EEV. T. P. KIEKMAN ON THE K-PAETITIONS OP THE E-GON AND E-ACE. 261 
For P( 8 , 2)4 we have ^^=2^, 
8=4-f-2.(l+l)=4-f2(2+0)=4-l-2(0+2) 
2= 4+2(0-l)=4-j-2(-l + 0) 
2Aa=2D(4, 0 )= 2 , and R^( 8 , 2 ) 4 = 1 , 
P(8,2)4=f{2-A.lj=0. 
For P( 8 , 4)4 we have ^^=2^, 
8=4+2(l-fl)=4+2(0+2)=4-f 2(2+0) 
4=4+2(0+0)=4+2(-l+l)=4+2(l-l) 
2A,=D(3, 0)D(3, 0)+2D(4, l)=l+4=5, 
and E^(8, 4 ) 4 = 1 , whence 
P(8, 4).=i{5-A.l}=l. 
which is the only form of P( 8 , ]c\, for A>1, ?^>2. 
LXVI. Problem w. To find P(r, k) 2 , the doubly irreversible {\-\-\i)-]yartitions of the 
r-gcm^ which have a nucleus line. 
It is proved in Theorems W and X, that these are all obtained by drawing a diameter 
b which is not an axis of reversion in one of the r-gons P’"(r, h — 1)„ and k — 1)„; 
and that any such diameter drawn in any one of these gives us one of I^(r, k)^. And it 
is evident that S is a diameter of the nucleus about which all is symmetrical ( LXIII.). 
When no diameter of the nucleus is a diagonal axis of reversion of the r-gon, every 
diameter that can be drawn is such a line Now in k—l)^ you may draw S from 
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any one ^ of ^ vertices of the nucleus, and from no more, because the configuration 
about 0 is repeated 2h times, about a vertex in each of the 2h repeated sequences. In a 
reversible partition no diameter of the nucleus is a diagonal axis of reversion, unless it 
be a clear axis. Therefore, in 
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whether the axes be clear or not in the last one, you can draw b from any one of ^ vertices 
of the r-gon; for none of the partitions in these classes have a clear diagonal axis, and 
each vertex g in one of the 47i intervals between two adjacent axes begins a different 
sequence, which does not occur reverted till you come to s', equidistant with s from the 
axis between them. Wherefore 
P(r, !“(»•, i-l),+|i(Ef*"+E2"'"+E+(r, /t-l),+ 
