262 EEV. T. P. KIEKMAif ON THE K-PAETITIONS OF THE E-GON AND E-ACE. 
After the + must come, first, 
because ^ cannot be drawn from any of the 2h terminations of clear diagonal axes, but 
may be drawn to begin a certain sequence AJi times firom other vertices, once in the 
interval between the terminations of every adjacent pair of axes ; and secondly, 
h-l), 
must be added, because, omitting the 4A terminations of clear diagonal axes, you can 
draw ^ to begin a certain sequence 4A times from the remaining vertices of the nucleus. 
Therefore, finally, 
P(r, !“(»•, k-l).+W^{r, yl'-l),) 
+^*(R2n»-. i-l).+Efn>-. lc-l):)+^-Rf^(r, A-1).}; 
where every value of A>0 and w>2 is included, that is found in our register, filled up 
by the formulae already given. 
LXVII. Hence, since we have found 
E“*'“(8, 0).=E|;»(8, 2).=E“‘(8. 2).=E;f(8, 4).=P(8, 4).=1, 
I’(8, 1). =0.E‘"''‘(8, 0),=0, 
P(8, 3), =tE“<'(8, 2).+5^R“(8, 2).=2, 
r(8, 5), E“(8, 4).+|.P(8, 4),=2. 
LXVIII. Problem x. To find I(r, k), the number of singly irreversible {\-\-^ypartitions 
of the x-gon. 
By the formula of XXIII. we have 
I(r, lc)=y^(r, i)-2.(£.E"(r, i))}, 
which completes the solution of our problem of the (I-j-k)-partitions of the polygon, or 
of the polyace ; for these are analytically and numerically the same. Hence follows, 
1(8, 0)=-MJ){S, 0)-I.R«(8, 0)} = 0, 
I(8,l)=i^{D(8,I)-f.K^(8,I)-8R(8,I)} =^{20-4-16} = 0. 
I(8,2)=^{H(8,2)-|K^(8,2)-8.E(8,2)} =^{120-4-2-8-4} = 5. 
1(8, 3)=^{H(8, 3)-fE^(8, 3)-8E(8, 3)-iH^(8, 3)}=J6-{300- 4-l-8-7-8-2}=I4, 
1(8, 4)-fE^(8, 4)-8.E(8, 4)-^P(8, 4)}=TtV{330-2-I-8-4-8-l}=18, 
I(8,5)=3L{H(8,5)-fE^(8,5)-8.E(8,5)-i^P(8,5)}=:3L{i32-4-I-8-4-8-2}= 5, 
which, along with the results in LXI., LXV., LXVII., complete the enumeration of 
the partitions of the octagon. 
