260 EEV. T, P. KIEKMAJ7 ON THE K-PAETITIONS OP THE E-GON AND E-ACE. 
is hi times repeated. And as the A^-ly irreversible has the same sequences reversed on 
71 
its other face, giving ^ different aspects which we have all constructed as read from the 
first side of the n-gon in one direction, we have had this same hi-lj irreversible 2n:M 
times before us among our results. 
Again, every A^-ly reversible (l-|-^)-partitionedr-gon has been ^ times brought before 
us in our 2A« constructions. For this has a different 2 -ple sequence reversible or 
h 
,71 7i , 
irreversible of ^ vertices of the %-gonal nucleus commencing at any one of ^ successive 
vertices, and we never see a sequence repeated in it till we have read over an interval 
equal to that between two alternate axes, that is an interv al of ^ sides of the w-gon 
(art. V.). This ^-ple sequence is h times repeated in the circuit of the r-gon ; so that 
we have counted this hi~\y reversible ^ times among those constructed from equation (E.). 
Wherefore 
and 
V(r, i:) k).)), (e>0). 
71 
LXV. The highest value of h in (E.) comes from putting the subindex ^ =1; or h=n. 
This gives 
nr, k)„=U-D 
r + n k—n 
n n 
a given number, since K”(r, is found in our register. 
71 
Next putting ^=2, or whatever next factor n may have, we can obtain P(r, Jc)„ for 
every value of /i> 1 that is a divisor of the three numbers r, h and n. 
Since 7i> 1, 2, r, k and n are none of them prime ; the only values of I*(8, k),^ that 
give them all divisible by h, and A:>>8— 3, are 
P(8, 2)a, P(8, 2)„ P(8, 4), and P(8, 4), 
By the last written formula. 
m 4),=:i{D(3, 0)-KX8, 4 )J=i(1-1)=0, 
for 1^(8, 2)6 we have three solutions, ^^=8^ 
8=6-f2(l + 0+0)=&c. 
2 = 6+2(0-l-l)=&c. 
giving 2A3=3D(3, 0) = 3, and as R"(8, 2)6=1, 
P(8, 2)6=f{3-fR^(8, 2)6}=K3-3)=0. 
