EEV. T. P. KIEKMAN ON THE K-PAETITIONS OF THE E-GON AND E-ACE. 247 
Thus we have completely determined 
^)„+R^*'(r, k)^. 
7i 
XL VI. When r— 8 in equation (C".), ^^=6, or ?^=4. If w=6, ^ = 3 and h—1 are 
g 
the only values making ^ integer. We have then the three solutions 
?'=8=64-2(l+0+0)=6+2(0+l+0)=6+2(0+0+l), 
^=2=6+2(0-l~l)=6-t-2(~l+0-l)=6+2(~l-l + 0); 
2;A3=;3D(3, 0)“3, and we obtain 
Rf(8, 2)e=i{3-R|f(8, 2)J==1. 
If w=4, h~2 or h~\ ; for the first of which we obtain by the formula for k)„^ 
Rf(8, 4),=:1{D(3, 0)-R|f(8, 4)J=:i(l-l)=0, 
k=4i being the only possible value. 
If A=l, we may have ^=4 or ^=2 ; for ^=4, we find the three solutions 
8=4-f2(2+0)=44-2(0+2)=4+2(l-fl) 
4=4+2(l-I)=4+2(-l+l)=:4-i-2(0-f0) ; 
2:A3=2D(4, 1)+D(3, 0)=4+1=5, 
whence Ef ( 8, 4)^ = i { 5 — E^f ( 8, 4)^ } = 2 . 
'For k=2, there are two solutions, 
8=4-f2(2+0)=4-f 2(0+2), 
2 = 4+2(0-l)=4+2(-l + 0); 
giving 2A2=2D(4, 0)=2, 
and Ef(8, 2),=i(2-0) = l; 
for there is nothing in our list under E(8, 2)^ to subtract in the formula for 2Rf *(f, k)^. 
XLVII. Our next step is to determine 
E*-’”“(r, ^)?^=E^'"'’(r, A:)„+E^’”"(r, k%. 
Problem m. To find Ec'““(r, k)„, the number of {l-\-\)^artitioned i-gons, built on the 
n-gonal nucleus to have h clear monogonal axes of reversion^ and none besides. 
The h clear axes (Theorem O) are monogonal axes of the %-gon, bisecting h equi- 
distant sides of it ; between which lie h vertices of the n-^orv upon those /^ axes ; and n is 
odd. 
The interval of 
n — h 
2h 
sides between such a vertex and bisected side is loaded the 
mih. from the vertex with a (l+e„j)-partitioned (2-\-a^)-gon, so that all is reversible about 
the h axes. The equations following are satisfied in the construction : 
