Received 14 January 2016; accepted 23 February 2016; published 26 February 2016

1. Introduction
The Wiener number was one of the oldest topological indices, which was introduced by Harry Wiener in 1947. About the recent reviews on matrices and topological indices related to Wiener number, refer to [1] - [4] . The RCW number is one of the hotest additions in the family of such descriptors. The notion of RCW number was first put forward by Ivanciuc and its applications were discussed in [5] - [8] .
Let G be a simple connected graph with vertex set
. For two vertices
, let
denote the distance between u and v in G. Then, the RCW number of G is defined by

where d is the diameter and the summation goes over all unordered pairs of distinct vertices of G. Some properties of the RCW number have been obtained in [9] [10] .
A tree is called a caterpillar if the removal of all pendant vertices makes it as a path. Otherwise, it is called a non-caterpillar.
For integers n and d satisfying
, let
be the tree obtained from the path
labelled as
by attaching the path
and
pendant vertices to vertex
for
(see Figure 1). Let ![]()
In this paper, we show that among all n-vertex non-caterpillars with given diameter d,
is the unique tree with minimum RCW number where
. Furthermore, we determine the non-caterpillars with the smallest, the second smallest and the third smallest RCW numbers.
2. RCW Numbers of Non-Caterpillars
All n-vertex trees with diameter 2, 3,
and
are caterpillars. Let n and d be integers with
and
. Let
be the class of non-caterpillars with n vertices and diameter d. Let
be
the class of non-caterpillars obtained by attaching the stars
at their centers and ![]()
pendant vertices to one center (fixed if it is bicentral) of the path
, where
,
and
for
(see Figure 2). Recall that
Obviously,
and
.
Let T be a tree. For
and
, let
be the degree of u in T and
be the
sum of all distances from u to the vertices in A, i.e.,
. Here and in the following
denotes the distance between vertices u and v in T.
Lemma 1 Let T be a tree with minimum RCW number in
, where
. Then,
.
Proof. Suppose that
Let
be a diametral path of T. If d is odd,
we require that
. Then at least one of
has degree at least three. There are two cases.
Case 1. One of
different from
has degree at least three. Let
be all the neighbors
outside
except those of
, where
is a neighbor of
. Let
be the subtree of
containing
.
be the tree formed from T by deleting edges
and adding edges
for
all
. Obviously,
. Let
and
. It is easily seen that
![]()
with equality if and only if
. Since
,
for
and
with
. We get
![]()
Then
![]()
with equality if and only if
(which is only possible for odd number d). But
, and thus if
then
. So
for
Thus
![]()
since
for
. It follows that
. This is a contradiction.
Case 2. Any verter
with
and
has degree two. Obviously,
. Let
be the (unique) path from x to
in T such that
. Since
, we have
. Let
be the neighbors of y in T, where
and
.
Let
be the tree obtained from T by deleting edges
and adding edges
for all
. Then
. Let
,
Since
for
, we get
![]()
This is a contradiction.
By combining Cases 1 and 2, we find that
is impossible. The result follows.
Lemma 2 Let
with
. Then
![]()
with equality if and only if
.
Proof. Let T be a tree with the minimum RCW number in
. Let
be a diametral path of T.
Suppose that there is a vertex
with
Let
be the neighbors of u different from
in T, where
. Clearly,
are pendant vertices for
. Let
be the tree
obtained from T by deleting edges
and adding edges
for
. Obviously,
Let
,
and
Since
for
, we get
![]()
and then
, this is a contradiction. Thus any vertex of T outside
has degree at most two.
Suppose that there are at least two vertices of T outside
with degree two. Let ![]()
with
and let x be the neighbor of y which is different from
in T. Let
be the tree formed from T by deleting edge yx and adding edge
. Obviously,
Let
Since
and
for
, we get
![]()
This is a contradiction. Thus there is exactly one vertex outside
with degree two and all other vertices of T outside
are pendant vertices. Then,
.
By a direct calculation, we get
![]()
Combining Lemmas 1 and 2, we get
Theorem 1 Let
, and
Then
![]()
with equality if and only if
.
Lemma 3 For
, there is
.
Proof. If d is even, then
![]()
If d is odd, then
![]()
The result follows.
Theorem 2 For
, there is
![]()
And
for any n-vertex non-caterpillar T different from
, ![]()
Proof. Let
, where
. If
, then T is a non-caterpillar
where
It follows that
![]()
and hence
is monotonically decreasing for
This implies
![]()
Now suppose that
. By Theorem 1 and Lemma 3, there is
![]()
where equality holds if and only if
We need only to show
![]()
Case 1. n is odd. Let
and
. Then there is
![]()
Case 2. n is even. Let
Then there is
![]()
Thus, the proof is finished.