EEY. T. P. KIEOIAN ON THE K-PAETITIONS OE THE ErGON AND E-ACE. 225 
the left members of which are to be calculated by the preceding article, we can deter- 
mine the coefficients A, B, . . LM. 
So far as I have pursued the inquiry, I find always one factorial form of the function 
r>(r, Ic ) ; but in order to prove that this is always the form, it is necessary to show that, if 
(3 + A-e-2)'l‘ 
CT? ’ 
and 
D(r-^-l, ^-£-1)= 
D(r, k) 
Ik—s ^ ik — £+ 1 
=f,2,2.{D(3-fA, g).D(r-A-l, A:~£-1)}=^X-^ ^ 
■fl+T fFi-2 
from A=0toA=r— 4, and from £=0 to z=ik — 1; which is the expression that I con- 
tinually find. 
This summation I must leave to the learned and industrious reader ; but, meanwhile, 
I shall venture to enunciate with the best demonstration, such as it is, that occurs to me, 
the following 
Theoeem T. The nuniier of {l-]r^)-divisi<ms of an x-gmi, i. e. of all the ways in which 
k diagonals can he drawn in it, none crossing another', is 
IT+T • 
For first, let the r-gon be divided into triangles, i. e. let k=r—S. Here the (3-1- A)- 
gon of the preceding article can have only h diagonals, or £=4; and the final result, the 
last equation of that article, becomes 
D(r, r— 3)=2^^:^.2;i{D(3-l-A, A).D(r— /i— 1, r—h—i)}, 
from A=0 to h—r—^L. 
Hence we obtain for r=4, r=5, r=6, &c., 
D(4, l)=f .D(3, 0).D(3, 0)=2=^, 
( 4-1 
D(5, 2)=|.{D(3, 0).H(4, 1)-1-D(4, 1).D(3, 0)} 
j5-3|l 
|5^’ 
D(6, 3)=t{D(3, 0}D(5, 2)-l-D(4, 1).D(4, l)-f-D(5, 2XD3, 0)} 
56-311 
= 14= 
(6-1 
D(7, 4)=i{D(3, 0)D(6, 3)+D(4, 1)D(5, 2)+D(5, 2)D(4, 1)+D(6, 3).D(3, 0)} 
=|{14+10+10+14} = 42 
77-3II ^ 
and as we have demonstrated that D(8, 5) is a sum of products of these continuous 
MDCCCLVII. 2 H 
