Improved Bounds on Diffsequences with Gaps in Powers of 2

TL;DR

This paper establishes a new lower bound on the minimum length needed to guarantee a monochromatic diffsequence with gaps in powers of two in 2-colorings, improving previous exponential bounds.

Contribution

It introduces an improved asymptotic lower bound for the minimal length of monochromatic diffsequences with gaps in powers of two, advancing prior exponential bounds.

Findings

New lower bound for (D,k) with asymptotic improvement

Enhanced understanding of diffsequence coloring thresholds

Refined exponential growth rate in bounds

Abstract

Let $D$ be a set of positive integers. A $D$-diffsequence of length $k$ is a sequence of positive integers $a_1 < \cdots < a_k$ such that $a_{i+1}-a_i\in D$ for $i=1,\ldots,k-1$. For $D=\{2^i\mid i\in \mathbb{Z}_{\ge 0}\}$, it is known that there exists a minimum integer $n$, denoted by $\Delta(D,k)$, such that every $2$-coloring of $\{1,\ldots n \}$ admits a monochromatic $D$-diffsequence of length $k$. In this work, we prove a new lower bound for $\Delta(D,k)$ to $\Delta(D,k)\ge \left(\sqrt{\frac{8k-5}{12}}-\frac12\right)2^{\left(\sqrt{\frac{8k-5}{3}}-3\right)}$, asymptotically improving the exponential constant in the bound proved by Clifton.

Equations

$$\Delta (D,k)\ge 2^{\sqrt{2k}}\left(\frac{(\sqrt{2}-1)k}{8}-\frac{\sqrt{k}}{8}\right)+\frac{\sqrt{k}}{2}.$$

$$\Delta (D,k)\ge \left(\sqrt{\frac{8k-5}{12}}-\frac{1}{2}\right)2^{\left(\sqrt{\frac{8k-5}{3}}-3\right)}.$$

$${\chi }^{(1)}(4i-j)=\{\begin{array}{ll}\chi (i),& \text{if }j\in \{2,3\},\\ 1-\chi (i),& \text{if }j\in \{0,1\}.\end{array}$$

$${\psi }_{{\mathbf{s}}^{(1)}}(h)\le {\psi }_{\mathbf{s}}(h+1)+3h+2.$$

$${\psi }_{{\mathbf{s}}^{(l-1)}}(0)\le l+\frac{3(l-1)(l-2)}{2}+2(l-1)=\frac{3l^2-3l+2}{2}.$$

Full text

1 Introduction

An $r$-coloring of a set $A$ is a function $\chi:A\to\{0,\ldots,r-1\}$, where each element of $A$ is assigned one of $r$ distinct colors. We say a subset $A^{\prime}\subseteq A$ is monochromatic under $\chi$ if $\chi$ is constant over $A^{\prime}$. In 1927, Van der Waerden [1] showed that for any $r$-coloring of the positive integers, at least one color is such that there are monochromatic arithmetic progressions of arbitrary lengths. Equivalently, for any positive integer $k$, there exists a minimum integer $n$, such that for any $r$-coloring of $\{1,\ldots,n\}$, there must be a monochromatic arithmetic progression of length $k$. Motivated by generalizing this property to other sequences, Landman and Robertson [3] considered the following question: Does an $r$-coloring of positive integers produce a monochromatic sequence of arbitrary length, whose gaps lie in some fixed set $D$? In particular, they introduced the notion of a $D$-diffsequence.

Definition 1**.**

Given a set $D$ of positive integers, a $D$-diffsequence of length $k$ is a sequence of positive integers $a_{1}<\dots<a_{k}$ such that $a_{i+1}-a_{i}\in D$ for $i\in\{1,\dots,k-1\}$.

Definition 2**.**

A set $D$ is called $r$-accessible if every $r$-coloring of the positive integers contains arbitrarily long monochromatic $D$-diffsequences.

Several works have sought to characterize the accessibility of canonical integer sets $D$ (see [2]), such as the set of Fibonacci numbers [3, 5], primes [3], and fixed translates of primes [6], and powers of integers. Consider the set of powers of some integer $t\in\mathbb{Z}_{>0}$, i.e. $D=\{t^{i}\mid i\in\mathbb{Z}_{\geq 0}\}$. For $t\geq 3$, there is a periodic coloring modulo $t-1$ that avoids arbitrarily long $D$-diffsequences. It is known that $D$ is only $2$-accessible when $t=2$ [2]. Analogous to Van der Waerden’s inquiry, the problem of $r$-accessibility of a set $D$ can be rephrased to finding the minimum integer $n$, denoted by $\Delta(D,k;r)$, such that every $r$-coloring of $\{1,\ldots,n\}$ admits a monochromatic $D$-diffsequence of length $k$.

Definition 3**.**

Let $\Delta(D,k;r)$ be the smallest positive integer $n$, such that for any $r$-coloring of $\{1,\ldots,n\}$, we can find a $k$ length monochromatic $D$-diffsequence. Let $\Delta(D,k)=\Delta(D,k;2)$.

For the set $D$ of powers of $2$, i.e., $D=\{2^{i}\mid i\in\mathbb{Z}_{\geq 0}\}$, Landman and Robertson [3] showed that $D$ is $2$-accessible, with $\Delta(D,k)\leq 2^{k}-1$. Chokshi et. al. [7] conjectured that $\Delta(D,k)$ can be lower bounded exponentially. Clifton [4] established a lower bound which is asymptotically $\mathcal{O}(2^{\sqrt{2k}})$. In particular,

[TABLE]

In this work, we establish a new lower bound for $\Delta(D,k)$, asymptotically improving the exponential constant in Clifton’s bound from $\mathcal{O}(2^{\sqrt{2k}})$ to $\mathcal{O}(2^{\sqrt{8k/3}})$. Our main result is stated as follows.

Theorem 1**.**

For $D=\{2^{i}\mid i\in\mathbb{Z}_{\geq 0}\}$ we have

[TABLE]

2 Construction

We will begin by formally constructing the coloring function $\kappa$ used to show the bound in Theorem 1. We define a mapping from some coloring of $l$ integers to a coloring of $4l$ integers, which we call an expansion. The process of expansion is then recursively used to construct the required coloring $\kappa$. To prove the lower bound in Theorem 1, we explicitly define a $2$-coloring scheme of $\{1,\ldots,l\cdot 4^{l-1}\}$ which avoids any monochromatic diffsequences of length $k=(3l^{2}-3l+2)/2$. For a coloring $\chi\colon\{1,\ldots,l\}\to\{0,1\}$, define its expansion $\chi^{(1)}\colon\{1,\ldots,4l\}\to\{0,1\}$ by setting, for $i\in\{1,\ldots,l\}$ and $j\in[0,3]$,

[TABLE]

The coloring scheme is constructed recursively using the process of expansion. Repeating this procedure $r$ times, we eventually obtain a coloring of the first $l\cdot 2^{2r}$ positive integers, which we call $\chi^{(r)}$. To show the lower bound in Theorem 1, begin with the trivial coloring: $\chi_{\mathrm{alt}}\colon\{1,\ldots,l\}\to\{0,1\}$ such that $\chi_{\mathrm{alt}}(i)=i\!\pmod{2}$. Then, we will show that the coloring $\chi_{\mathrm{alt}}^{(l-1)}\colon[l\,\cdot\,4^{l-1}]\to\{0,1\}$ avoids monochromatic diffsequences of length $k$, i.e. $\kappa\equiv\chi_{\mathrm{alt}}^{(l-1)}$.

3 Preliminaries

In this section, we will rephrase $2$-coloring functions, and the process of expansion to their more intuitive analogue, binary strings. For positive integers $N$, by $[N]$ we mean the set $\{1,2,3,\ldots,N\}$. By a 2-coloring of $[N]$ we mean a mapping $\chi:[N]\to\{0,1\}$. For the rest of this work, we will use the notation $D$ to denote the set of all non-negative integer powers of two, i.e., $D=\{2^{i}\mid i\in\mathbb{Z}_{\geq 0}\}$. Note that any $2$-coloring of $[n]$ can be represented as a binary string $\mathbf{s}=s_{1}\ldots s_{n}$, where $s_{i}=\chi(i)$. For a string $\mathbf{s}$ corresponding to a given $2$-coloring of $[n]$, a monochromatic diffsequence $\mathcal{A}=\{a_{i}\}_{i=1}^{k}$ is such that $a_{i+1}-a_{i}\in D$ and $s_{a_{i+1}}=s_{a_{i}}$, i.e., $\chi(a_{i+1})=\chi(a_{i})$, for $i\in[k-1]$.

Definition 4**.**

Let $\mathbf{s}$ be a binary string of length $l$. The expansion of $\mathbf{s}$ is defined as a binary string $\mathbf{s}^{(1)}$ of length $4l$, obtained by replacing each [math] in $\mathbf{s}$ by $0011$ and each $1$ in $\mathbf{s}$ by $1100$. The string $\mathbf{s}^{(r)}$ is recursively defined as the expansion of $\mathbf{s}^{(r-1)}$, where $r\geq 1$ and $\mathbf{s}^{(0)}=\mathbf{s}$.

Example 1**.**

Consider the string $\mathbf{t}=1011$, its expansion is $\mathbf{t}^{(1)}=\overbrace{1100}_{\phantom{.}}\,\overbrace{0011}_{\phantom{.}}\,\overbrace{1100}_{\phantom{.}}\,\overbrace{1100}_{\phantom{.}}$.

In Lemma 1, we will bound the length of certain diffsequences of an expanded string. Therefore, we wish to understand how diffsequences behave under expansion. Let $\mathbf{s}^{(1)}$ be the expansion of some string $\mathbf{s}$. The process of expansion is such that the $i^{th}$ bit of $\mathbf{s}^{(1)}$ is determined uniquely by the $\lceil i/4\rceil^{th}$ bit of the string $\mathbf{s}$. This lets us determine the block, either $0011$ or $1100$, to which the $i^{th}$ bit of $\mathbf{s}^{(1)}$ belongs. For convenience, we call this mapping the position of $i$.

Definition 5**.**

Let the position of $i\in\mathbb{Z}_{+}$ be defined as the function $\text{pos}(i)=\left\lceil i/4\right\rceil$.

The choice of $0011$ and $1100$ as the types of blocks used in expansion is motivated by the following two reasons: a) The length of blocks being $4$ implies that for any two bit indices $i$ and $i^{\prime}$ in the expanded string $\mathbf{s}^{(1)}$ that differ by a power of $2$, i.e. $i^{\prime}-i\in D$, their corresponding positions in the string $\mathbf{s}$ are such that $\mathrm{pos}(i^{\prime})-\mathrm{pos}(i)\in D\,\cup\,\{0\}$; and b) For a monochromatic diffsequence $a_{1}<\ldots<a_{k}$ in the expanded string which is such that some of its consecutive elements lie in different blocks i.e. $s_{pos(a_{j+1})}\neq s_{pos(a_{j})}$, for some $j\in[k-1]$, we must have that $a_{j+1}-a_{j}\leq 2$, i.e. they must lie in consecutive blocks. This is enabled by the fact that the blocks $0011$ and $1100$ are conjugates of each other. This will let us reduce monochromatic diffsequences in the expanded string $\mathbf{s}^{(1)}$ to two contiguous diffsequences of opposite colors in $\mathbf{s}$, partitioned by a hop. A hop is defined to be an index in a diffsequence, at which the assigned color of the diffsequence changes.

Definition 6**.**

Let $\mathbf{s}\in\{0,1\}^{n}$ and $\mathcal{A}=\{a_{i}\}_{i=1}^{k}$ a $D$-diffsequence. A hop is defined as an index $j\in[k-1]$ such that $a_{j+1}-a_{j}=1$ and $s_{a_{j+1}}\neq s_{a_{j}}$. Define $H_{\mathcal{A}}$ to be the set of all hops in $\mathcal{A}$.

Definition 7**.**

For a string $\mathbf{s}\in\{0,1\}^{n}$, define $\Psi_{\mathbf{s}}(h)$ to be the set of all diffsequences with at most $h$ hops. Formally, $\Psi_{\mathbf{s}}(h):=\{\mathcal{A}=\{a_{i}\}_{i=1}^{k}:\mathcal{A}\text{ is a }D\text{-diffsequence with }|H_{\mathcal{A}}|\leq h\}$. Let $\psi_{\mathbf{s}}(h)$ denote the maximum length of any diffsequence in $\Psi_{\mathbf{s}}(h)$, and $\psi_{\mathbf{s}}(h)=0$ if $\Psi_{\mathbf{s}}(h)=\emptyset$.

Remark 1**.**

$\Psi_{\mathbf{s}}(0)$* is the set of all monochromatic diffsequences of $\mathbf{s}$. Therefore, if the length of $\mathbf{s}$ is greater than $\Delta(D,k)$, the maximum length of elements of $\Psi_{\mathbf{s}}(0)$ is at least $k$.*

Example 2**.**

Consider the string $\mathbf{s}=1100001100111100$ of length $16$. The sequence of integers $\mathcal{A}=\{1,2,3,5,9\}$ is an example of a $D$-diffsequence. The corresponding subsequence of $\mathbf{s}$ is $s_{1}s_{2}s_{3}s_{5}s_{9}=11000$, and $H_{\mathcal{A}}=\{2\}$. Since this is the only hop in the diffsequence, $\mathcal{A}\in\Psi_{\mathbf{s}}(1)$.

The following proposition and its corollaries give us a recipe to reduce diffsequences with $h$ hops in some expanded string $\mathbf{s}^{(1)}$, to diffsequences with $h+1$ hops in $\mathbf{s}$. Proposition 1 shows that in the expanded string $\mathbf{s}^{(1)}$, if two positions $\{x,y\}$ of some diffsequence have the same color $c\in\{0,1\}$, and their corresponding positions $\{\text{pos}(x),\text{pos}(y)\}$ in $\mathbf{s}$ have different colors, then $\text{pos}(y)=\text{pos}(x)+1$ i.e. $x$ and $y$ lie in blocks next to each other. Moreover, we must have $s_{\text{pos}(x)}=1-c$, and $s_{\text{pos}(y)}=c$.

Proposition 1**.**

For a binary string $\mathbf{s}$, let $x,y$ be positive integers such that $y-x\in D$ and $s^{(1)}_{x}=s^{(1)}_{y}$. If $s_{\text{pos}(x)}\neq s_{\text{pos}(y)}$, we must have $s_{\text{pos}(x)}=1-s^{(1)}_{x}$, $s_{\text{pos}(y)}=s^{(1)}_{y}$, and $\text{pos}(y)=\text{pos}(x)+1$.

Proof.

The condition $s_{\text{pos}(x)}\neq s_{\text{pos}(y)}$ means that $s_{x}^{(1)}$ and $s_{y}^{(1)}$ belong to different types of blocks, $0011$ and $1100$. Without loss of generality, assume that $s_{x}^{(1)}=s_{y}^{(1)}$ = 1. The proof is divided into two cases according to the value of $y-x$. (a) For $y-x\in\{1,2\}$, we have $\text{pos}(y)-\text{pos}(x)\in\{0,1\}$ which means that $s_{x}^{(1)}$ and $s_{y}^{(1)}$ belong either to the same block or consecutive blocks. However, since $s_{x}^{(1)}$ and $s_{y}^{(1)}$ must belong to different types of blocks, these must be consecutive. Therefore $\text{pos}(y)-\text{pos}(x)=1$. The condition $s_{x}^{(1)}=s_{y}^{(1)}=1$ is only possible when $s_{x}^{(1)}\in\{0011\}$, and $s_{y}^{(1)}\in\{1100\}$, which implies $s_{\text{pos}(x)}=0$ and $s_{\text{pos}(y)}=1$. (b) For $y-x=2^{m}$, $x\equiv y\!\pmod{4}$ and therefore, $s_{x}^{(1)}$ and $s_{y}^{(1)}$ occupy the same position within their respective blocks. Since the two types of blocks, $0011$ and $1100$ are conjugates of each other, $s^{(1)}_{x}=s^{(1)}_{y}$ is only possible when the blocks are identical. This would mean that $s_{\text{pos}(x)}$ and $s_{\text{pos}(y)}$ are identical, which is a contradiction. ∎

Let $x_{1}<\dots<x_{k}$ be a monochromatic diffsequence of color $c\in\{0,1\}$ in the string $\mathbf{s}^{(1)}$, which is the expansion of some string $\mathbf{s}$. As noted earlier, $y-x\in D$, implies $\text{pos}(y)-\text{pos}(x)\in D\ \cup\{0\}$; the sequence characterized by the set $\{\text{pos}(x_{i})\}_{i=1}^{k}$ is a $D$-diffsquence in $\mathbf{s}$. A consequence of Proposition 1 is that the substring $s_{\text{pos}(x_{i})}$ can be split into two contiguous monochromatic substrings, the first with the color $1-c$ and the other with $c$. This result is proved formally in Corollary 1. Therefore, a monochromatic diffsequence in $\mathbf{s}^{(1)}$ can be mapped to a diffsequence with one switch in color in $\mathbf{s}$, i.e., with one hop (Definition 6). This is a mapping between elements of $\Psi_{\mathbf{s^{(1)}}}(0)$ to elements of $\Psi_{\mathbf{s}}(1)$. In Corollary 2, we will establish that the set $\{\text{pos}(x_{i})\}_{i=1}^{k}$ has at least $k-1$ distinct elements under certain conditions. These two corollaries together help in bounding $\psi_{\mathbf{s^{(1)}}}(0)$ in terms of $\psi_{\mathbf{s}}(1)$ (Definition 7) as $\psi_{\mathbf{s^{(1)}}}(0)\leq\psi_{\mathbf{s}}(1)+2$.

Corollary 1**.**

Let $\mathbf{s}^{(1)}$ be an expansion of a binary string $\mathbf{s}$ and $x_{1}<\dots<x_{k}$ be a monochromatic $D$-diffsequence in $\mathbf{s}^{(1)}$ with color $c\in\{0,1\}$. Then, there exists a non negative integer $m\leq k$ such that $s_{\text{pos}(x_{j})}=1-c$ if and only if $j\leq m$.

Proof.

From Proposition 1 we have the following possibilities for each $i\leq k-1$: $s_{\text{pos}(x_{i})}=s_{\text{pos}(x_{i+1})}$ or $s_{\text{pos}(x_{i})}=1-c$ and $s_{\text{pos}(x_{i+1})}=c$. Now, if there exists an index $j$ for which $s_{\text{pos}(x_{j})}=1-c$ and $s_{\text{pos}(x_{j+1})}=c$ (if not, then we are already done), then pick the smallest such $j$. Thus, $s_{\text{pos}(x_{i})}=1-c$ for all $i\leq j$. We must have $s_{\text{pos}(x_{i})}=c$ for all $i>j$, as otherwise there would exist an index $j^{\prime}$ such that $s_{\text{pos}(x_{j^{\prime}})}=c$ and $s_{\text{pos}(x_{j^{\prime}+1})}=1-c$, which is not possible. Hence, proving our claim. ∎

Corollary 2**.**

Let $\mathbf{s}^{(1)}$ be an expansion of a binary string $\mathbf{s}$ and $x_{1}<\dots<x_{k}$ be a monochromatic $D$-diffsequence in $\mathbf{s}^{(1)}$ with color $c\in\{0,1\}$. Suppose that there exists $d\in\{0,1\}$ such that $s_{\text{pos}(x_{i})}=d$ for all $1\leq i\leq k$. Then the set $\{\text{pos}(x_{i})\}^{k}_{i=1}$ has at least $k-1$ distinct elements.

Proof.

Without loss of generality, assume $c=1$. The proofs for $d=0$ and $d=1$ are analogous; therefore, we will further assume $d=0$. Since $\{\text{pos}(x_{i})\}^{k}_{i=1}$ is a non-decreasing sequence, to prove it has at least $k-1$ distinct elements, it suffices to show that $\text{pos}(x_{j})=\text{pos}(x_{j+1})$ is true for at most one index. If it exists, let $j$ be such an index. For each $1\leq i\leq k$, the conditions $s^{(1)}_{x_{i}}=1$ and $s_{\text{pos}(x_{i})}=0$ imply that $s^{(1)}_{x_{i}}$ corresponds to the third or the fourth position in the block $0011$. Furthermore, if $s^{(1)}_{x_{i}}$ corresponds to the fourth position in the block $0011$, then $s^{(1)}_{x_{i+1}}$ must also correspond to the fourth position. This is because if it corresponds to the third position, then the difference between $x_{i+1}$ and $x_{i}$ would be odd, implying it to be $1$, but this is clearly not possible. For an index $j$ to satisfy $\text{pos}(x_{j})=\text{pos}(x_{j+1})$, we must have $s^{(1)}_{x_{j}}$ and $s^{(1)}_{x_{j+1}}$ correspond to the third and fourth position in the block $0011$ respectively. This means that all indices at least $j+1$ must correspond to the fourth position in the block, thus enforcing the impossibility of another index $i$ satisfying $\text{pos}(x_{i})=\text{pos}(x_{i+1})$. ∎

Corollaries 1 and 2 show that the process of expansion naturally induces a mapping from monochromatic diffsequences in an expanded string and diffsequences with one hop in the original string. Motivated by this approach, in Lemma 1, we inductively extend this reasoning to obtain a mapping from elements in $\Psi_{\mathbf{s}^{(1)}}(h)$ to $\Psi_{\mathbf{s}}(h+1)$ for some string $\mathbf{s}$. This will help us obtain a bound on the maximum length of a monochromatic diffsequence in our explicit construction.

4 Main result

To prove the main result of this work, Theorem 1, we explicitly need to show that the construction in Section 2 does not admit any monochromatic $D$-diffsequences of length $k$. For any string $\mathbf{s}$ and its expansion $\mathbf{s}^{(1)}$, Lemma 1 gives a bound on the difference between the maximum length of a diffsequence with at most $h+1$ hops in $\mathbf{s}$, $\psi_{\mathbf{s}}(h+1)$, and the maximum length of a diffsequence with $h$ hops in the expanded string $\mathbf{s}^{(1)}$, $\psi_{\mathbf{s}^{(1)}}(h)$ (Definition 7). Formally, for any binary string $\mathbf{s}$, Lemma 1 shows that $\psi_{\mathbf{s}^{(1)}}(h)\leq\psi_{\mathbf{s}}(h+1)+3h+2$. Recall that in our explicit construction, we begin with the initial coloring $\chi_{\text{alt}}\colon[l]\to\{0,1\}$ such that $\chi_{\text{alt}}(i)=i\!\pmod{2}$. This corresponds to the binary string $\mathbf{s}=1010\ldots$ of length $l$. This string is such that the diffsequence $\{i\}_{i=1}^{l}$ has $l-1$ hops, i.e. $\{i\}_{i=1}^{l}\in\Psi_{\mathbf{s}}(l-1)$ and $\psi_{\mathbf{s}}(l-1)=l$. Therefore, using Lemma 1 recursively, we can show that $\psi_{\mathbf{s}^{(l-1)}}(0)\leq(3l^{2}-3l+2)/2$. That is, we show that the maximum possible length of a monochromatic diffsequence in $\mathbf{s}^{(l-1)}$ is at most $(3l^{2}-3l+2)/2$. The proof of Lemma 1 uses induction on the number of hops, exploiting the fact that diffsequences with $h$ hops, i.e., diffsequences in $\Psi_{\mathbf{s}^{(1)}}(h)$, are a concatenation of diffsequences with $h-1$ hops and monochromatic diffsequences. For an expanded string, we can therefore inductively apply Corollaries 1 and 2.

Lemma 1**.**

Let $\mathbf{s}$ be a binary string and let $\mathbf{s}^{(1)}$ denote the expansion of $\mathbf{s}$. Then

[TABLE]

Proof.

Using induction on $h$, we will show that for any diffsequence in $\Psi_{\mathbf{s^{(1)}}}(h)$ of length $\ell$, there is a corresponding diffsequence in $\Psi_{\mathbf{s}}(h+1)$ with length at least $\ell-3h-2$. To prove the base case, $h=0$, let $\mathcal{P}_{0}=\{p_{i}\}_{i=1}^{\ell}\in\Psi_{\mathbf{s}^{(1)}}(0)$ be a monochromatic diffsequence in $\mathbf{s}^{(1)}$ of length $\ell$. Without loss of generality, assume that $\mathcal{P}_{0}$ has color $c=1$ in $\mathbf{s}^{(1)}$, i.e., $s^{(1)}_{p_{i}}=1$, for $1\leq i\leq\ell$. Now, consider the diffsequence $\text{pos}(\mathcal{P}_{0})=\{\text{pos}(p_{i})\}_{i=1}^{\ell}$ in $\mathbf{s}$. By Corollary 1, we know that the substring $\{s_{\text{pos}(x_{i})}\}_{i=1}^{\ell}$ can be split into at most two contiguous monochromatic substrings, the first with the color $1-c=0$ and the other with $c=1$. Let $\mathcal{P}_{\text{left}}$ and $\mathcal{P}_{\text{right}}$ be (possibly empty) subsequences of $\mathcal{P}_{0}$ such that $\text{pos}(\mathcal{P}_{\text{left}})$ has color [math] in $\mathbf{s}$, and $\text{pos}(\mathcal{P}_{\text{right}})$ has color $1$ in $\mathbf{s}$. Using Corollary 2, note that $|\text{pos}(\mathcal{P}_{\text{left}})|\geq|\mathcal{P}_{\text{left}}|-1$ and $|\text{pos}(\mathcal{P}_{\text{right}})|\geq|\mathcal{P}_{\text{right}}|-1$. Because $\text{pos}(\mathcal{P}_{0})$ is the union of $\text{pos}(\mathcal{P}_{\text{left}})$ and $\text{pos}(\mathcal{P}_{\text{right}})$, and both $\text{pos}(\mathcal{P}_{\text{left}})$ and $\text{pos}(\mathcal{P}_{\text{right}})$ are disjoint, we get that $|\text{pos}(\mathcal{P}_{0})|=|\text{pos}(\mathcal{P}_{\text{left}})|+|\text{pos}(\mathcal{P}_{\text{right}})|\geq|\mathcal{P}_{\text{left}}|-1+|\mathcal{P}_{\text{left}}|-1=|\mathcal{P}|-2$.

To prove the inductive hypothesis, assume that the result is true for all $k\leq h-1$. Let $\mathcal{P}_{h}=\{p_{i}\}^{\ell}_{i=1}$ be a diffsequence in $\mathbf{s}^{(1)}$ with at most $h$ hops. If $\mathcal{P}_{h}$ has at most $h-1$ hops, i.e. $\mathcal{P}_{h}\in\Psi_{\mathbf{s}^{(1)}}(h-1)$, then using the inductive hypothesis for $k=h-1$, we are done. Now, let us consider the case where $\mathcal{P}_{h}$ has exactly $h\geq 1$ hops. Let $p_{i}\to p_{i+1}$ be the first time there is a hop in $\mathbf{s}^{(1)}$. Let $\mathcal{P}_{\text{left}}$ be the first $i$ elements of $\mathcal{P}_{h}$ and $\mathcal{P}_{\text{right}}$ be the remaining part. Analogous to the proof of the base case, we split $\mathcal{P}_{h}$ into two contiguous subsequences $\mathcal{P}_{\text{left}}$ and $\mathcal{P}_{\text{right}}$ such that the hop between $\mathcal{P}_{\text{left}}$ and $\mathcal{P}_{\text{right}}$ is the first hop in $\mathcal{P}_{h}$. This means that $\mathcal{P}_{\text{left}}$ is monochromatic in $\mathbf{s}^{(1)}$ and $\mathcal{P}_{\text{right}}$ has at most $h-1$ hops in $\mathbf{s}^{(1)}$. Notice that $\text{pos}(\mathcal{P}_{h})$ is the union of $\text{pos}(\mathcal{P}_{\text{left}})$ and $\text{pos}(\mathcal{P}_{\text{right}})$. Furthermore, $\text{pos}(\mathcal{P}_{\text{left}})$ and $\text{pos}(\mathcal{P}_{\text{right}})$ intersect in at most one element. Hence, we can lower bound $|\mathcal{P}_{h}|$ by $|\text{pos}(\mathcal{P}_{\text{left}})|+|\text{pos}(\mathcal{P}_{\text{right}})|-1$. As $\mathcal{P}_{\text{left}}\in\Psi_{\mathbf{s}^{(1)}}(0)$ and $\mathcal{P}_{\text{right}}\in\Psi_{\mathbf{s}^{(1)}}(h-1)$, we can apply the induction hypothesis to both $\mathcal{P}_{\text{left}}$ and $\mathcal{P}_{\text{right}}$ to obtain the following. $|\text{pos}(\mathcal{P}_{h})|\geq|\text{pos}(\mathcal{P}_{\text{left}})|+|\text{pos}(\mathcal{P}_{\text{right}})|-1\geq|(\mathcal{P}_{\text{left}})|-2+|(\mathcal{P}_{\text{right}})|-3(h-1)-2-1\geq|\mathcal{P}_{h}|-3h-2.$ ∎

Proof of Theorem 1. We now explicitly construct a coloring of length $l\cdot 4^{l-1}$ with the length of the maximum diffsequence at most $(3l^{2}-3l+2)/2$. We start with the binary string $\mathbf{s}$ of length $l$ consisting of alternating ones and zeros, i.e, the one corresponding to the coloring $\chi(i)\equiv i\pmod{2}$. This ensures $\psi_{\mathbf{s}}(l-1)=l$. By Lemma 1, $\psi_{\mathbf{s}^{(i)}}(l-1-i)\leq\psi_{\mathbf{s}^{(i-1)}}(l-1-i+1)+3(l-i-1)+2$ for all $1\leq i\leq l-1$. So, summing over all $1\leq i\leq l-1$ we get

[TABLE]

For any positive integer $k>\frac{3l^{2}-3l+2}{2}$, we obtain a coloring of length $l\cdot 4^{l-1}$ that avoids a monochromatic diffsequence of length $k$, therefore, implying that $\Delta(D,k)\geq l\cdot 4^{l-1}$. Hence, choosing the largest such $l=\left\lfloor\sqrt{\frac{2(k-1)}{3}+\frac{1}{4}}+\frac{1}{2}\right\rfloor$, we obtain $\Delta(D,k)\geq\left(\sqrt{\frac{8k-5}{12}}-\frac{1}{2}\right)2^{\left(\sqrt{\frac{8k-5}{3}}-3\right)}$.