EEV. T. P. KIEKMAN ON THE K-PAETITIONS OF THE E-GON AND E-ACE. 269 
1(10, 2)=2.i{M(8, 2, 0 ).m( 4, 1, 0)}-fl.w(8, 2, 0) 
=2.i{4-2}+4=6, 
he will find the sum 
1(10, 2)+l(10, 3)+l(10, 4)=6-f 13+2 = 21, 
r(10, 3)=1, 
and R(10, 2)+R(10, 4)=5, 
the 7-partitions of the nonagon, in that notation, as we have just found them: Aide (1) 
of this memou*. 
LXXl. We shall for a final example determine the 8-partitions of the decagon. 
There is no nucleus but 3 and 2 ; we want then 
R'^(10,7)3R^(10,7)„ R'^'(10, 7 ) 3 , Ro(10,7), P(10, 7)3 r(10, 7),. 
By Theorems M, E (XIV., IX), no axis can be agonal, or monogonal ; we seek then 
R^-^^IO, 7 ) 3 , R^‘''(10, 7 ) 3 , R‘''(10, 7 ) 3 , and Rf(10, 7). 
Ro*’(10, 7 ) 3=0 by Problem i, because n is odd; and R^f(10, 7 ) 3=0 for the same 
reason, as is also R**(10, 7 ) 3 . Therefore R**(10, 7 ) 3 = 0 . 
By Problem p, 
R^^'^'^IO, 7)3=2,2„{2Rf"'+R|f+R+^'^‘+R^,f^"')(10, 6)„} = 0; for 
R 2 A<ii because 10 is not a multiple of 4/i (Problem 1) ; 
R|“' (10, 6)„=0,.for the same reason (Problem k) ; 
Rf“^‘'*(10, 6)„=0, because 10 is no multiple of &i (Problem b) ; 
R4Aogdi(io, because 10 — n, (w>2) is no multiple of 8A (Problem c). 
R''* (10, 7)3 = 2/i2„R^^*^‘’''‘(10, 6)„=Rf(10, 6)„, because 10 is a multiple of 4/i+2 (Pro- 
blems 1, q) ; 
Rf (10, 7) =2,2,S„ 
R<2. + l,tf,(10, 
?+l 'I + &C.J (Problem t), 
all the remaining part of the expression vanishing; for we shall presently see that %=4 
only in both these equations; and R[,^*‘^®^*'(10, 7) has been proved above to be =0, as 
have Rf*’(10, 7 — c)„ and R|“'(r, 7 — c)„ also ; and Rf“^'''(10, 7 — e)„=0, because 1 — e and 
10— w are not multiples of 4A, unless /i=l, w=6 ; (w> 2), which makes equations (A.) 
/n — 2h .. \ 
10=6 + 4 . 1 , and 7-^=6-2 + 4.-l, 
e=7 ; but R^''^‘'‘( 10 , 6)0 has no existence, for if there are no diagonals the nucleus is 
the 10-gon itself. Rrf*“^‘^'(10, 7— e)„=0; for this can only be R^“^*’(10, 7 — for a 
reason just given ; but (Problem c) 7 —e — 6 is even, therefore e=l, and as /?=!, 
equations (A'.) become 
10=6+.2(2 + 0)+4.0, and A:=6=6+2.2 + 4.-l, 
i. e. in H, Uo=0, So=2, or a 4-gon has two diagonals not crossing, which is absurd. 
