EEV. T. P. KIEKMAN ON THE K-PAETITIONS OE THE E-OON AND E-ACE. 235 
n — Ah 
XXXIII. In equations (A".) ■ may have any value >0. When n=-^h, we have 
r— %=^.(2a,-fl), 
‘2 
n n 
fC 2 ~ 2 ^^ ’ 
hence 
and as Ao=I, 
2r 2k-n . 
3+2ai— 
2r 2k — 
which of course =0 if the fractions are reducible, and also (IX.), if r is divisible by n. 
,1 . , , . ... .1 , . n — Ah , 
From this we obtain, puttmg the submdex ^ =1, or 2A=-? 
k!"V. 
Tl 7t 
and thus k) can be found in succession for every whole value of ^ from 2A=^ 
to 2A = 2, which gives 
r— 2A^2a^-f Ij and 7^— 2A^£^— I^ 
divisible by 4A. 
W'e proceed with our partitions of the octagon, by finding ^)„. 
R^“^*(8, ^)g=0, because the submdex is not integer. 
4 4^ 
If 7? = 4, = 0 and h=\ ; when (A".) becomes 8=4-{-2(2a,-j-I), which is absurd; 
.-. R^“^'^'(8, ^)=0, for any nucleus. 
XXXIV. Problem e. To find Ef^‘^(r, k)„, the number of {l-\-^)-fartitioned v-go7is 
having an n-gonal nucleus and h diagonal and h agonal axes of reversion, of which none 
are clear. 
The construction is like the preceding, with the exception that 27i equidistant sides, viz. the 
nth, ^th, ||th, . . . ^^wth. 
2h 
are all loaded with the same (I+2ao)‘goii, having at least one agonal 
axis of reversion, £„ and being any numbers in our register of agonally reversibles, 
£_» and being found as in XXXII. 
4A Ah 
The equations to be satisfied are. 
where and I 
r — W-1-2A. (24-2ao) + 4A^ff,+a2H“ • • 4* a n—ih 
k^n-k-lh.iQ -j-IA/ei -1-^2 + . • + €n~Ak\ “f“2A . £» 
\ Ah } Ah " 
■ (A'".) 
2 I 2 
