 # Using Graph Theory to Derive Inequalities for the Bell Numbers

The Bell numbers count the number of different ways to partition a set of n elements while the graphical Bell numbers count the number of non-equivalent partitions of the vertex set of a graph into stable sets. This relation between graph theory and integer sequences has motivated us to study properties on the average number of colors in the non-equivalent colorings of a graph to discover new non trivial inequalities for the Bell numbers. Example are given to illustrate our approach.

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## 1 Introduction

The Bell numbers count the number of different ways to partition a set that has exactly elements. Starting with , the first few Bell numbers are 1, 1, 2, 5, 15, 52, 203 (sequence A141390). The integer can be defined as the sum

 Bn=n∑k=0{nk}

where is the Stirling number of the second kind, with parameters and (i.e., the number of partitions of a set of elements into blocks). Dobiński’s formula  gives

 Bn=1e∞∑k=0knk!.

The 2-Bell numbers count the total number of blocks in all partitions of a set of elements. Starting with and , the first few 2-Bell numbers are 1, 3, 10, 37, 151, 674 (sequence A005493). More formally, the integer is defined as

 Tn=n+1∑k=0k{n+1k}=Bn+2−Bn+1.

Odlyzko and Richmond  have studied the average number of blocks in a partition of a set of elements, which can be defined as

 An=Tn−1Bn.

A concept very close to the Bell numbers is also defined in graph theory. More precisely, a coloring of a graph is an assignment of colors to its vertices such that adjacent vertices have different colors. The chromatic number is the minimum number of colors in a coloring of . Two colorings are equivalent if they induce the same partition of the vertex set into color classes. For an integer , we define as the number of proper non-equivalent colorings of a graph that use exactly colors. Since for or , the total number of non-equivalent colorings of a graph is defined as

 B(G)=n∑k=0S(G,k)=n∑k=χ(G)S(G,k).

In other words, is the number of partitions of the vertex set of whose blocks are stable sets (i.e., sets of pairwise non-adjacent vertices). This invariant has been studied by several authors in the last few years [1, 6, 7, 9, 10, 11] under the name of (graphical) Bell number of .

Let be the total number of stable sets in the set of non-equivalent colorings of a graph . More precisely, we define

 T(G)=n∑k=χ(G)kS(G,k).

We are interested in computing the average number of colors in the non-equivalent colorings of , that is

 A(G)=T(G)B(G).

Clearly, where is the empty graph with vertices. As another example, consider the cycle on 5 vertices. As shown in Figure 1, there are five colorings of with 3 colors, five with 4 colors, and one with 5 colors, which gives , and

This close link between Bell numbers and graph colorings indicates that it is possible to use graph theory to derive inequalities for the Bell numbers. This is the aim of this article. The next section gives values of for some families of graphs and basic properties involving . We give in Section 3 several examples of inequalities for the Bell numbers that can be deduced from relations involving .

Let and be two vertices in a graph . We denote by the graph obtained by identifying (merging) the vertices and and, if and are adjacent vertices, by removing the edge that links and . If parallel edges are created, we keep only one. Also, if is adjacent to , we denote by the graph obtained from by removing the edge that links with , while if is not adjacent to , we denote by the graph obtained by linking with . In what follows, we let , and be the complete graph of order , the path of order , and the cycle of order , respectively. We denote the disjoint union of two graphs and by . We refer to Diestel  for basic notions of graph theory that are not defined here.

## 2 Some values and properties of A(G)

The deletion-contraction rule (also often called the Fundamental Reduction Theorem ) is a well known method to compute [7, 11]. More precisely, let and be any pair of distinct vertices of . We have,

 S(G,k)=S(G−uv,k)−S(G∣uv,k) for every pair u,v of adjacent vertices in G, (1) S(G,k)=S(G+uv,k)+S(G∣uv,k) for every pair u,v of non-adjacent vertices in G. (2)

It follows that

 B(G)=B(G−uv)−B(G∣uv)T(G)=T(G−uv)−T(G∣uv)} for every pair u,v of adjacent vertices in G, (3) B(G)=B(G+uv)+B(G∣uv)T(G)=T(G+uv)+T(G∣uv)} for every pair u,v of non-adjacent vertices in G. (4)

Let be a vertex in a graph . We denote by the graph obtained from by removing and all its incident edges. A vertex of a graph is dominating if is adjacent to all other vertices of .

###### Proposition 1.

If has a dominating vertex , then .

###### Proof.

Clearly, for all , which implies

 B(G) =n∑k=χ(G)S(G,k) =n∑k=χ(G)S(G−v,k−1)=n−1∑k=χ(G−v)S(G−v,k)=B(G−v) and T(G) =n∑k=χ(G)kS(G,k) =n∑k=χ(G)kS(G−v,k−1)=n−1∑k=χ(G−v)(k+1)S(G−v,k)=T(G−v)+B(G−v).

Hence,

Duncan  has proved that if is a tree, then for all . This leads to our second Proposition.

###### Proposition 2.

Let be a tree of order . Then and .

###### Proof.

Since , we immediately get

 B(G)=n∑k=0{n−1k}=Bn−1.

Let us now compute . Since for all trees , we have , where is the star with a center of degree linked to vertices of degree 1. Hence, and . Since is a dominating vertex in , we have

 A(G)=A(K1,n−1)=1+A(¯¯¯¯¯Kn−1)=1+Bn−Bn−1Bn−1=BnBn−1

which implies . ∎

###### Proposition 3.

Let be the graph obtained from a tree of order by adding isolated verices. Then and .

###### Proof.

For , the result follows from Proposition 2. For larger values of , we proceed by induction. Let be the tree obtained from by adding a new vertex and linking it to exactly one vertex in . Equations (4) give

 B(T∪pK1) =B(T′∪(p−1)K1)+B(T∪(p−1)K1) =p−1∑i=0(p−1i)Bn+i+p−1∑i=0(p−1i)Bn+i−1 =p∑i=1(p−1i−1)Bn+i−1+p−1∑i=0(p−1i)Bn+i−1 =Bn+p−1+p−1∑i=1((p−1i−1)+(p−1i))Bn+i−1+Bn−1 =p∑i=0(pi)Bn+i−1.

The proof for is similar. ∎

###### Proposition 4.

Let be a cycle of order . Then,

 B(Cn)=n−1∑j=1(−1)j+1Bn−j % and T(Cn)=n−1∑j=1(−1)j+1Bn−j+1.
###### Proof.

Duncan  proved that . It is therefore sufficient to prove that .

If , then . If , Equations (3) together with the fact that is a tree give , and the result follows by induction. ∎

###### Proposition 5.

Let be the graph obtained from a cycle of order by adding isolated verices. Then

 B(Cn∪pK1)=n−1∑j=1(−1)j+1p∑i=0(pi)Bn+i−j and T(Cn∪pK1)=n−1∑j=1(−1)j+1p∑i=0(pi)Bn+i−j+1.
###### Proof.

For , the result follows from Proposition 4. For larger values of , we proceed by induction. If then Equations (3) and Proposition (3) give

 B(C3∪pK1) =B(P3∪pK1)−B(P2∪pK1) =p∑i=0(pi)B3+i−1−p∑i=0(pi)B2+i−1 =2∑j=1(−1)j+1p∑i=0(pi)B3+i−j.

Hence, the result is valid for . So assume and that the statement holds for smaller values of :

 B(Cn∪pK1) =B(Pn∪pK1)−B(Cn−1∪pK1) =p∑i=0(pi)Bn+i−1−n−2∑j=1(−1)j+1p∑i=0(pi)Bn+i−j−1 =p∑i=0(pi)Bn+i−1+n−1∑j=2(−1)j+1p∑i=0(pi)Bn+i−j =n−1∑j=1(−1)j+1p∑i=0(pi)Bn+i−j.

The proof for is similar. ∎

###### Proposition 6.

Let be a graph with a vertex of degree 1. Then .

###### Proof.

Since has degree 1, we have . Assuming that is of order , we have

 B(G) =n∑k=0S(G,k)=n−1∑k=0kS(G−v,k)−n−1∑k=0S(G−v,k)+n−1∑k=0S(G−v,k) =n−1∑k=0kS(G−v,k) and T(G) =n∑k=0kS(G,k)=n∑k=0(k2−k)S(G−v,k)−n∑k=0kS(G−v,k−1) =(n−1∑k=0k2S(G−v,k)−n−1∑k=0kS(G−v,k))+(n−1∑k=0(k)S(G−v,k)+n−1∑k=0S(G−v,k)) =n∑k=0k2S(G−v,k)+n−1∑k=0S(G−v,k).

We therefore have

 B(G−v)T(G)−T(G−v)B(G) = n−1∑k=0S(G−v,k)(n∑k′=0k′2S(G−v,k′)+n−1∑k′=0S(G−v,k′))−n−1∑k=0kS(G−v,k)n−1∑k′=0k′S(G−v,k′) = n−1∑k=0(S(G−v,k))2(k2+1−k2)+n−2∑k=0n−1∑k′=k+1S(G−v,k)S(G−v,k′)(k′2+1+k2+1−2kk′) = n−1∑k=0(S(G−v,k))2+n−2∑k=0n−1∑k′=k+1S(G−v,k)S(G−v,k′)((k−k′)2+2)>0

which implies . ∎

###### Proposition 7.

Let and be graphs, and let be positive numbers such that

• for all .

Then .

###### Proof.

Since , we have for . Hence,

 A(G) =T(G)B(G)=T(H)+∑ri=1αiT(Fi)B(H)+∑ri=1αiB(Fi)

## 3 Inequalities for the Bell numbers

In this section, we show how to derive inequalities for the Bell numbers, using properties related to the average number of colors in non-equivalent colorings of . We start by analyzing paths. As already mentioned, is the graph obtained by adding isolated vertices to a path on vertices.

for all and .

###### Proof.

It follows from Equations (4) that

 B(Pn∪(p+1)K1)=B(Pn+1∪pK1)+B(Pn∪pK1)

and

 T(Pn∪(p+1)K1)=T(Pn+1∪pK1)+T(Pn∪pK1).

Also, we know from Proposition 6 that . Hence, it follows from Proposition 7 that . ∎

Proposition 3 immediately gives the following Corollary.

###### Corollary 9.

If and then

 (p+1∑i=0(p+1i)Bn+i))(p∑i=0(pi)Bn+i+1)<(p+1∑i=0(p+1i)Bn+i−1))(p∑i=0(pi)Bn+i).
###### Examples 10.

For and , Corollary 9 provides the following inequalities for the Bell numbers:

 (Bn+Bn+1)Bn<(Bn−1+Bn)Bn+1⟺B2n

These inequalities also follow from Proposition 6. Indeed, is obtained from by removing a vertex of degree 1, which implies

 A(Pn)

Note that Engel  has shown that the sequence is log-convex, which implies (with a non-strict inequality) for . Recently, Alzer  has proved that the sequence is strictly log-convex by showing that

 Bn−1Bn+1−B2n=12e2∞∑k=2k−1∑j=1jn−1(k−j)n−1j!(k−j)!(k−2j)2

for all . Since , this also implies for all .

As a second example, assume and . Corollary 9 gives

 (Bn+Bn+1+Bn+2)(Bn+Bn+1)<(Bn−1+Bn+Bn+1)(Bn+1+Bn+2) ⟺ Bn(Bn+Bn+1)

For and , we denote the graph obtained by linking one extremity of to one vertex of (see Figure 2). Also, is the graph obtained from by adding isolated vertices. We now compare with to derive new inequalities involving the Bell numbers.

for all and .

###### Proof.

Note first that Equations (3) give and . Hence,

 A(Pn+1∪pK1)−A(H3,n−3∪pK1)=T(Pn+1∪pK1)B(Pn+1∪pK1)−T(H3,n−3∪pK1)B(H3,n−3∪pK1) = T(Pn+1∪pK1)B(Pn+1∪pK1)−T(Pn∪pK1)−T(Pn−1∪pK1)B(Pn∪pK1)−B(Pn−1∪pK1) = T(Pn+1∪pK1)(B(Pn∪pK1)−B(Pn−1∪pK1))−B(Pn+1∪pK1)(T(Pn∪pK1)−T(Pn−1∪pK1))B(Pn+1∪pK1)B(H3,n−3∪pK1).

Let be the numerator of the above fraction. It follows from Proposition 3 that

 f(n,p)=p∑i=0(pi)Bn+i+1(p∑ℓ=0(pℓ)(Bn+ℓ−1−Bn+ℓ−2))−p∑i=0(pi)Bn+i(p∑ℓ=0(pℓ)(Bn+ℓ−Bn+ℓ−1)).

It remains to prove that for all and . Since , we have

 Bn−Bn−1=1e∞∑k=1(knk!−kn−1k!)=1e∞∑k=2kn−1k!(k−1).

Hence,

 e2f(n,p)= p∑i=0p∑ℓ=0∞∑j=1∞∑k=2(pi)(pℓ)jn+i+1j!kn+ℓ−2k!(k−1)−p∑i=0p∑ℓ=0∞∑j=1∞∑k=2(pi)(pℓ)jn+ij!kn+ℓ−1k!(k−1) = p∑i=0p∑ℓ=0∞∑j=1∞∑k=2(pi)(pℓ)jn+ij!kn+ℓ−2k!(k−1)(j−k) = p∑i=0p∑ℓ=0∑j>k≥1(pi)(pℓ)(jn+ikn+ℓ−2j!k!(k−1)(j−k)−jn+i−2kn+ℓj!k!(j−1)(j−k)) = p∑i=0p∑ℓ=0∑j>k≥1(pi)(pℓ)jn−2kn−2j!k!(j−k)(ji+2kℓ(k−1)−jikℓ+2(j−1)) = ∑j>k≥1jn−2kn−2j!k!(j−k)((k−1)j2p∑i=0(pi)jip∑ℓ=0(pℓ)kℓ−(j−1)k2p∑i=0(pi)jip∑ℓ=0(pℓ)kℓ) = ∑j>k≥1jn−2kn−2j!k!(j−k)(j+1)p(k+1)p((k−1)j2−(j−1)k2).

Let . We have proved that

 e2f(n,p)=∑j>k≥1jn−2kn−2j!k!(j−k)(j+1)p(k+1)pg(j,k).

Note that , , and

 g(j,k)=j2k−j2−jk2+k2=(j−k)(jk−j−k)=(j−k)(k(j2−1)+j(k2−1)).

Hence, for , and it remains to prove that