EEV. T. P. KFRKM AN ON THE K-PAETITIONS OE THE E-OON AND E-ACE. 255 
w-gonal nucleus in any 2A-ly reversible (1+A:— e)-partitioned y-gon, across a clear agonal 
axis. 
When the axes are all clear and agonal, we can score across either configuration, i. e. 
upon either of two adjacent axes, and thus obtain twice this number towards k )" ; 
but if only half are clear and agonal, we have only one configuration to score, and can 
obtain this number only. Hence we see that the product under of the number 
above written, into 
is the entire sum Eo^(y, k)’\ of y-gons in Eo^(y, k) obtainable by scoring one of an even 
number of axes, by the construction of Theorem E. 
LVIII. Adding, then, to this entire sum the portion of Eo^(y, k) obtained by scoring 
one of an odd number of axes (LVI.), we express the complete result thus : — 
+i{2E^*“^(y, ^_e)„+E|f^(y, ^-e}„+Ef-“^*-(y, k-e),} 
X 
comprising all the values of e, h and n on our register of (l-j-A: — e)-partitioned rever- 
sibles specified in the right member: (^^>2, e>0, A>0). 
When k=e^ the y-gon scored is the simple y-gon, or %=y. 
The only numbers of those in the right member of this expression that our list con- 
tains for y=8, are 
0)3=1, E:^(8, 2 )e=l, E:^(8, 4 ),= 1 , E|f(8, 2),=1 
(XXVm. XXXVIII.), of which the two last disappear by the multiplier 4 , since e 
always > 0 . Therefore 
K:»(8, l)=i('E»«'''(8, 0).x|^^)=l 
R?'(8, 2)=i|E; «'»(8, 0),x ((f)’'-'-^3‘'-A)}=0 
E:>(8, 3)=R;<'(8, 2)„x (?^Ai) = 1 
E;»(8, 4)=Ef(8, 2).x(|f'-'.(3"=0. 
LIX. Problem t. To find Ef(r, k), the number of {l-\-\)-fartitioned x-gons singly 
reversible about a scored diagonal axis. 
We write again in the sense of art. LVI., 
Ef(y, ^)=Ef(y, ^)'-f Eo^'(y, kf. 
