EEV. T. P. KIEKMAN ON THE K-PAETITIONS OP THE E-OON AND E-ACE. 233 
r-\-n k—'U 
n ’ n 
)}’ 
unity; wherefore 
which is a number given in our register, or else =0. 
The next step is to put the subindex ■ — or -=:2/i, and find 
4ft 3 
"^agdi 
K (r, kX=0, 
because 2A= | does not give an integer 
— . agdi 
Ei ('■. k)., 
a number readily obtained, and so on through all integer values of from 2h~^ to 
. 2A=2. 
XXXI. For an example take 
r=22, ^=12, n~Q, h~l, and seek E^“^*(22, 12)g. 
By equations (A'.) we have 
2 2 = 6 -f- 4 -f- 4ceo -j- 4ffl, 
or ao>2, 4+2a„>8. 
Our register as far as r=8 contains, under 
0) = 1, 0)=1, 1)=1, 2),=1, E^''^‘**(4, 0)=1, 
where ao=2 ci^ — l ao=l ao=2 a~0 
Sq 0, 2g 0, £g 1 £q 2 Sg=0. 
But as 
12-6-2.S0 . . , 
7 is integer, s,= l=zoio; 
and the only solution of (A'.) is 
where 
22=6+2. (2+2. l)-f4. 2 
12 = 6+2.1 +4.1, 
c,=2, ^1=1. 
D(2+2, 1)=2 = 2A., 2H=E^“^"‘(6, 1)=1 
Er'^^(22, 12)6=2HA.-EL^n22, 12)6 
= 2 - 0 , 
T -\-7t 22 -|- 6 • f y. , . 
because ~~^= “g — ’ iii ^dcV-> f^)n^ is not integer. 
We proceed with our partitions of the octagon, and seek Edr^'*'(8, k)^. 
T ~^7l 7i 
Now E^'^*(8, >^:)6=0, because — ^ here is not integer: let w=4; then, — =0, the 
8 -f-4 
only integer value, and E^“^'''(8, ^) 4 = 0 , because -^=3, a value of r not found in our 
register under E“^ or E“^*. Therefore E^*'“^*'(8, k)—0. 
MDCCCLVII. 2 I 
