BEV. T. P. KTRKMAlf ON THE K-PAETITIONS OE THE E-GON AND E-ACE. 245 
n—h 
Next putting =1, we find (r, and so on for every even value of A>0 that 
makes n—h^ T—n—]i{l-\-2a^), and k—n—h^, 
all divisible by 2A. 
The greatest value of A in equation (C.), XLI., is h=^i. Those equations become 
r — 2n 
n -ao+a. 
2k — 2n 
n -®o4-h 
and 
© 
<1 
11 
rH 
11 
o 
<1 
wherefore. 
R^‘"(r, A)=0, 
and because (XXXIII., XXXV.) 
or ErCn ^),=42. {24R™'(8+2<r., i.) ■ 2.R-”-(3+2a., £,))} ) ; 
the number of products MM' under the first 2 being that of the solutions of the above 
equations in «!, £j, every order of values counting for a solution. From this Ro*(r, k)„ 
can be found for every value of A>1 in equation (C.), XLI., that makes n, r, and 
divisible by 2A, conditions imposed by those equations. But we cannot 
attempt by these formulae to find Rf(r, k)n, (k=l), because (XIII.) this subclass reduces 
to no nucleus. 
XLIV. We proceed with our partitions of the 8-gon, to find R^/'^S, k)„ and Ro-*(8, k)^. 
g ^ 
In equations (C'.), n=Q or w=4, for 7’=8 ; and -^=integer gives only Ii=.2. Hence 
r=8=6-f2.(l-f0)+4.0, 
;^=2 = 6+2.0-l-4.-I, 
the only value of k, because go>l+2ao— 1. Wherefore 
R^f(8, 2)e=R-(3, 0)A,=:I. 
4—h 
Next, if w=4, -^= integer, =0, gives A =4, 
7’=8=4-f4(l-f0) 
A=4=4-l-4.0, 
the only value of k ; 
R^f(8, 4),=R-(3, 0)A,=1. 
Next, in equations (C.), r=8 gives n—Q, or w=4, of which the first, putting - 
4 2/i 
ger, gives only A=l, the value above forbidden. But — = gives A=2; 
whence A=4, and by the formula for Rf •*(r, k)„ we obtain {a,Q=:s^=0), 
Rr(8, 4),=i{R(3, 0).R(3, 0)-R|f(8, 4)J=i{l-l}=0. 
