A Class of Exponential Sequences with Shift-Invariant Discriminators

TL;DR

This paper introduces a class of exponential sequences whose discriminators remain unchanged under any shift, revealing a unique invariance property in the structure of these sequences.

Contribution

It identifies and characterizes exponential sequences with shift-invariant discriminators, a novel property not previously documented.

Findings

Discovered a class of exponential sequences with shift-invariant discriminators

Proved the invariance property holds for these sequences

Provided theoretical framework for analyzing discriminator invariance

Abstract

The discriminator of an integer sequence s = (s(i))_{i>=0}, introduced by Arnold, Benkoski, and McCabe in 1985, is the function D_s(n) that sends n to the least integer m such that the numbers s(0), s(1), ..., s(n-1) are pairwise incongruent modulo m. In this note we present a class of exponential sequences that have the special property that their discriminators are shift-invariant, i.e., that the discriminator of the sequence is the same even if the sequence is shifted by any positive constant.

Equations

$$(\mathrm{ex}(n){)}_{n\ge 0}={\left(a\frac{(t^2{)}^n-1}{2^b}\right)}_{n\ge 0}$$

$$t^2$$

$$=2^{b+1}\left(2^{b-1}{c}^2+2^{b-3}+2^{b-1}c\pm c\right)\pm 2^b+1,$$

$$⟹t^2$$

$$t^{2^k}\equiv 2^{k+b-1}+1\text{mbox}(\mathrm{mod}{2}^{k+b}\text{mbox})$$

$${\left(t^{2^k}\right)}^2=t^{2^{k+1}}$$

$$=2^{k+b+1}\left(2^{k+b-1}{c}^2+2^{k+b-3}+2^{k+b-1}c+c\right)+2^{k+b}+1,$$

$$⟹t^{2^{k+1}}$$

$$(t^2{)}^{2^{\ell -1}}=t^{2^{\ell }}$$

$$⟹(t^2{)}^{2^{\ell -1}}$$

$$⟹(t^2{)}^{2^k}$$

$$(t^2{)}^{2^{k-1}}=t^{2^k}$$

$$⟹(t^2{)}^{2^{k-1}}$$

$$(t^2{)}^i,(t^2{)}^{i+1},\dots ,(t^2{)}^{i+2^k-1}$$

$$(t^2{)}^i-1,(t^2{)}^{i+1}-1,\dots ,(t^2{)}^{i+2^k-1}-1$$

$$\left\{\frac{(t^2{)}^i-1}{2^b},\frac{(t^2{)}^{i+1}-1}{2^b},\dots ,\frac{(t^2{)}^{i+2^k-1}-1}{2^b}\right\}$$

$$m$$

$$(t^2{)}^i$$

$$\text{and }(t^2{)}^i$$

$$i$$

$$j$$

$$\le \underset{1\le \ell \le u}{\max }\lceil \frac{y_{\ell }}{2}\rceil +\frac{2^x}{2^v}\prod _{1\le \ell \le v}{q}_{\ell }^{z_{\ell }}=\underset{1\le \ell \le u}{\max }\lceil \frac{y_{\ell }}{2}\rceil +\frac{m}{2^v{\prod }_{1\le \ell \le u}{p}_{\ell }^{y_{\ell }}}.$$

$$j\le \underset{1\le \ell \le u}{\max }\lceil \frac{y_{\ell }}{2}\rceil +\frac{m}{2^v{\prod }_{1\le \ell \le u}{p}_{\ell }^{y_{\ell }}}=\frac{m}{2^v}\le \frac{m}{2}<\frac{2^{k+1}}{2}=2^k.$$

$$j\le \underset{1\le \ell \le u}{\max }\lceil \frac{y_{\ell }}{2}\rceil +\frac{m}{2^v{\prod }_{1\le \ell \le u}{p}_{\ell }^{y_{\ell }}}\le \lceil \frac{y_r}{2}\rceil +\frac{m}{p_r^{y_r}}\le \frac{y_r+1}{2}+\frac{m}{3}\le \frac{y_r}{2}+\frac{1}{2}+\frac{m}{3}.$$

$$j\le \frac{y_r}{2}+\frac{1}{2}+\frac{m}{3}\le \frac{m}{6}+\frac{1}{2}+\frac{m}{3}=\frac{m}{2}+\frac{1}{2}=\frac{m+1}{2}.$$

$$j\le \lceil \frac{m}{2}\rceil \le \lceil \frac{2^{k+1}}{2}\rceil \le 2^k.$$

$$m|a(s(j)-s(i))=as(j)-as(i)=s^{\prime }(j)-s^{\prime }(i).$$

$$m\nmid a(s(j)-s(i))=as(j)-as(i)=s^{\prime }(j)-s^{\prime }(i)$$

$$\{s^{\prime }(0),s^{\prime }(1),\dots ,s^{\prime }(n-1)\}$$

$$D_{\mathrm{exs}}(n)=2^{\lceil {\log }_{2}n\rceil }.$$

$$\{\mathrm{ex}(c),\mathrm{ex}(c+1),\dots ,\mathrm{ex}(c+2^{k+1}-1)\},$$

$$\{\mathrm{exs}(0,c),\mathrm{exs}(1,c),\dots ,\mathrm{exs}(n-1,c)\}.$$

$$(t^2{)}^i\equiv (t^2{)}^j\text{mbox}(\mathrm{mod}{2}^{b}m\text{mbox})$$

Full text

1 Discriminators

Let $m$ be a positive integer. If $S$ is a set of integers that are pairwise incongruent modulo $m$, we say that $m$ discriminates $S$. Now let $\textbf{s}=(s(i))_{i\geq 0}$ be a sequence of distinct integers. For all integers $n\geq 1$, we define $D_{\textbf{s}}(n)$ to be the least positive integer $m$ that discriminates the set $\{s(0),s(1),\ldots,s(n-1)\}$. The function $D_{\textbf{s}}(n)$ is called the discriminator of the sequence s.

The discriminator was first introduced by Arnold, Benkoski, and McCabe [1]. They derived the discriminator for the sequence $1,4,9,\ldots$ of positive integer squares. More recently, discriminators of various sequences were studied by Schumer and Steinig [11], Barcau [2], Schumer [10], Bremser, Schumer, and Washington [3], Moree and Roskam [8], Moree [6], Moree and Mullen [7], Zieve [13], Sun [12], Moree and Zumalacárrequi [9], and Haque and Shallit [5].

In all of these cases, however, the discriminator is based on the first $n$ terms of a sequence, for $n\geq 2$. Therefore, the discriminator can depend crucially on the starting point of a given sequence. For example, although the discriminator for the first three positive squares, $\{1,4,9\}$, is $6$, we can see that the number $6$ does not discriminate the length-3 “window” into the shifted sequence, $\{4,9,16\}$, since $16\equiv 4\ \mbox{{\rm(mod}}\ 6\mbox{\rm)}$.

Furthermore, there has been very little work on the discriminators of exponential sequences. Sun [12] presented some conjectures concerning certain exponential sequences, while in a recent tour de force, Moree and Zumalacárrequi [9] computed the discriminator for the sequence $\left(\frac{|(-3)^{j}-5|}{4}\right)_{j\geq 0}$.

We say that the discriminator of a sequence is shift-invariant if the discriminator for the sequence is the same even if the sequence is shifted by any positive integer $c$, i.e., for all positive integers $c$ the discriminator of the sequence $(s(n))_{n\geq 1}$ is the same as the discriminator of the sequence $(s(n+c))_{n\geq 0}$. In this paper, we present a class of exponential sequences whose discriminators are shift-invariant.

We define this class of exponential sequences as follows:

[TABLE]

for odd positive integers $a$ and $t$, where $b$ is the smallest positive integer such that $t\not\equiv\pm 1\ \mbox{{\rm(mod}}\ 2^{b}\mbox{\rm)}$. A typical example is the sequence $\left(\frac{9^{n}-1}{8}\right)_{n\geq 0}$. We show that the discriminator for all sequences of this form is $D_{\operatorname{ex}}(n)=2^{\lceil\log_{2}n\rceil}$. Furthermore, we show that this discriminator is shift-invariant, i.e., it applies to every sequence $(\operatorname{ex}(n+c))_{n\geq 0}$ for $c\geq 0$.

The outline of the paper is as follows. In Section 2 we obtain an upper bound for the discriminator of $\left(\frac{(t^{2})^{n}-1}{2^{b}}\right)_{n\geq 0}$ and all of its shifts. In Section 3 we prove some lemmas that are essential to our lower bound proof. Finally, in Section 4 we put the results together to determine the discriminator for $\left(a\frac{(t^{2})^{n}-1}{2^{b}}\right)_{n\geq 0}$ and all of its shifts.

2 Upper bound

In this section, we derive an upper bound for the discriminator of the sequence $\left(\frac{(t^{2})^{n}-1}{2^{b}}\right)_{n\geq 0}$ and all of its shifts. We start with some useful lemmas.

Lemma 1**.**

Let $t$ be an odd integer, and let $b$ be the smallest positive integer such that $t\not\equiv\pm 1\ \mbox{{\rm(mod}}\ 2^{b}\mbox{\rm)}$. Then $t^{2}\equiv 2^{b}+1\ \mbox{{\rm(mod}}\ 2^{b+1}\mbox{\rm)}$.

Proof.

Note that since every odd integer equals $\pm 1$ modulo 4, we must have $b\geq 3$. From the definition of $b$, we have $t\equiv 2^{b-1}\pm 1\ \mbox{{\rm(mod}}\ 2^{b}\mbox{\rm)}$. Hence $t=2^{b}c+2^{b-1}\pm 1$ for some integer $c$. By squaring both sides of the equation, we get

[TABLE]

∎

Lemma 2**.**

Let $t$ be an odd integer, and let $b$ be the smallest positive integer such that $t\not\equiv\pm 1\ \mbox{{\rm(mod}}\ 2^{b}\mbox{\rm)}$. Then we have

[TABLE]

for all integers $k\geq 1$.

Proof.

By induction on $k$.

Base case:

From Lemma 1, we have $t^{2}\equiv 2^{b}+1\ \mbox{{\rm(mod}}\ 2^{b+1}\mbox{\rm)}$.

Induction:

Suppose Eq. (1) holds for some $k\geq 1$, i.e., $t^{2^{k}}\equiv 2^{k+b-1}+1\ \mbox{{\rm(mod}}\ 2^{k+b}\mbox{\rm)}$. This means that $t^{2^{k}}=2^{k+b}c+2^{k+b-1}+1$ for some integer $c$. Once again, by squaring both sides of the equation, we get

[TABLE]

This shows that Eq. (1) holds for $k+1$ as well, thus completing the induction.

∎

This gives the following corollary.

Corollary 3**.**

Let $t$ be an odd integer, and let $b$ be the smallest positive integer such that $t\not\equiv\pm 1\ \mbox{{\rm(mod}}\ 2^{b}\mbox{\rm)}$. Then for $k\geq 1$, the powers of $t^{2}$ form a cyclic subgroup of order $2^{k}$ in $({\mathbb{Z}}/2^{k+b})^{*}$.

Proof.

Let $\ell=k+1$. Since $\ell\geq 1$, we can apply Eq. (1) to get

[TABLE]

Furthermore, by applying Eq. (1) directly, we get

[TABLE]

Therefore, the order of the subgroup generated by $t^{2}$ in $({\mathbb{Z}}/2^{k+b})^{*}$ is $2^{k}$. ∎

Lemma 4**.**

Let $t$ be an odd integer, and let $b$ be the smallest positive integer such that $t\not\equiv\pm 1\ \mbox{{\rm(mod}}\ 2^{b}\mbox{\rm)}$. Then for $k\geq 0$, the number $2^{k}$ discriminates every set of $2^{k}$ consecutive terms of the sequence $\left(\frac{(t^{2})^{n}-1}{2^{b}}\right)_{n\geq 0}$.

Proof.

For every $i\geq 0$, it follows from Corollary 3 that the numbers

[TABLE]

are distinct modulo $2^{k+b}$. By subtracting 1 from every element, we have that the numbers

[TABLE]

are distinct modulo $2^{k+b}$. Furthermore, these numbers are also congruent to 0 modulo $2^{b}$ because $t^{2}\equiv 1\ \mbox{{\rm(mod}}\ 2^{b}\mbox{\rm)}$ from Lemma 1. It follows that the set of quotients

[TABLE]

consists of integers that are distinct modulo $\frac{2^{k+b}}{2^{b}}=2^{k}$.

Such a set of quotients coincides with every set of $2^{k}$ consecutive terms of the sequence $\left(\frac{(t^{2})^{n}-1}{2^{b}}\right)_{n\geq 0}$. Since the numbers in each set are distinct modulo $2^{k}$, the desired result follows. ∎

3 Lower bound

In this section, we establish some results useful for the lower bound on the discriminator of the sequence $\left(\frac{(t^{2})^{n}-1}{2^{b}}\right)_{n\geq 0}$. We start with an easy technical lemma, whose proof is omitted.

Lemma 5**.**

Let $m$ be a positive integer. Then $\log_{3}m\leq\frac{m}{3}$.

The main lemma for proving the lower bound is as follows:

Lemma 6**.**

Let $t$ be an odd integer, and let $b$ be the smallest positive integer such that $t\not\equiv\pm 1\ \mbox{{\rm(mod}}\ 2^{b}\mbox{\rm)}$. Then for all $k\geq 0$ and $1\leq m\leq 2^{k+1}$, there exists a pair of integers, $i$ and $j$, where $0\leq i<j\leq 2^{k}$, such that $(t^{2})^{i}\equiv(t^{2})^{j}\ \mbox{{\rm(mod}}\ 2^{b}m\mbox{\rm)}$.

Proof.

Let the prime factorization of $m$ be

[TABLE]

where $u,v,x,y_{\ell},z_{\ell}\geq 0$, while $p_{1},p_{2},\ldots,p_{u}$ are the prime factors of $m$ that also divide $t$, and $q_{1},q_{2},\ldots,q_{v}$ are the odd prime factors of $m$ that do not divide $t$. For each $\ell\leq u$, let $e_{\ell}$ be the integer such that $p_{\ell}^{e_{\ell}}||t$, i.e., we have $p_{\ell}^{e_{\ell}}|t$ but $p_{\ell}^{e_{\ell}+1}\nmid t$.

We need to find a pair $(i,j)$ such that $(t^{2})^{i}\equiv(t^{2})^{j}\ \mbox{{\rm(mod}}\ 2^{b}m\mbox{\rm)}$. From the Chinese remainder theorem, we know it suffices to find a pair $(i,j)$ such that

[TABLE]

For the first of these equations, we know from Corollary 3 that $(t^{2})^{i}\equiv(t^{2})^{i+2^{x}}\ \mbox{{\rm(mod}}\ 2^{x+b}\mbox{\rm)}$. In other words, it suffices to have $2^{x}|(j-i)$ to satisfy $(t^{2})^{i}\equiv(t^{2})^{j}\ \mbox{{\rm(mod}}\ 2^{x+b}\mbox{\rm)}$.

Next, we consider the $u$ equations of the form $(t^{2})^{i}\equiv(t^{2})^{j}\ \mbox{{\rm(mod}}\ p_{\ell}^{y_{\ell}}\mbox{\rm)}$. Since $p_{\ell}^{e_{\ell}}$ is a factor of $t$, it follows that $(t^{2})^{y_{\ell}/2e_{\ell}}$ is a multiple of $(p_{\ell}^{2e_{\ell}})^{y_{\ell}/2e_{\ell}}=p_{\ell}^{y_{\ell}}$. Therefore, $(t^{2})^{y_{\ell}/2e_{\ell}}\equiv 0\ \mbox{{\rm(mod}}\ p_{\ell}^{y_{\ell}}\mbox{\rm)}$. Any further multiplication by $t^{2}$ also yields 0 modulo $p_{\ell}^{y_{\ell}}$. Thus, it suffices to have $j>i\geq\frac{y_{\ell}}{2e_{\ell}}$ in order to ensure that $(t^{2})^{i}\equiv(t^{2})^{j}\ \mbox{{\rm(mod}}\ p_{\ell}^{y_{\ell}}\mbox{\rm)}$.

Finally, there are $v$ equations of the form $(t^{2})^{i}\equiv(t^{2})^{j}\ \mbox{{\rm(mod}}\ q_{\ell}^{z_{\ell}}\mbox{\rm)}$. In each case, $q_{\ell}$ is co-prime to $t$, which means that $(t^{2})^{\varphi(q_{\ell}^{z_{\ell}})/2}=t^{\varphi(q_{\ell}^{z_{\ell}})}\equiv 1\ \mbox{{\rm(mod}}\ q_{\ell}^{z_{\ell}}\mbox{\rm)}$, where $\varphi(n)$ is Euler’s totient function. Now $\frac{\varphi(q_{\ell}^{z_{\ell}})}{2}=\frac{q_{\ell}^{z_{\ell}-1}(q_{\ell}-1)}{2}$. Thus, it is sufficient to have $\frac{q_{\ell}^{z_{\ell}-1}(q_{\ell}-1)}{2}|(j-i)$ in order to ensure that $(t^{2})^{i}\equiv(t^{2})^{j}\ \mbox{{\rm(mod}}\ q_{\ell}^{z_{\ell}}\mbox{\rm)}$.

Merging these ideas together, we choose the following values for $i$ and $j$:

[TABLE]

to ensure that $(t^{2})^{i}\equiv(t^{2})^{j}\ \mbox{{\rm(mod}}\ 2^{b}m\mbox{\rm)}$. It is clear that $0\leq i<j$. In order to show that $j\leq 2^{k}$, we first observe that

[TABLE]

We now consider the following two cases:

Case 1: $u=0$.

If $v=0$ as well, then $j=2^{x}=m<2^{k+1}$, which means that $x\leq k$ and thus $j\leq 2^{k}$. Otherwise, if $v\geq 1$, then we have

[TABLE]

Case 2: $u\geq 1$.

Let $r$ be such that $y_{r}=\max_{1\leq\ell\leq u}y_{\ell}$, and thus, $p_{r}$ is the corresponding prime number with exponent $y_{r}$. Since $p_{r}^{y_{r}}\geq p_{r}\geq 3$, we have

[TABLE]

Note that $y_{r}\leq\log_{p_{r}}m\leq\log_{3}m\leq\frac{m}{3}$ from Lemma 5, which means that

[TABLE]

Since both $m$ and $j$ are integers, this implies that

[TABLE]

In both cases, we have $j\leq 2^{k}$, thus fulfilling the required conditions. ∎

4 Discriminator of $(\operatorname{ex}(n))_{n\geq 0}$ and its shifted counterparts

In this section, we combine the results of the previous sections to determine the discriminator for $(\operatorname{ex}(n))_{n\geq 1}$, as well as its shifted counterparts. We first prove a general lemma about the discriminator of some scaled sequences.

Lemma 7**.**

Given a sequence $s(0),s(1),\ldots,$ and a non-zero integer $a$, let $s^{\prime}(0),s^{\prime}(1),\ldots,$ denote the sequence such that $s^{\prime}(i)=as(i)$ for all $i\geq 0$. Then, for every $n$ such that $\gcd(|a|,D_{s}(n))=1$, we have $D_{s^{\prime}}(n)=D_{s}(n)$.

Proof.

From the definition of the discriminator, we know that for every $m<D_{s}(n)$, there exists a pair of integers $i$ and $j$ with $i<j<n$, such that $m|s(j)-s(i)$. Thus, for this same pair of $i$ and $j$, we have

[TABLE]

Therefore, $m$ cannot discriminate the set $\{s^{\prime}(0),s^{\prime}(1),\ldots,s^{\prime}(n-1)\}$ and so $D_{s^{\prime}}(n)\geq D_{s}(n)$.

But for $m=D_{s}(n)$, we know that for all $i$ and $j$ with $i<j<n$, we have $m\nmid s(j)-s(i)$. Since $\gcd(m,|a|)=1$, it follows that

[TABLE]

for all $i$ and $j$ with $i<j<n$. Therefore, $m=D_{s}(n)$ discriminates the set

[TABLE]

and so $D_{s^{\prime}}(n)\leq D_{s}(n)$.

Putting these results together, we have $D_{s^{\prime}}(n)=D_{s}(n)$. ∎

We now compute the discriminator for $(\operatorname{ex}(n))_{n\geq 0}=\left(a\frac{(t^{2})^{n}-1}{2^{b}}\right)_{n\geq 0}$, and also for its shifted counterparts, which we denote by $(\operatorname{exs}(n,c))_{n\geq 0}=(\operatorname{ex}(n+c))_{n\geq 0}$ for some integer $c\geq 0$.

Theorem 8**.**

Let $t$, $a$, $b$, and $c$ be integers such that $a$ and $t$ are odd, $c\geq 0$, and let $b$ be the smallest integer such that $t\not\equiv\pm 1\ \mbox{{\rm(mod}}\ 2^{b}\mbox{\rm)}$. Then the discriminator for the sequence $(\operatorname{exs}(n,c))_{n\geq 0}=\left(a\frac{(t^{2})^{n+c}-1}{2^{b}}\right)_{n\geq 0}$ is

[TABLE]

Proof.

First we compute the discriminator for $a=1$, where the sequence is of the form $(\operatorname{exs}(n))_{n\geq 0}=\left(\frac{(t^{2})^{n+c}-1}{2^{b}}\right)_{n\geq 0}$.

The case for $n=1$ is trivial. Otherwise, let $k\geq 0$ be such that $2^{k}<n\leq 2^{k+1}$. We show that $D_{\operatorname{exs}}(n)=2^{k+1}$.

From Lemma 4, we know that $2^{k+1}$ discriminates the set,

[TABLE]

as well as every smaller subset of these numbers. Therefore, $2^{k+1}$ discriminates

[TABLE]

In other words, $D_{\operatorname{exs}}(n)\leq 2^{k+1}$.

Now let $m$ be a positive integer such that $m<2^{k+1}$. By Lemma 6, we know that there exists a pair of integers, $i$ and $j$, such that

[TABLE]

Note that since $(t^{2})\equiv 1\ \mbox{{\rm(mod}}\ 2^{b}\mbox{\rm)}$ from Lemma 1, we have $(t^{2})^{i+c}-1\equiv(t^{2})^{j+c}-1\equiv 1-1\equiv 0\ \mbox{{\rm(mod}}\ 2^{b}\mbox{\rm)}$. Therefore,

[TABLE]

In other words, $\operatorname{exs}(i,c)\equiv\operatorname{exs}(j,c)\ \mbox{{\rm(mod}}\ m\mbox{\rm)}$ while both numbers are in the set

[TABLE]

since $i<j\leq 2^{k}<n$. Therefore, $m$ fails to discriminate this set. Since this applies for all $m<2^{k+1}$, we have $D_{\operatorname{exs}}(n)\geq 2^{k+1}$.

Since we have $2^{k+1}\leq D_{\operatorname{exs}}\leq 2^{k+1}$, this means that $D_{\operatorname{exs}}(n)=2^{k+1}$ and thus $D_{\operatorname{exs}}(n)=2^{\lceil\log_{2}n\rceil}$, provided that $a=1$.

Even for $a\neq 1$, we observe that the value of $2^{\lceil\log_{2}n\rceil}$ is a power of 2 for all $n$, and so it is co-prime to all odd $a$. Therefore, we can apply Lemma 7 to prove that the discriminator remains unchanged for odd values of $a$, thus proving that the discriminator for the sequence, $(\operatorname{exs}(n,c))_{n\geq 0}=\left(a\frac{(t^{2})^{n+c}-1}{2^{b}}\right)_{n\geq 0}$ is $D_{\operatorname{exs}}(n)=2^{\lceil\log_{2}n\rceil}$. ∎

5 Final remarks

We have considered sequences of the form $(\operatorname{ex}(n))_{n\geq 0}=\left(a\frac{(t^{2})^{n}-1}{2^{b}}\right)_{n\geq 0}$ for odd integers $a$ and $t$, where $b$ is the smallest positive integer such that $t\not\equiv\pm 1\ \mbox{{\rm(mod}}\ 2^{b}\mbox{\rm)}$. We showed that the discriminator for this sequence is characterized by $D_{\operatorname{ex}}(n)=2^{\lceil\log_{2}n\rceil}$ and that the discriminator is shift-invariant, i.e., all sequences of the form $(\operatorname{ex}(n+c))_{n\geq 0}$ for $c\geq 0$ share the same discriminator.

This raises the question of what other sequences have shift-invariant discriminators. It is easy to show that sequences defined by a linear equation, i.e. of the form $(an+b)_{n\geq 0}$, have shift-invariant discriminators. Furthermore, the first author has recently shown [4] that the sequence $(2^{k}cn^{2}+bcn)_{n\geq 0}$, for a positive integer $k$ and odd integers $b,c$, also has a shift-invariant discriminator $2^{\lceil\log_{2}n\rceil}$.

It is an open problem as to whether there are any sequences, other than those mentioned here, whose discriminators are shift-invariant. Futhermore, all sequences whose discriminators are known to be shift-invariant have discriminators with linear growth, but we do not know if this is true of all shift-invariant discriminators.

6 Acknowledgments

We are grateful to Pieter Moree for introducing us to this interesting topic of discriminators. He also suggested the idea of generalizing $t$ to be any positive odd integer, thus broadening the class of exponential sequences presented in this paper.