The collected mathematical papers of Arthur Cayley.

610]

APPLICATION TO THE THEORY OF CHEMICAL COMBINATIONS.

429~

To count the trees on the principle first referred to, we require the notions of
"centre" and " bicentre," due, I believe, to Sylvester; and to establish these we
require the notions of "main branch" and "altitude": viz. in a tree, selecting any
knot at pleasure as a root, the branches which issue from the root, each with all
the branches that belong to it, are the main branches, and the distance of the furthest
knot, measured by the number of intermediate branches, is the altitude of the main
A B
A
branch. Thus in the left-hand figure, taking A as the root, there are 3 main branches
of the altitudes 3, 3, 1 respectively: in the right-hand figure, taking A as the root,
there are 4 main branches of the altitudes 2, 2, 1, 3 respectively; and we have
then the theorem that in every tree there is either one and only one centre, or else
one and only one bicentre; viz. we have (as in the left-hand figure) a centre A
which is such that there issue from it two or more main branches of altitudes equal
to each other and superior to those of the other main branches (if any); or else
(as in the right-hand figure) a bicentre AB, viz. two contiguous knots, such that
issuing from A (but not counting AB), and issuing from B (but not counting BA),
we have two or more main branches, one at least from A and one at least from B,
of altitudes equal to each other and superior to those of the other main branches in
question (if any). The theorem, once understood, is proved without difficulty: we
consider two terminal knots, the distance of which, measured by the number of
intermediate branches, is greater than or equal to that of any other two terminal
knots; if, as in the left-hand figure, the distance is even, then the central knot A
is the centre of the tree; if, as in the right-hand figure, the distance is odd, then
the two central knots AB form the bicentre of the tree.
In the former case, observe that if G, H are the two terminal knots, the distance
of which is = 2, then the distance of each from A is =X, and there cannot be
giving the values
mn= 1, 1, 3, 13, 75, 541, 4683, 47293,...
for
m= 1, 2, 3, 4, 5, 6, 7, 8,...
But if from each non-terminal knot there ascend two and only two branches, then in this case im = coefficient
1-V1 -4x
of x'-l in -, viz. we have the very simple form
1.3.5... 2m-3
1.2.3...m
giving
mn== 1, 1, 2, 5, 14, 42,
for
ml= 1, 2, 3, 4, 5, 7,...