences we get that

Thus, taking norms and using the fact that $p_1$ is inert in ${\mathbb K}$ and so has norm $p_1^2$, we get $p_1^{2b_1}\mid N_{\mathbb K}((\alpha ^j-\alpha ^{-j})-(\alpha ^i-\alpha ^{-i}))$, that is

$$\begin{aligned} p_1^{2b_1}\mid 32(U_{j}-U_i)^2, \end{aligned}$$

giving

Since $i<j$ and by assumption $U_0,\ldots ,U_{n-1}$ are pairwise distinct modulo $p_1^{b_1}$, it follows that $j\ge n$ and hence, by (7),

$$\begin{aligned} n\le \frac{p_1^{b_1-1}(p_1+1)}{4}+\frac{7}{4}. \end{aligned}$$

We check when the right hand side is less than m / 2. This gives

$$\begin{aligned} \frac{p_1^{b_1-1}(p_1+1)}{4}+\frac{7}{4}<\frac{p_1^{b_1}}{2}, \end{aligned}$$

or $2p_1^{b_1}>7+p_1^{b_1}+p_1^{b_1-1}$, which is equivalent to $p_1^{b_1-1}(p_1-1)>7$. This holds whenever $p_1^{b_1}\ge 11$. Thus, only the cases $m=p_1^{b_1}\le 9$ need to be checked, so $n<9$. We check that in this range there is no odd discriminant. Thus, indeed $n<m/2$, and by the previous argument we can now replace m by a power of two in the interval [m / 2, m), and get a contradiction. $\square$

Lemma 8

Assume that $m=2^ap_1^{b_1}={\mathcal D}_1(n)$ for some $n\ge 1$ and that $b_1\ge 1$. If $b_1>1$, then $p_1\Vert U_{z(p_1)}$.

Proof

This is trivial. Indeed, if $b_1>1$ and $p_1^2\mid U_{z(p_1)}$, then $z(p_1^{b_1})\mid p_1^{b_1-2}(p_1+1)/2$ by Lemma 2 iii. Thus, in this case

$$\begin{aligned} z(m)\mid {\text { lcm}}[2^a, p_1^{b_1-2}(p_1+1)/2]\mid 2^{a-1}p_1^{b_1-2} (p_1+1), \end{aligned}$$

Since $2^{a-1}p_1^{b_1-2} (p_1+1)=m (p_1+1)/(2p_1^2)<m/2$, we have obtained a contradiction. $\square$

Lemma 9

Assume that $m=2^ap_1^{b_1}$ is such that $a\ge 1$, and $z(p_1)=(p_1+1)/2$. Then holds if and only if .

Proof

Since , it follows that . It remains to understand what happens modulo $p_1^{b_1}$. Since $e_1=-1$, $p_1$ is a prime in ${\mathbb Z}[{\sqrt{2}}]$. Let $\lambda$ denote the common value of $U_i$ and $U_j$ modulo $p_1^{b_1}$. Then $\alpha ^i$ and $\alpha ^j$ are both roots of

in ${\mathbb Z}[{\sqrt{2}}]/(p_1^{b_1}{\mathbb Z}[{\sqrt{2}}])$. Taking the difference and factoring we get that

(8)

Now various things can happen. Namely, $p_1^{b_1}$ can divide the first factor or the second factor of (8). If $b_1>1$, some power of $p_1$ may divide the first factor and some power of $p_1$ can divide the second factor. We investigate each of these options.

(i) $p_1^{b_1}\mid (\alpha ^i-\alpha ^j)$.

Then . Since i and j are of the same parity, it follows that . By Lemma 5 we then infer that $z(p_1^{b_1})|(i-j)/2$. By Lemma 4 we have $z(p_1^{b_1})|p_1^{b_1-1}(p_1+1)/2$. Since by assumption it follows that $z(p_1^{b_1})$ is odd and so divides $i-j$. Since $i-j$ is also divisible by $2^a=z(2^a)$, it is divisible by ${\text { lcm}}[z(2^a), z(p_1^{b_1})]=z(m)$.

(ii) $p_1^{b_1}$ does not divide $\alpha ^i-\alpha ^j$.

We want to show that this case does not occur. If it does, then $p_1$ divides

$$\begin{aligned} \alpha ^i+\alpha ^j-4{\sqrt{2}}\lambda . \end{aligned}$$

(9)

Assume first that $p_1\mid \lambda$. Then $p_1\mid U_i$ and $p_1\mid U_j$ so both i and j are divisible by the odd number $z(p_1)=(p_1+1)/2$. Also, . Since $i=z(p_1)i_1$ and $j=z(p_1)j_1$, where and , it follows that $\alpha ^i$ and $\alpha ^j$ are both congruent either to 1 (if $i_1$ and $j_1$ are even) or to $-1$ (if $i_1$ and $j_1$ are odd) modulo $p_1$. Thus, modulo $p_1$ the expression (9) is in fact congruent to $\pm 2$ modulo $p_1$, which is certainly not zero. Thus, $\lambda \ne 0$. Then

(10)

The prime $p_1$ is inert so we can conjugate the above relation to get

(11)

Multiplying the second congruence by $\alpha ^{i+j}$ and subtracting (11) from (10), we get $4$. Thus, $\alpha ^{i+j}\equiv -1\mod p_1$. But the smallest k such that is $k=z(p_1)=(p_1+1)/2$ which is odd. Hence, $i+j$ is an odd multiple of $z(p_1)$, therefore an odd number itself, which is a contradiction since . Thus, this case does not appear. This implies that if $U_i\equiv U_j\pmod m$.

Conversely, assume $i>j$ and . We need to show that . Since , it follows that $i-j$ is even and hence $U_i-U_j=U_{(i-j)/2}V_{(i+j)/2}$. Since $2^{a-1}\mid (i-j)/2$, we get, by iteratively applying the formula $U_{2n}=U_nV_n$, that

$$\begin{aligned} U_i-U_j=U_{(i-j)/2^{a}} V_{(i-j)/2^{a}} V_{(i-j)/2^{a-1}}\cdots V_{(i-j)/4} V_{(i+j)/2}. \end{aligned}$$

In the right–hand side we have a factors from the V sequence and each of them is a multiple of 2. Hence, $2^a\mid (U_i-U_j)$. As for the divisibility by $p_1^{b_1}$, note that since $z(p_1^{b_1})\mid (i-j)$ and $i-j$ is even, it follows that

$$\begin{aligned} i-j=p_1^{b_1-1}(p_1+1)\ell \end{aligned}$$

for some positive integer $\ell$. Since , it follows that . The same holds if we replace $\alpha$ by $\alpha ^{-1}$. Thus,

and the same congruence holds if $\alpha$ is replaced by $\alpha ^{-1}$. Subtracting these two congruences we get $p_1^{b_1}\mid ((\alpha ^i-\alpha ^{-i})-(\alpha ^j-\alpha ^{-j}))$. Computing norms in ${\mathbb K}$ and using the fact that $p_1$ is inert, we get $p_1^{2b_1}\mid N_{\mathbb K}((\alpha ^i-\alpha ^{-i})-(\alpha ^j-\alpha ^{-j}))$ and so

$$\begin{aligned} p_1^{2b_1}\mid 32(U_i-U_j)^2. \end{aligned}$$

Thus, . Hence, . $\square$

5 The end of the proof or why 5 and not 37?

We need a few more results before we are prepared well enough to establish Theorem 1.

Lemma 10

For $n\ge 2^{24}\cdot 5^3$ the interval [5n / 3, 37n / 19) contains a number of the form $2^a\cdot 5^{b}$ with $a\ge 1$ and $b\ge 0$.

Proof

It is enough to show that there exists an strictly increasing sequence of integers $\{m_i\}_{i=1}^{\infty }$ of the form $m_i=2^{a_i+1}\cdot 5^{b_i}$ with $a_1=23$ and $b_1=3$, $a_i,b_i\ge 0$, having the property that

$$\begin{aligned} 1<\frac{m_{i+1}}{m_i}<\frac{111}{95}. \end{aligned}$$

Since both $2^7/5^3$ and $5^{10}/2^{23}$ are in (1, 111 / 95), the idea is to use the substitutions $5^3\rightarrow 2^7$ and $2^{23}\rightarrow 5^{10}$ to produce a strictly increasing sequence starting from $m_1$. Note that we can at each stage make one of these substitutions as otherwise we have reached a number dividing $2\cdot 2^{22}\cdot 5^{2}<m_1$, a contradiction. $\square$

Corollary 3

Suppose that $m=2^a\cdot p^{b}$, $p>5$, $a,b\ge 1$. If $m\ge \frac{37}{19}\cdot 2^{24}\cdot 5^{3}$, then m is not a discriminator value.

Proof

Suppose that $\mathcal {D}_1(n)=m$, then we must have

$$\begin{aligned} z(m)=2^a\cdot p^{b-1}(p+1)/(2k)\ge 19m/37\ge n, \end{aligned}$$

that is $m\ge 37n/19$. By Lemma 10 in the interval [5n / 3, 37n / 19) there is an integer of the form $m=2^c\cdot 5^{d}$ with $c\ge 1$. This integer discriminates the first n terms of the sequence and is smaller than m. This contradicts the definition of the discriminator. $\square$

Thus we see that in some sense there is an abundance of numbers of the form $m=2^{a}\cdot 5^b$ that are in addition fairly regularly distributed. Since they discriminate the first n terms provided that $m\ge 5n/3$, rather than the weaker $m\ge 2np/(p+1)$ for $p>5$, they remain as values, whereas numbers of the form $m=2^{a}\cdot p^b$ with $p>5$ do not.

Lemma 11

If $n>1,$ then ${\mathcal D}_1(n)=2^a\cdot 5^b$ for some $a\ge 1$ and $b\ge 0$.

Proof

By Lemma 7 we have $a\ge 1$. If $m={\mathcal D}_1(n)\ne 2^a$ for some $a\ge 1$, then by Lemma 6 iii it is of the form $m=2^a\cdot p_1^{b_1}$ for some . Let us assume for the sake of contradiction that we have discriminators of the form $m=2^ap_1^{b_1}$ for some odd $p_1> 5$. Then $z(p_1^{b_1})=p_1^{b_1-1}(p_1+1)/2$. Let A be minimal with $p_1^{b_1}<2^{A+8}$. Then $A\ge 2$ by our calculation because we did not find any such $p_1$ on calculating ${\mathcal D}_1(n)$ for $n\le 2^{10}$ (cf. Sect. 1). Then

$$\begin{aligned} 2^{A+7}<p_1^{b_1-1}\frac{(p_1+1)}{2}. \end{aligned}$$

Consider the numbers

$$\begin{aligned} 2^{A+7},\quad 2^{A+1}\cdot 3\cdot 5^2,\quad 2^{A+1}\cdot 5^3,\quad 2^{A+8}. \end{aligned}$$

Assuming that $p_1>50$, $2^{A+8}>p_1^{b_1}$ and that $p_1^{b_1}+p_1^{b_1-1}>2^{A+8}$, we obtain

$$\begin{aligned} 0<2^{A+8}-p_1^{b_1}<p_1^{b_1-1}<\frac{2^{A+8}}{50}. \end{aligned}$$

Hence, $p_1^{b_1}$ sits in the last $2\%$ of the interval ending at $2^{A+8}$. Since

$$\begin{aligned} 2^{A+8}-2^{A+1} 5^3=2^{A+1} 3=2^{A+8} (3/128)>2^{A+8}/50, \end{aligned}$$

it follows that $p_1^{b_1}>2^{A+1} 5^3$. We now claim that $p_1^{b_1-1}(p_1+1)/2<2^{A+1} \cdot 3\cdot 5^2$. Indeed, for that to happen it suffices that

$$\begin{aligned} 2^{A+8} \left( \frac{p_1+1}{2p_1}\right) <2^{A+1}\cdot 3\cdot 5^2, \end{aligned}$$

so $2p_1/(p_1+1)>128/75$, which is equivalent to $22p_1>128$, which is true for $p_1>50$.

Since $2^a\cdot p_1^{b_1}$ discriminates the first $2^a\cdot p_1^{b_1-1}(p_1+1)/2$ terms of the sequence (but not more) and the same integers are discriminated by the smaller number $2^{a+A+1}\cdot 5^3$, the number $2^a\cdot p_1^{b_1}$ is not a discriminator value.

So, it remains to check the primes $p_1<50$. Since , we just need to check $p_1\in \{13,29,37\}$. Fortunately, 13 is a Wiefrich prime for $\alpha$ in that

so we cannot use powers $13^{b_1}$ with $b_1>1$, while for $b_1=1$ the interval $[z(p_1),p_1]=[7,13]$ contains 8 which is a power of 2. For 29, we have that $z(29)=5$ (instead of $(29+1)/2$), so 29 is not good either.

It remains to deal with $p=37$. We will show that for $k_i=2\cdot 37^i$ and $1\le i\le 5$, there is a power of the form $2^{e_i}<k_i$ that discriminates the same terms of the sequence as $k_i$ does, thus showing that $k_i$ cannot be a discriminator. By the same token, any potential value $2^{\alpha }\cdot 37^i$, $1\le i\le 6$, is outdone by $2^{\alpha +e_i}$. Any remaining value of the form $2^{\alpha }\cdot 37^i$ has $i\ge 6$ and $\alpha \ge 1$ and cannot be a value by Corollary 3.

The numbers $2^{e_i}$ we are looking for must satisfy

$$\begin{aligned} 2\cdot 37^{i-1}\cdot 19\le 2^{e_i}< 2\cdot 37^i\text { for }1\le i\le 5. \end{aligned}$$

(Recall that the number $2\cdot 37^i$ discriminates the first $2\cdot 37^{i-1}\cdot 19$ terms of the sequence and not more terms.) Note that the numbers $e_i$ are unique if they exist. Some simple computer algebra computations yield $e_1=6,e_2=11,e_3=16,e_4=21$ and $e_5=27$. $\square$

Lemma 12

We say that m discriminates $U_0,\ldots ,U_{n-1}$ if these integers are pairwise distinct modulo m.

• (i) The integer $m=2^a$ discriminates $U_0,\ldots ,U_{n-1}$ if and only if $m\ge n$.

• (ii) The integer $m=2^a\cdot 5^b$ with $a,b\ge 1$ discriminates $U_0,\ldots ,U_{n-1}$ if and only if $m\ge 5n/3$.

Proof

Case i follows from Lemma 1. Now suppose that $a,b\ge 1$. By Lemma 9 the integer m discriminates $U_0,\ldots ,U_{z(m)-1}$, but not $U_0,\ldots ,U_{z(m)}$. It follows that m discriminates $U_0,\ldots ,U_{n-1}$ iff $n\le z(m)$. As it is easily seen that $z(m)=3m/5$, the result follows. $\square$

At long last we are ready to prove Theorem 1.

Proof of Theorem 1

As the statement is correct for $n=1$, we may assume that $n>1$. By Lemma 11 it then follows that either $m=2^a$ for some $a\ge 1$ or $m=2^a\cdot 5^b$ with $a,b\ge 1$. On invoking Lemma 12 we infer that the first assertion holds true.

It remains to determine the image of the discriminator ${\mathcal D}_1$. Let us suppose that $m=2^a\cdot 5^b$ with $a,b\ge 1$ occurs as value. Let $\alpha$ be the unique integer such that $2^{\alpha }<2^a\cdot 5^b<2^{\alpha +1}$. By Lemma 12 it now follows that we must have $z(m)>2^{\alpha }$, that is $2^a\cdot 5^{b-1}\cdot 3>2^{\alpha }$. It follows that m occurs as value iff

$$\begin{aligned} \frac{5}{3}\cdot 2^{\alpha }<2^a\cdot 5^b<2^{\alpha +1}. \end{aligned}$$

(12)

(Indeed, under these conditions we have ${\mathcal D}_1(n)=2^a\cdot 5^b$ for $n\in [2^a+1,2^a\cdot 5^{b-1}\cdot 3]$.) Inequality (12) can be rewritten as $5/6<2^{a-\alpha -1}<1$ and, after taking logarithms, is seen to have a solution iff $b\in {\mathcal M}$. If it has a solution, then we must have $\alpha -a=\lfloor b\log 5/\log 2\rfloor$. In particular for each $a\ge 1$ and $b\in {\mathcal M}$, the number $2^a\cdot 5^b$ occurs as value. $\square$

6 General k

6.1 Introduction

What is happening for $k>1$? It turns out that the situation is quite different.

For $k=2$ we have the following result.

Theorem 4

Let $e\ge 0$ be the smallest integer such that $2^e\ge n$ and $f\ge 1$ the smallest integer such that $3\cdot 2^f\ge n$. Then ${\mathcal D}_2(n)=\min \{2^e,3\cdot 2^f\}$.

Proof

If $z(m)=m$, then m divides $3\cdot 2^a$ for some $a\ge 0$. For the other integers m we have $z(m)\le 3m/5$ (actually even $z(m)\le 7m/13$). It follows that if m discriminates the first n values of the sequence U(2), then we must have $m\ge 5n/3$. It is easy to check that for every $n\ge 2$ there is a power of two or a number of the form $3\cdot 2^a$ in the interval [n, 5n / 3). As ${\mathcal D}_2(1)=1$ we are done. $\square$

For $k>2$ the situation is rather more complex and described in Theorem 3. In our proof of this result the rank of appearance plays a crucial role. Its most important properties are summarized in Lemma 14. After some further preparatory work we will finally present a proof of Theorem 3 in Sect. 6.6.

6.2 The index of appearance

6.2.1 The case where $p\mid k(k+1)$

The index of appearance for primes p dividing $k(k+1)$ is determined in Lemma 3 for $p>2$. By Lemma 1 we have $z(2^b)=2^b$. In general $z(p^b)=p^b$ for these primes, but for a prime p which we call special a complication can arise giving rise to $z(p^b)\mid p^{b-1}$ for $b\ge 2$.

Definition 1

A prime p is said to be special if $p|k(k+1)$ and $p^2|U_p$.

The special feature of a special prime p is that $p^b$ with $b\ge 2$ cannot divide a discriminator value. Recall that $z(p^a)=p^{\max \{a-c,0\}} z(p)$, where $c=\nu _p(U_{z(p)})$ by Lemma 2.

Lemma 13

Let $p\ge 3$ be an odd prime. If $z(p^{b})|p^{b-1}$, then $m=p^bm_1$ with $p\not \mid m_1$ is not a discriminator value.

Proof

Taking $i=0$ and $j=p^{b-1}z(m_1)$ we have It follows that $n\le p^{b-1} z(m_1)\le m/p$ so any power of 2 in [m / 3, m) (and such a power exists) is a better discriminator than m. $\square$

By Lemma 3 only the prime 3 can be special.

6.2.2 The case where $p\not \mid k(k+1)$

Let us now look at odd prime numbers p such that $p\not \mid k(k+1)$. These come in two types according to the sign of

$$\begin{aligned} e_p=\Big (\frac{k(k+1)}{p}\Big ). \end{aligned}$$

(13)

Suppose that $e_p=1$. Then either

$$\begin{aligned} \left( \frac{k}{p}\right) =\left( \frac{k+1}{p}\right) =1\quad {\text { or}}\quad \left( \frac{k}{p}\right) =\left( \frac{k+1}{p}\right) =-1. \end{aligned}$$

In the first case,

In the second case, a similar calculation shows that $\beta ^p\equiv -\beta$. Thus, and since $\alpha =\beta ^2$, we get that . Since the last congruence implies that we obtain on subtracting these two congruences that $p\mid U_{(p-1)/2}$. Thus, $z(p)\mid (p-1)/2$. In case $e_p=-1$, a similar calculation shows that , so . This shows that , which leads to $z(p)\mid (p+1)/2$. There is one more observation which is useful here. Assume that $e_p=-1$, which implies that $z(p)\mid (p+1)/2$. Suppose that . Then $(p+1)/2$ is even. Assume further that

$$\begin{aligned} \left( \frac{k+1}{p}\right) =1\quad {\text { and }}\quad \left( \frac{k}{p}\right) =-1. \end{aligned}$$

In this case, by the above arguments, we have that , so . This gives . Since $(p+1)/2$ is even we conclude that

$$\begin{aligned} p\mid (\alpha ^{(p+1)/4}-1)(\alpha ^{(p+1)/4}+1). \end{aligned}$$

Since p is inert in ${\mathbb K}$, we get that , from which we deduce that $p\mid U_{(p+1)/4}$, Hence, $z(p)\mid (p+1)/4$ in this case.

6.2.3 General m

Lemma 14

Let $k\ge 1$. We have $z(m)=m$ if and only if

$$\begin{aligned} {\left\{ \begin{array}{ll} m\in {\mathcal P}(k(k+1)),~9\not \mid m;\\ m\in {\mathcal P}(k(k+1)),~9\mid m,~ \text {and }3{~is~not~special.} \end{array}\right. } \end{aligned}$$

For the remaining integers m we have

$$\begin{aligned} z(m)\le \alpha _k m, \end{aligned}$$

with

$$\begin{aligned} \alpha _k:=\lim \sup _{m\rightarrow \infty }\Big \{\frac{z_k(m)}{m}:z_k(m)<m\Big \}. \end{aligned}$$

(14)

One has

$$\begin{aligned} \alpha _k=\lim \sup _{p\rightarrow \infty }\Big \{\frac{z_k(p)}{p}:z_k(p)<p\Big \}. \end{aligned}$$

(15)

Furthermore, we have $\alpha _k=2/3$ if and $\alpha _k\le 3/5$ otherwise.

Corollary 4

We have $z(m)\le m$.

Corollary 5

We have

$$\begin{aligned} {\mathcal A}_k=\{m~{\text { odd}}:z(m)=m\text { and } m\in {\mathcal P}(k)\}, \end{aligned}$$

and

$$\begin{aligned} {\mathcal B}_k=\{m~\text { even}:z(m)=m\}. \end{aligned}$$

Proof of Lemma 14

By the above discussion if $p\not \mid k(k+1)$, then $z(p^b)<p^b$. Thus if $z(m)=m$, then $m\in {\mathcal P}(k(k+1))$. The first assertion now follows by Lemma 1 (which shows that $z(2^b)=2^b$) and Lemma 3 and the observation that if $m=\prod _{i=1}^s p_i^{b_i}$ is the factorization of m with $z(p_i^{b_i})=p_i^{b_i}$, then

$$\begin{aligned} z(m)=\text { lcm}(z(p_1^{b_1}),\ldots ,z(p_s^{b_s}))= \prod _{i=1}^s p_i^{b_i}=m. \end{aligned}$$

If $m=\prod _{i=1}^s p_i^{b_i}$ is the factorization of any integer, then

$$\begin{aligned} \frac{z(m)}{m}\le \prod _{i=1}^s \frac{z(p_i^{b_i})}{p_i^{b_i}} \le \prod _{i=1}^s \frac{z(p_i)}{p_i}. \end{aligned}$$

From these inequalities we infer the truth of (15). The proof is concluded on noting that

and that $(p+1)/2p$ is a decreasing function of p. $\square$

It is easy to see that if there is a prime p with $z(p)=(p+1)/2$, then

$$\begin{aligned} \alpha _k=\frac{q+1}{2q}, \end{aligned}$$

where q is the smallest prime such that $z(q)=(q+1)/2$.

6.3 The congruence $U_i(k)\equiv U_j(k)\pmod m$

In this subsection we study the congruence . By the Chinese Remainder Theorem it suffices to study it modulo prime powers $p^b$. For powers of 2, this has been done at the beginning of Sect. 2. Recall that the discriminant $\Delta (k)$ equals $16k(k+1)$. It turns out that primes p dividing $\Delta (k)$ are easier to understand than the others. From now on, we eliminate the index k from $U_n(k),~\alpha (k),~\Delta (k)$ and so on. We treat the case when $p\mid k(k+1)$. In case m is even, there are two subcases, one easy and one harder, according to whether $p\mid k$ or $p\mid (k+1)$.

Lemma 15

Assume $p\mid k$ is odd. Let $b\ge 1$ be arbitrary. Then if and only if .

Proof

We prove the only if assertion. We let a be such that $p^a\Vert k$. We put $k(k+1)=du^2$, and let ${\mathbb K}={\mathbb Q}[{\sqrt{d}}]$. We let $\pi$ be any prime ideal diving p and let e be such that $\pi ^e\Vert p$. For example, $e=2$ if $p\mid d$. Let $\lambda$ be the residue class of the number $U_i$ modulo $p^b$. Then implies that

The same holds for $\alpha ^i$ replaced by $\alpha ^j$. Hence, these numbers both satisfy the quadratic congruence

Taking their difference we get

(16)

In case $p\mid k$, we have that . Thus, the second factor above is congruent to . In particular, $\pi$ is coprime to that factor. Thus,

This leads to . Changing $\alpha$ to $\alpha ^{-1}$ and taking the difference of the above expressions we arrive at . Thus,

Clearly, the exponent of $\pi$ in $2{\sqrt{k(k+1)}}$ is exactly ae / 2. Thus, $\pi ^{eb}\mid U_{i-j}$. Since this is true for all prime power ideals $\pi ^e$ dividing p, we get that $p^b\mid U_{i-j}$. Thus, .

For the if assertion, assume that . Then the congruence holds which implies . In turn this gives so . Since , the factor $\alpha ^{i-j}+1$ is congruent to , so coprime to $\pi$. So , giving . Since $\alpha$ is a unit we also get . Taking the difference of the last two congruences, we get

Simplifying the square-root which contributes a power $\pi ^{ae/2}$ to the left–hand side of the above congruence, we get

and since this is true for all $\pi \mid p$, we get that . $\square$

Now we treat the more delicate case $p\mid (k+1)$. Here we have the following analogue of Lemma 15.

Lemma 16

Assume that p is odd and $p\mid (k+1)$. Let $b\ge 1$ be arbitrary. Then is equivalent to one of the following:

• (i) If , then .

• (ii) If , then .

Proof

The proof is similar to the previous lemma. Let $p^a\mid (k+1)$ and let $\pi$ be some prime ideal in ${\mathbb K}$ such that $\pi ^e\mid p$. Then

Let again $\lambda$ be the value of . The same argument as before leads us to the congruence (16). The first factor is congruent to

The second one is congruent to . Thus, $\pi$ never divides both factors, and $\pi ^{ae/2}$ divides $\alpha ^i-\alpha ^j$ in case , and it divides $\alpha ^i+\alpha ^j-4{\sqrt{k(k+1)}} \lambda$ in case .

In case , we have . Thus, . Arguing as in the proof of the preceding lemma yields and hence .

Assume now that . Multiply both sides of the congruence

by $\alpha ^j$ and rewrite it as

Since $\pi ^{ae/2}\mid 4{\sqrt{k(k+1)}} \alpha ^j$, it follows that the value of the right–hand side is determined by , which is $(\alpha ^j-\alpha ^{-j})/(4{\sqrt{k(k+1)}})$. Thus,

So we get that . The same holds with $\alpha$ replaced by $\alpha ^{-1}$. Subtracting both congruences we get that

$$\begin{aligned} \pi ^{be+ae/2}\mid (\alpha ^{i+j}-\alpha ^{-i-j})=4{\sqrt{k(k+1)}} U_{i+j}, \end{aligned}$$

leading to $(\pi ^e)^b\mid U_{i+j},$ and thus to $z(p^b)\mid (i+j).$

We now have to do the if parts. They are pretty similar to the previous analysis. We start with $i\equiv j\pmod 2$. Then $i-j\equiv 0\pmod {z(p^b)}$, so $U_{i-j}\equiv 0\pmod {p^b}$. This gives as in the previous case $\alpha ^{i-j}\equiv \alpha ^{-(i-j)}\pmod {\pi ^{eb+ae/2}}$, so $\alpha ^{2(i-j)}\equiv 1\pmod {\pi ^{eb+ae/2}}$. Thus, $(\alpha ^{i-j}-1)(\alpha ^{i-j}+1) \equiv 0\pmod {\pi ^{be+ae/2}}$. Since $i-j$ is even, $\alpha ^{i-j}\equiv (-1)^{i-j}\pmod \pi \equiv 1\pmod {\pi }$, so the second factor is congruent to $2\pmod {\pi }$, so it is coprime to $\pi$. So, $\alpha ^{i-j}-1\equiv 0\pmod {\pi ^{be+ae/2}}$. Now the argument continues as in the last part of the proof of the preceding lemma to get to the conclusion that $U_i\equiv U_j\pmod {p^b}$.

A similar argument works when $i\not \equiv j\pmod 2$. With the same argument we get from $i+j\equiv 0\pmod {z(p^b)}$ to the relation $U_{i+j}\equiv 0\pmod {p^b}$, which on its turn leads to $(\alpha ^{i+j}-1)(\alpha ^{i+j}+1)\equiv 0\pmod {\pi ^{be+ae/2}}$. Since $i+j$ is odd, the factor $\alpha ^{i+j}-1$ is congruent to is $-2\pmod {\pi }$, so it is coprime to $\pi$. So, $\alpha ^{i+j}+1\equiv 0\pmod {\pi ^{eb+ae/2}}$ and multiplying with a suitable power of $\alpha$ and rearranging we get $\alpha ^i\equiv -\alpha ^{-j}\pmod {\pi ^{be+ae/2}}$, and also $\alpha ^{-i}\equiv -\alpha ^j\pmod {\pi ^{be+ae/2}}$. Taking the difference of these last two congruences, we get $\alpha ^i-\alpha ^{-i}-\alpha ^j+\alpha ^{-j}\equiv 0\pmod {\pi ^{be+ae/2}}$, which leads to the congruence $2\sqrt{k(k+1)}(U_i-U_j)\equiv 0\pmod {\pi ^{be+ae/2}}$. Simplifying $2\sqrt{k(k+1)}$, we get that $\pi ^{be}$ divides $U_i-U_j$, and since $\pi$ is an arbitrary prime ideal of p, we conclude that $U_i\equiv U_j\pmod {p^b}$. $\square$

Definition 2

We write ${\mathcal P}(r)$ for the set of positive integers composed only of prime factors dividing r.

Lemma 17

We have

$$\begin{aligned} i\equiv j\pmod {m} \iff U_i\equiv U_j\pmod {m}, \end{aligned}$$

(17)

precisely when

$$\begin{aligned} m\in {\mathcal A}_k\cup {\mathcal B}_k. \end{aligned}$$

Proof

Since $0=U_0(k)\equiv U_{z(m)}(k)\pmod m$, we must have $z(m)\ge m$. As $z(m)\le m$ by Corollary 12 it follows that $z(m)=m$.

First subcase: m is odd.

Since $z(m)=m$ all prime divisors of m must divide $k(k+1)$. Now suppose that m has an odd prime divisor p dividing $k+1$. Thus $m=p^am_1$ with $m_1$ coprime to p and odd. Note that $z(p^a)=p^a$. Consider $i=(p^{a}-1)m_1/2$ and $j=(p^{a}+1)m_1/2$. Then $i\not \equiv j\pmod 2$ and $p^{a}\mid (i+j)$. Thus, $U_i\equiv U_j\pmod {p^{a}}$ by Lemma 16. Since $m_1\mid i$ and $m_1\mid j$ and $m_1$ is composed of primes dividing $\Delta (k)=16k(k+1)$, it follows that $U_i\equiv U_j\equiv 0\pmod {m_1}$ and hence we have $U_i\equiv U_j\pmod {m}$ with $m\not \mid (j-i)$. It follows that (17) is not satisfied. Thus we conclude that if an odd integer m is to satisfy (17), then it has to be in ${\mathcal P}(k)$. For such an integer, by Lemma 15 and the Chinese remainder theorem, (17) is always satisfied. It follows that the solution set of odd m satisfying (17) is $\{m~{\text { odd}}:z(m)=m\text { and } m\in {\mathcal P}(k)\},$ which by Corollary 5 equals ${\mathcal A}_k$.

Second subcase: m is even. Both the left and the right side of (17) imply that $i\equiv j\pmod {2}$. On applying Lemmas 15 and 16 and the Chinese remainder theorem we see that in this case the solution set is $\{m~\text { even}:z(m)=m\},$ which by Corollary 5 equals ${\mathcal B}_k$. $\square$

6.4 A Diophantine interlude

The prime 3 sometimes being special leads us to solve a very easy Diophantine problem (left to the reader).

Lemma 18

If $k>2$, then $k(k+1)$ has an odd prime factor that is not special.

Proof

If $k(k+1)$ only has an odd prime factor that is special, then it must be 3 and $k\equiv 2,6\pmod {9}$. It follows that for such a k there are a, b for which the Diophantine equation

$$\begin{aligned} k(k+1)=2^a\cdot 3^b, \end{aligned}$$

(18)

has a solution. However, this is easily shown to be impossible for $k>2$. $\square$

It is slightly more challenging to find all solutions $k\ge 1$ of (18). In that case one is led to the Diophantine equation

$$\begin{aligned} 2^a-3^b\equiv \pm 1, \end{aligned}$$

which was already solved centuries ago by Levi ben Gerson (alias Leo Hebraeus), who lived in Spain from 1288 to 1344, cf. Ribenboim [12, p. 5]. It has the solutions $(a,b)=(1,0),(0,1),(2,1)$ and $(a,b)=(3,2)$, corresponding to, respectively, $k=1,2,3$ and $k=8$.

6.5 Bertrand’s postulate for S-units

Before we embark on the proof of our main result we make a small excursion in Diophantine approximation.

Lemma 19

Let $\alpha >1$ be a real number and p be an arbitrary odd prime. Then there exists a real number $x(\alpha )$ such that for every $n\ge x(\alpha )$ the interval $[n,n\alpha )$ contains an even integer of the form $2^a\cdot p^b$.

Proof

Along the lines of the proof of Lemma 10. If $\beta$ is irrational, then the sequence of integers $\{m\beta \}_{m=1}^{\infty }$ is uniformly distributed. This allows one to find quotients $2^c/p^d$ and $p^r/2^s$ that are in the interval $(1,\alpha )$. Then proceed as in the proof of Lemma 10. $\square$

The result also holds for S-units of the form $\prod _{i=1}^s p_i^{b_i}$ with $p_1<\ldots < p_s$ primes and $s\ge 2$.

6.6 Proof of the main result for general k

Finally we are in the position to prove our main result for $k>1$.

Proof of Theorem 3

Let $k>2$.

First case: $m\in {\mathcal A}_k\cup {\mathcal B}_k.$ (Note that $z(m)=m$ for these m.)

By Lemma 17 we infer that the inequality (2) holds true and moreover the equivalence (3). The “$\Leftarrow$” implication in (3) yields ${\mathcal A}_k\cup {\mathcal B}_k \subseteq {\mathcal D}_k$.

Second case: $z(m)=m$ and $m\not \in {\mathcal A}_k\cup {\mathcal B}_k$.

In this case m has a odd prime divisor p that also divides $k+1$. Now write $m=p^a\cdot m_1$ with $p\not \mid m_1$ and $m_1$ odd. Note that $z(p^a)=p^a$. Consider $i=(p^a-1)m_1/2$ and $j=(p^a+1)m_1/2$. Then $i\not \equiv j\pmod 2$ and $p^{a}\mid (i+j)$. Thus, $U_i\equiv U_j\pmod {p^{a}}$ by Lemma 16. Since $m_1\mid i$ and $m_1\mid j$ and $m_1$ is composed of primes dividing $\Delta (k)$, it follows that $U_i\equiv U_j\equiv 0\pmod {m_1}$. This shows that if m discriminates the numbers $U_0(k),\ldots ,U_{n-1}(k),$ then

$$\begin{aligned} n\le \left( \frac{p^{a}+1}{2}\right) m_1. \end{aligned}$$

The interval $[(p^a+1)/2,p^{a})$ contains a power of 2, say $2^b$. Then $2^{b}m_{1}$ is a better discriminator than $p^{a}m_1=m 1$. Thus if $z(m)=m$ and $m\not \in {\mathcal A}_k\cup {\mathcal B}_k,$ then m is not a discriminator value.

Third case: $z(m)<m$.

Here it follows by Lemma 14 that $z(m)\le \alpha _k m\le 2m/3$. In order for m to discriminate the first n terms we must have $n\le z(m)\le 2m/3$, that is $m\ge 3n/2$. Now if in the interval [n, 3n / 2) there is an element from ${\mathcal A}_k\cup {\mathcal B}_k,$ this will discriminate the first n terms too and is a better discriminator than m. Thus in this case in (2) we have equality.

Since by assumption $k>2$, by Lemma 18 there exists a non-special odd prime p dividing $k(k+1)$ and hence if $a,b\ge 0,$ then $2^{1+a}\cdot p^b \in {\mathcal A}_k\cup {\mathcal B}_k.$ It now follows by Lemma 19 that for every n large enough the interval [n, 3n / 2) contains an element from ${\mathcal A}_k\cup {\mathcal B}_k$ and so there are at most finitely many n for which in (2) strict inequality holds. $\square$

6.7 The set ${\mathcal F}_k$

As was remarked in the introduction a consequence of Theorems 2 and 3 is that for $k>1$ there is a finite set ${\mathcal F}_k$ such that

$$\begin{aligned} {\mathcal D}_{k}= {\mathcal A}_k\cup {\mathcal B}_k\cup {\mathcal F}_k. \end{aligned}$$

The set ${\mathcal F}_k$ is not a figment of our proof of this result, as the following result shows.

Lemma 20

There are infinitely many k for the finite set ${\mathcal F}_k$ is non-empty. It can have a cardinality larger than any given bound.

Proof

Let N be large and $k\equiv 1\pmod {N!}$. Then $U(k)\pmod m$ is the same as $U(1)\pmod m$ for all $m\le N$. In particular, if $N>2\cdot 5^{m_s}$, where $m_s$ is the $s^\text {th}$ element of the set ${\mathcal M}$, then certainly ${\mathcal D}_1\cap [1,N]$ will contain the numbers $2\cdot 5^{m_i}$ for $i=1,\ldots ,s$, and $5\not \mid k(k+1)$ (in fact, $k\equiv 1\pmod {5}$, so $5\not \mid k(k+1)$), therefore all such numbers are in the set ${\mathcal F}_k$ for such values of k. $\square$

Thus it is illusory to want to describe ${\mathcal F}_k$ completely for every $k\ge 1$. Nevertheless, in part II [5] we will explore how far we can get in this respect.

7 Analogy with the polynomial discriminator

In our situation for $k\ge 1$ on the one hand there are enough integers m with $z(m)=m$ and ${\mathcal D}_k(m)=m$, on the other hand for the remaining m either $z(m)=m$ and m is not a discriminator value or we have $z(m)\le \alpha _km$ with $\alpha _k<1,$ a constant not depending on m. Thus the distribution of $\{z(m)/m:m\ge 1\}$ shows a gap directly below 1 (namely $(\alpha _k,1)$).

For polynomial discriminators the analogue of z(p) is V(p), the number of values assumed by the polynomial modulo p. If on the one hand there are enough integers m such that f permutes $\mathbb Z/m\mathbb Z$, and on the other hand V(p) / p with $V(p)<p$ is bounded away from 1 (thus also shows a gap directly below 1), then the polynomial discriminator can be easily described for all n large enough. See Moree [9] and Zieve [16] for details.