## 1 Introduction

Given a class of graphs $${\mathcal {C}}$$ its $$\chi$$-bounding function is the function $$\chi _{\mathcal {C}}:\mathbb {N}\rightarrow \mathbb {N}\cup \{\infty \}$$ defined as

\begin{aligned} \chi _{\mathcal {C}}(n)=\sup \{\chi (G):G\in {\mathcal {C}}\text { and } \omega (G)=n\} \text {,} \end{aligned}

where $$\chi (G)$$ and $$\omega (G)$$ denote, respectively, the chromatic number and the clique number of G. A class of graphs $${\mathcal {C}}$$ is $$\chi$$-bounded if there is a function $$f:\mathbb {N}\rightarrow \mathbb {N}$$ such that $$\chi (G)\leqslant f(\omega (G))$$ for every graph $$G\in {\mathcal {C}}$$, or equivalently if $$\chi _{\mathcal {C}}(n)$$ is finite for every $$n\in \mathbb {N}$$. A class $${\mathcal {C}}$$ is polynomially $$\chi$$-bounded if such a function f can be chosen to be a polynomial. A class $${\mathcal {C}}$$ is hereditary if it is closed under taking induced subgraphs.

A well-known and fundamental open problem, due to Esperet [6], has been to decide whether every hereditary class of graphs which is $$\chi$$-bounded is polynomially $$\chi$$-bounded. We provide a negative answer to this question. More generally, we prove that $$\chi$$-bounding functions may be arbitrary, so long as they are bounded from below by a certain cubic function.

### Theorem 1

Let $$f:\mathbb {N}\rightarrow \mathbb {N}\cup \{\infty \}$$ be such that $$f(1)=1$$ and $$f(n)\geqslant \left( {\begin{array}{c}3n+1\\ 3\end{array}}\right)$$ for every $$n\geqslant 2$$. Then there exists a hereditary class of graphs $${\mathcal {G}}$$ such that $$\chi _{\mathcal {G}}(n)=f(n)$$ for every $$n\in \mathbb {N}$$.

On the other hand, $$\chi$$-bounding functions are not entirely arbitrary. For instance, Scott and Seymour [11] proved that every hereditary class of graphs $${\mathcal {C}}$$ with $$\chi _{\mathcal {C}}(2)=2$$ satisfies $$\chi _{\mathcal {C}}(n)\leqslant 2^{2^{n+1}}$$.

The proof of Theorem 1 is heavily based on the idea used by Carbonero, Hompe, Moore, and Spirkl [2] in their very recent solution to another well-known problem attributed to Esperet [12]. They proved that for every $$k\in \mathbb {N}$$, there is a graph G with $$\omega (G)=3$$ and $$\chi (G)\geqslant k$$ such that every triangle-free induced subgraph of G has chromatic number at most 4. Their proof, in turn, relies on an idea by Kierstead and Trotter [8], who proved in 1992 that the class of oriented graphs excluding a directed path on four vertices as an induced subgraph is not $$\chi$$-bounded. We further generalise the aforesaid result of Carbonero, Hompe, Moore, and Spirkl [2] to higher clique numbers. Specifically, we prove the following general bound, which we use to derive Theorem 1.

### Theorem 2

For every pair of integers n and k with $$k\geqslant n\geqslant 2$$, there exists a graph G with clique number n and chromatic number k such that every induced subgraph of G with clique number $$m<n$$ has chromatic number at most $$\left( {\begin{array}{c}3m+1\\ 3\end{array}}\right)$$.

In case that n is a prime number, we prove a better bound, which matches the bound of 4 from [2] when $$n=3$$.

### Theorem 3

For every pair of integers p and k with p a prime and $$k\geqslant p$$, there exists a graph G with clique number p and chromatic number k such that every induced subgraph of G with clique number $$m<p$$ has chromatic number at most $$\left( {\begin{array}{c}m+2\\ 3\end{array}}\right)$$.

In the first version of this paper [1], we proved a weaker version of Theorem 3 with $$\left( {\begin{array}{c}m+2\\ 3\end{array}}\right)$$ replaced by $$m^{m^2}$$. Despite the worse bound obtained, that alternative proof may still be of interest. The mere qualitative statement that for every prime p, there are graphs with clique number p and arbitrarily large chromatic number whose induced subgraphs with clique number less than p have bounded chromatic number suffices to imply the negative answer to Esperet’s question.

After [1] appeared, Girão et al. [7] proved another generalisation of the aforesaid qualitative version of Theorem 3. Namely, they proved that for every graph F with at least one edge, there are graphs of arbitrarily large chromatic number and the same clique number as F in which every F-free induced subgraph has chromatic number at most some constant $$c_F$$ depending only on F. They also showed the analogous statement where clique number is replaced by odd girth.

See [12] and [9] for recent surveys on $$\chi$$-boundedness and polynomial $$\chi$$-boundedness.

## 2 Proof

First, we show that Theorem 2 implies Theorem 1.

### Proof of Theorem 1 Assuming Theorem 2

Fix a function $$f:\mathbb {N}\rightarrow \mathbb {N}\cup \{\infty \}$$ such that $$f(1)=1$$ and $$f(n)\geqslant \left( {\begin{array}{c}3n+1\\ 3\end{array}}\right)$$ for every $$n\geqslant 2$$. By Theorem 2, for every pair of integers n and k with $$k\geqslant n\geqslant 2$$, there exists a graph $$H_{n,k}$$ with clique number n and chromatic number k such that every induced subgraph of $$H_{n,k}$$ with clique number $$m<n$$ is $$\left( {\begin{array}{c}3m+1\\ 3\end{array}}\right)$$-colourable.

We now consider two cases. If f(n) is finite, we put $${\mathcal {H}}_n=\{H_{n,f(n)}\}$$. Otherwise $$f(n)=\infty$$, and we put $${\mathcal {H}}_n=\{H_{n,k}:k\geqslant n\}$$. Finally, we let $${\mathcal {H}}=\bigcup _{n=2}^\infty {\mathcal {H}}_n$$, and we let $${\mathcal {G}}$$ be the hereditary closure of $${\mathcal {H}}$$, i.e., the family of all induced subgraphs of the graphs in $${\mathcal {H}}$$.

We now argue that $$\chi _{\mathcal {G}}(n)=f(n)$$ for all $$n\in \mathbb {N}$$. The claim holds trivially for $$n=1$$, so assume $$n\geqslant 2$$. If $$f(n)=\infty$$, then the sequence of graphs $$\{H_{n,k}:k\geqslant n\}\subseteq {\mathcal {G}}$$ all have clique number equal to n and have unbounded chromatic number, thus showing that $$\chi _{\mathcal {G}}(n)=\infty$$, as claimed. Otherwise, f(n) is finite. The graph $$H_{n,f(n)}\in {\mathcal {G}}$$ shows that $$\chi _{\mathcal {G}}(n)\geqslant f(n)$$. For the reverse inequality, let $$G\in {\mathcal {G}}$$ be such that $$\omega (G)=n$$. Then there exist integers k and $$n^*$$ with $$k\geqslant n^*\geqslant n$$ such that G is an induced subgraph of $$H_{n^*,k}\in {\mathcal {H}}$$. The unique graph of $${\mathcal {H}}$$ with clique number n is $$H_{n,f(n)}$$. So if $$n^*=n$$, then $$\chi (G)\leqslant \chi (H_{n,f(n)})=f(n)$$, and if $$n^*>n$$, then $$\chi (G)\leqslant \left( {\begin{array}{c}3n+1\\ 3\end{array}}\right)$$. Combining these inequalities, we conclude that

\begin{aligned} f(n) \leqslant \chi _{\mathcal {G}}(n) \leqslant \max \{\genfrac(){0.0pt}1{3n+1}{3},\,f(n)\} = f(n) \text {,} \end{aligned}

and the theorem follows. $$\square$$

The rest of the paper is devoted to proving Theorem 2. We begin with the following lemma.

### Lemma 4

For every positive integer k, there is a graph $$G_k$$ and an acyclic orientation of its edges with the following properties:

1. (1)

$$\chi (G_k)=k$$;

2. (2)

for every pair of vertices u and v, there is at most one directed path from u to v in $$G_k$$;

3. (3)

there is a directed path in $$G_k$$ on k vertices;

4. (4)

there is a k-colouring $$\phi$$ of $$G_k$$ such that $$\phi (u)\ne \phi (v)$$ for any two distinct vertices u and v such that there is a directed path from u to v in $$G_k$$.

Various well-known constructions of triangle-free graphs with arbitrarily high chromatic number, such as Zykov’s [13] and Tutte’s [3, 4], satisfy the condition of Lemma 4 once the edges are oriented in a way that follows naturally from the construction. See [2] and [8] for an explicit construction of the graphs $$G_k$$ with the appropriate acyclic orientations, based on Zykov’s construction. It is only implicit that the acyclic orientations of the graphs in [2] and [8] satisfy all of the properties in the conclusion of Lemma 4, so for the sake completeness we provide a proof based on Tutte’s construction.

### Proof of Lemma 4

We proceed by induction on k. The base case $$k=1$$ follows by taking a single-vertex graph as $$G_1$$. For the induction step, assume $$G_{k-1}$$ is an acyclically oriented graph satisfying conditions (1)–(4) for $$k-1$$. To construct $$G_k$$, begin with a stable set S with $$|S|=(k-1)(|V(G_{k-1})|-1)+1$$, and for every subset X of S with $$|X|=|V(G_{k-1})|$$, add an isomorphic copy $$G_X$$ of $$G_{k-1}$$ (with the same orientation as in $$G_{k-1}$$) and an arbitrary perfect matching between the vertices in X and the vertices of $$G_X$$, oriented from X to $$G_X$$. This clearly preserves acyclicity of the orientation. Since the vertices in S have in-degree zero, either every directed path is contained in some copy $$G_X$$ of $$G_{k-1}$$, or the starting vertex u is contained in S and every other vertex is contained in some copy $$G_X$$ of $$G_{k-1}$$. As every vertex in S has at most one edge to each copy $$G_X$$ of $$G_{k-1}$$, the induction hypothesis implies that condition (2) is preserved. Any directed path on $$k-1$$ vertices in $$G_X$$ extends to a directed path on k vertices in $$G_k$$ by adding a vertex from S, so (3) holds. Any colouring of the copies $$G_X$$ of $$G_{k-1}$$ with a common palette of $$k-1$$ colours extends to a k-colouring of $$G_k$$ by using a single new colour on S, which shows that $$\chi (G_k)\leqslant \chi (G_{k-1})+1$$ and condition (4) is preserved. Finally, suppose there exists a $$(k-1)$$-colouring of $$G_k$$. Then, since $$|S|>(k-1)(|V(G_{k-1})|-1)$$, there is a monochromatic set $$X\subset S$$ with $$|X|=|V(G_{k-1})|$$. Since X and $$G_X$$ are connected by a perfect matching, at most $$k-2$$ colours are used on $$G_X$$, which contradicts the fact that $$\chi (G_X)=\chi (G_{k-1})=k-1$$. Hence $$\chi (G_k)=k$$, as claimed in (1). $$\square$$

For the rest of the argument, we fix an arbitrary sequence $$(G_k)_{k\in \mathbb {N}}$$ of graphs given by Lemma 4. Now, for every pair of positive integers k and p, where p is a prime number, we construct a graph $$G_{k,p}$$ by adding edges to $$G_k$$ as follows.

Let $$\leqslant$$ be the directed reachability order of the vertices of $$G_k$$, that is, $$u\leqslant v$$ if and only if there is a (unique) directed path from u to v in $$G_k$$. Since the orientation of $$G_k$$ given by Lemma 4 is acyclic, $$\leqslant$$ is indeed a partial order. For every pair of vertices u and v in $$G_k$$ such that $$u\leqslant v$$, let d(uv) be the length of the unique directed path from u to v in $$G_k$$ (i.e., the number of edges in that path). The graph $$G_{k,p}$$ has the same vertex set as $$G_k$$ and has the set $$\{uv:u<v$$ and $$d(u,v)\not \equiv 0\pmod {p}\}$$ as the edge set. We consider each such edge uv as oriented from u to v. Since the original (oriented) edges uv of $$G_k$$ satisfy $$u<v$$ and $$d(u,v)=1$$, the graph $$G_{k,p}$$ contains $$G_k$$ as a subgraph. Furthermore, every edge of $$G_{k,p}$$ connects vertices with different colours in a k-colouring $$\phi$$ of $$G_k$$ claimed in Lemma 4. Therefore, $$\chi (G_{k,p})=k$$. Furthermore, $$G_{k,p}$$ is acyclic since $$G_k$$ is acyclic.

Next, we examine cliques in $$G_{k,p}$$ (and its induced subgraphs). Since $$G_{k,p}$$ is acyclic, every clique of $$G_{k,p}$$ induces a transitive tournament. Given a clique C of an acyclic oriented graph, we let t(C) be the unique in-degree zero vertex of the transitive tournament induced by C. We call t(C) the tail of C. Given a clique C of $$G_{k,p}$$, we let r(C) be the subset of $$\mathbb {Z}_p$$ such that $$r(C)\equiv \{d(t(C),v):v\in C\}\pmod {p}$$. We call r(C) the residue of the clique C. Note that 0 is always contained in r(C) since $$t(C)\in C$$. Furthermore, $$|C|=|r(C)|$$, otherwise there would exist two distinct vertices $$u,v\in C$$ such that $$d(t(C),u)\equiv d(t(C),v)\pmod {p}$$, and so $$d(u,v)\equiv 0\pmod {p}$$, which would contradict the fact that u and v are adjacent. This observation allows us to determine the clique number of $$G_{k,p}$$.

### Lemma 5

For every positive integer k and every prime $$p\leqslant k$$, the graph $$G_{k,p}$$ has clique number p.

### Proof

Since $$G_k$$ contains a directed path on k vertices and $$p\leqslant k$$, the graph $$G_{k,p}$$ contains a clique of size p. Conversely, if C is a clique in $$G_{k,p}$$, then $$|C|=|r(C)|\leqslant |\mathbb {Z}_p|=p$$. $$\square$$

A rotation of a subset X of $$\mathbb {Z}_p$$ is a subset of $$\mathbb {Z}_p$$ of the form $$X+a=\{x+a:x\in X\}$$ for any $$a\in \mathbb {Z}_p$$. A subset of $$\mathbb {Z}_p$$ is rooted if it contains 0. The rotation $$X+a$$ of a rooted subset X of $$\mathbb {Z}_p$$ is rooted if and only if $$-a\in X$$. Let $$\sim _p$$ be the equivalence relation on the rooted subsets of $$\mathbb {Z}_p$$ such that $$X\sim _pY$$ whenever Y is a rotation of X. Let $$[X]_p$$ denote the equivalence class of X in $$\sim _p$$. For every proper rooted subset X of $$\mathbb {Z}_p$$ (such that $$X\ne \mathbb {Z}_p$$), since p is a prime, all rotations $$X+a$$ of X with $$a\in \mathbb {Z}_p$$ are distinct. (Indeed, if $$X+a=X$$, then $$\sum _{x\in X}x\equiv \sum _{x \in X}(x+a)\equiv \sum _{x\in X}x+a\cdot |X|\pmod {p}$$, so $$a\cdot |X|\equiv 0\pmod {p}$$, which yields $$a\equiv 0\pmod {p}$$.) In particular, we have $$|[X]_p|=|X|$$. Order every equivalence class arbitrarily, and for every proper rooted subset X of $$\mathbb {Z}_p$$, let $$c(X)\in \{1,\ldots ,|X|\}$$ denote the position of X in this ordering.

### Lemma 6

For every positive integer k, every prime p, and every induced subgraph G of $$G_{k,p}$$ with clique number $$m<p$$, we have $$\chi (G)\leqslant \left( {\begin{array}{c}m+2\\ 3\end{array}}\right)$$.

### Proof

We will colour the vertices of G by triples of integers (abc) with $$m\geqslant a\geqslant b\geqslant c\geqslant 1$$. Since there are $$\left( {\begin{array}{c}m+2\\ 3\end{array}}\right)$$ choices for such a triple, this will be a $$\left( {\begin{array}{c}m+2\\ 3\end{array}}\right)$$-colouring of G.

For each vertex v of G, let a(v) be the maximum size of a clique in G with tail v. Thus $$m\geqslant a(v)\geqslant 1$$. Let B(v) be the intersection of the residues of all cliques of size a(v) with tail v in G. Since 0 belongs to the residue of every clique, we have $$0\in B(v)$$. Let $$b(v)=|B(v)|$$, so that $$a(v)\geqslant b(v)\geqslant 1$$. Let $$c(v)=c(B(v))$$, so that $$b(v)\geqslant c(v)\geqslant 1$$, as $$|[B(v)]_p|=|B(v)|=b(v)$$. Finally, let $$\psi (v)=(a(v),b(v),c(v))$$. We have $$m\geqslant a(v)\geqslant b(v)\geqslant c(v)\geqslant 1$$ for every v, so it remains to show that $$\psi$$ is a proper colouring of G.

Suppose for the sake of contradiction that some two vertices u and v of G with $$\psi (u)=\psi (v)$$ are connected by an edge of G oriented from u to v. Let $$d\in \mathbb {Z}_p$$ be such that $$d(u,v)\equiv d\pmod {p}$$. Since u and v are adjacent in G, we have $$d\ne 0$$. Observe that if C is a clique in G with residue X and tail v, then prepending u to C and possibly removing the unique vertex w in C with $$d(v,w)\equiv -d\pmod {p}$$ (if it exists) gives us a clique with residue $$(X+d)\cup \{0\}$$ and tail u. Therefore, since $$a(u)=a(v)$$, the residue of every clique of size a(v) with tail v must contain $$-d$$. Thus $$-d\in B(v)$$, and if X is the residue of a clique of size a(v) with tail v, then $$X+d$$ is the residue of a clique of the same size with tail u. Hence $$B(u)\subseteq B(v)+d$$, and since $$b(u)=b(v)$$, we further conclude that $$B(u)=B(v)+d$$. Since 0 belongs to the residue of every clique, both B(u) and B(v) are rooted and $$B(u)\sim _pB(v)$$. Thus $$B(u)=B(v)$$, as $$c(u)=c(v)$$. However, since $$b(u)=b(v)\leqslant m<p$$ and $$d\ne 0$$, we have $$B(u)=B(v)+d\ne B(v)$$, which is a contradiction. This shows that $$\psi$$ is a proper colouring of G, as desired. $$\square$$

By combining Lemmas 5 and 6, we have so far proven Theorem 3. Next, we extend the construction to non-primes in order to prove Theorem 2.

For every triple of positive integers knp with p prime and $$p>2n$$, we construct a subgraph $$G_{k,n,p}$$ of $$G_{k,p}$$ by including the edge uv if and only if $$d(u,v)\equiv x\pmod {p}$$ where $$x\in \{\pm 1,\pm 2,\ldots ,\pm (n-1)\}$$. We will now determine the clique number and the chromatic number of $$G_{k,n,p}$$.

### Lemma 7

Let k, n, and p be positive integers with p prime, $$p>2n$$, and $$k\geqslant n$$. Then $$G_{k,n,p}$$ has clique number n and chromatic number k.

### Proof

We have $$\chi (G_k)=\chi (G_{k,p})=k$$. Since $$G_k$$ is a subgraph of $$G_{k,n,p}$$ and $$G_{k,n,p}$$ is a subgraph of $$G_{k,p}$$, it follows that $$\chi (G_{k,n,p})=k$$. Next, we determine the clique number of $$G_{k,n,p}$$.

Let $$I=\{n,n+1,\ldots ,p-n\}\subset \mathbb {Z}_p$$. Note that uv is an edge of $$G_{k,n,p}$$ if and only if $$u<v$$ and $$d(u,v)\notin \{0\}\cup I\pmod {p}$$. Since $$G_k$$ contains a directed path on k vertices and $$n\leqslant k$$, the graph $$G_{k,n,p}$$ has clique number at least n. It remains to show that $$G_{k,n,p}$$ has clique number at most n.

Let C be a clique in $$G_{k,n,p}$$, and let $$v=t(C)$$. If there are vertices $$x,y\in C$$ with $$d(v,y)\in I+d(v,x)\pmod {p}$$, then either $$y<x$$ and $$d(y,x)=d(v,x)-d(v,y)=-(d(v,y)-d(v,x))\in -I=I\pmod {p}$$, or $$x<y$$ and $$d(x,y)=d(v,y)-d(v,x)\in I\pmod {p}$$. In either case, xy is not an edge of $$G_{k,n,p}$$, contradicting the assumption that C is a clique. Thus the set r(C) is disjoint from the set $$\bigcup _{x\in C}(I+d(v,x))=I+r(C)$$, which implies that r(C) contains at most one of i and $$p-n+i$$ for each $$i\in \{1,2,\ldots ,n-1\}$$. Since $$r(C)\subset \{0,1,\ldots ,n-1,p-n+1,p-n+2,\ldots ,p-1\}$$, we conclude that $$|C|=|r(C)|\leqslant n$$. $$\square$$

Next we examine the maximum size of a clique in an induced subgraph of $$G_{k,n,p}$$ that is induced by the vertices of a clique in $$G_{k,p}$$. This will allow us to compare the chromatic number of induced subgraphs of $$G_{k,p}$$ and $$G_{k,n,p}$$ that have the same vertex set.

### Lemma 8

Let k, n, and p be positive integers with p prime, $$p>2n$$, and $$k\geqslant n$$. Then for every clique C of $$G_{k,p}$$, the induced subgraph $$G_{k,n,p}[C]$$ of $$G_{k,n,p}$$ contains a clique of size at least $$\frac{n}{p}|C|$$.

### Proof

Let C be a clique in $$G_{k,p}$$. For each $$i\in \mathbb {Z}_p$$, let $$J_i=\{i,i+1,\ldots ,i+n-1\}\subset \mathbb {Z}_p$$. Since each $$i\in \mathbb {Z}_p$$ is contained in exactly n of the p sets $$J_0,\ldots ,J_{p-1}$$, by the pigeon-hole principle, there exists $$i\in \mathbb {Z}_p$$ such that $$|r(C)\cap J_i|\geqslant \frac{n}{p}|r(C)|$$. Let $$C_i=\{v\in C:d(t(C),v)\in J_i\pmod {p}\}$$. It follows that $$|C_i|=|r(C)\cap J_i|\geqslant \frac{n}{p}|r(C)|=\frac{n}{p}|C|$$. It remains to show that $$C_i$$ is a clique in $$G_{k,n,p}$$.

Let x and y be distinct vertices in $$C_i$$. Since $$x,y\in C$$, they are adjacent in $$G_{k,p}$$, so $$d(t(C),x)\not \equiv d(t(C),y)\pmod {p}$$, and we can assume without loss of generality that $$x<y$$. It follows that $$d(x,y)=d(t(C),y)-d(t(C),x)\in \{\pm 1,\pm 2,\ldots ,\pm (n-1)\}\pmod {p}$$, as $$d(t(C),x),\,d(t(C),y)\in J_i\pmod {p}$$. Hence x and y are adjacent in $$G_{k,n,p}$$. We conclude that $$C_i$$ is indeed a clique in $$G_{k,n,p}$$. $$\square$$

### Lemma 9

Let k, n, and p be positive integers with p prime, $$p>2n$$, and $$k\geqslant n$$, and let G be an induced subgraph of $$G_{k,n,p}$$ with $$m=\omega (G)<n$$. Then $$\chi (G)\leqslant \left( {\begin{array}{c}\lfloor mp/n\rfloor +2\\ 3\end{array}}\right)$$.

### Proof

Let $$G'=G_{k,p}[V(G)]$$. Lemma 8 yields $$\omega (G')\leqslant \lfloor mp/n\rfloor$$. The fact that G is a subgraph of $$G'$$ and Lemma 6 yield $$\chi (G)\leqslant \chi (G')\leqslant \left( {\begin{array}{c}\lfloor mp/n\rfloor +2\\ 3\end{array}}\right)$$. $$\square$$

Theorem 2 now follows from Lemma 7, Lemma 9, and the following theorem of Schur [10] on the gaps between prime numbers.

### Theorem 10

For every integer $$n\geqslant 2$$, there is a prime p such that $$2n<p<3n$$.

## 3 Concluding Remarks

To better understand $$\chi$$-bounding functions, it is of course of interest to improve the bound of $$\left( {\begin{array}{c}3m+1\\ 3\end{array}}\right)$$ in Theorem 2 (and equivalently this same lower bound function for f in Theorem 1).

A slight tweak to the last step of the proof improves this bound slightly to $$\left( {\begin{array}{c}2m\\ 3\end{array}}\right) +o(m^3)$$. To do this, instead of using Theorem 10, we can use the fact that for any $$\epsilon >0$$, there exists a $$n_\epsilon$$ such that for every $$n\geqslant n_\epsilon$$, there is always a prime p with $$2n<p<(2+\epsilon )n$$. This follows from the prime number theorem that the number of primes at most n is asymptotically equal to $$n/\ln n$$. For a more recent and explicit result on the gaps between primes, see [5].

One may hope that another way to further improve this bound would be to improve the bound of $$\left( {\begin{array}{c}m+2\\ 3\end{array}}\right)$$ in Lemma 6. However, in our construction, Lemma 6 is in some sense best possible. For every prime p, we have been able to construct a graph $$G_k'$$ (with k large enough) that satisfies the conclusion of Lemma 4, and such that for every positive integer $$m<p$$, the graph $$G_{k,p}'$$ (as constructed from $$G_k'$$) contains an induced subgraph with clique number m and chromatic number $$\left( {\begin{array}{c}m+2\\ 3\end{array}}\right)$$. So any improvements would require an entirely new construction.

In the other direction, the only result restricting $$\chi$$-bounding functions is that of Scott and Seymour [11] stating that if a hereditary class of graphs $${\mathcal {C}}$$ satisfies $$\chi _{\mathcal {C}}(2)\leqslant 2$$, then $${\mathcal {C}}$$ is $$\chi$$-bounded. We conjecture the following generalisation.

### Conjecture 11

For every integer $$k\geqslant 2$$, if $${\mathcal {C}}$$ is a hereditary class of graphs such that $$\chi _{\mathcal {C}}(n)\leqslant k$$ for every positive integer $$n\leqslant k$$, then the class $${\mathcal {C}}$$ is $$\chi$$-bounded.