268 EEV. T. P. KTEKMAJf ON THE K-PAETITIONS OF THE E-GON AND E-ACE. 
Problem x. To find I(r, h) (LXVIII.), 
I(r, ^)=i{D(r, ,!:))}. 
LXX. As a further illustration of these formulae and their use, and for a comparison 
of methods, it will be worth our while to deduce the seven partitions of the 9-gon, and 
the eight partitions of the 10 -gon, which are obtained by way of example in the memoir 
referred to, in the first article. 
First, to find R*(9, 6 ) and P(9, 6 ). 
As the 9-gon is divided into triangles, there can be only two nuclei, a triangle or a 
line. We ask, then, what are R*(9, 6)3 and R'‘(9, 6)2 % 
By Theorem E, the axes are monogonal only, and R*’"°(9, 6)3 is what we are seeking. 
By Problem m, 
Rr( 9 , 6)3=0, 
R--(9, 6)3=D(5, 2)=5; 
and by Problem n, 
Rf"(9, 6)3=(2R’"“^-1-R"'"“^‘'‘)(4, 1)=0, 
R 2 dt( 4 , 1)^=1 being the only entry in our register under (4, 1). 
For R™"(9, 6)3, we have, by equations (D'.), ^^^=1=A. 
9 = 3 -j- 2 -|- 2 Kq -|- 2 < 2 i , 
6=3+ £ 9 + 2 ^ 1 , 
4+2ao=8 — 2 <Zj, A,=D(2+a,, 
£ 0 = 3 — 2^1, 
Rr(9, 6)3=2.{5:(2R’”“^+R^’"“^''‘)(8-2a„ ^-2e,)}.'D{2+a„ e,). 
The only values of are 0 or 1 ; if ei = l, a,<j:2, if ^ 1 = 0 , And as we have 
nothing in our register under (R'"“«’ or R’”“^*)((4, 1 ) or ( 6 , 3)^, 
Rr(9, 6)3=0, 
and (Problems o, p, q) R(9, 6)2 =0 ; 
wherefore R(9, 6) =R™"(9, 6)3=5. 
Next to find P(9, 6 ) 3 ; by Problem u, 
P(9, 6)3=i{D(4, 1)-R^(9, 6)3}=i(2-0)=l, 
the subtracted term having just been proved to be zero. And as 9 is odd (Problem w). 
P(9, 6 ) 2 = 0 , wherefore (Problem x), 
1(9, 6)=^{D(9, 6)-9R(9, 6)-i^P(9, 6)=-5i8{429-9-5-6-l} = 21. 
This agrees with the results at p. 409 of my former memoir, saving an error of 
transcription in the seventh line from the bottom, in which +l.w( 8 , 2 , 0 ) is omitted 
from the value of 1(10, 2) as it stands above (line 14). If the indulgent reader will 
kindly correct the sixth and seventh lines thus. 
