270 EEV. T. P. KIEKMAN ON THE K-PAETITIONS OF THE E-GON AXD E-ACE. 
Therefore 7— e)„ is what we have to determine for every value of h. e. 
and n. 
Now in R*‘''(10, 7 — e)„ (Problem 1), when h is odd, 10, 7 — e and n are not all divisible 
by 2^ if ?^>6; for w<10. And in Rf'(10, 7— (9)e, h is either 1 or 3. For A=3 the 
formula for R^ gives R^*'(10, 7— e)e=0, because ^ is not integer; and for A=l, 
— — 3 
10=6-4-2.(ai+a2+«3) 
7 — e= 6 +2, ( 6 ,- 1 - 62 - 1 - 63 ) ; 
the solutions are 
^-^=2=l-l-l-l-0, or = 24 - 0 - 1-0 
1~6=2(0-1-0-1), or =2(1-1-1), or =2(0-l-l), 
( 6 — 2\ 3i-i 
~~ 2 ' ) 
Therefore n is not 6. 
Let then w=4, and seek R*‘^'(10, 7 — 6 ) 4 , for h odd. If ^ =1, the formula for R^* 
10 + 4 . 4 
gives Rf*(10, 7— 6 ) 4 = 0 , because — ^ is not integer. Then ^=2, A=l, and we have 
for 6=1, 
10 = 44 - 2 . ( 2 - 1 - 1 )= 44 - 2 ( 34 - 0 ) 
6=4-f2 (14-0)=44-2(2~l), 
( 4 
j vanishes. These four are 
then the only solutions, and 
2(A2)=2D(4, 1)D(3, 0)4-2D(5, 2)=2(2-k5)=14 ; 
whence Rf(10, 6 ) 4 =i{ 14 - 2 X 2 R'*+‘)‘'' 4 -Rif 6 ) 4 } 
where (^> 0 ). 
Now 
6 ) 4=0 (Problem 1), because 10 and 6 and 4 have no common measure 2(^4-l)• 
Rff (10, 6 ) 4=0 (Problem k), because 10 is not divisible by 4. 
R*' ^'(10, 6 ) 4=0 (Problem b), because 10 is no multiple of M. 
R 2 iaff<i! 6 ) 4=0 (Problem c), because 10 — 4 is no multiple of 47. 
Therefore Rf(10, 6 ) 4 = 414=7 = Rf(10, 7), 
and R(10, 7)=R'^'(10, 7 ) 2 -l-Rf( 10 , 7)=2Rf(10, 6 ) 4 = 14 . 
LXXII. We have next to determine P(10, 7), A>1. 
The formulae for 1™ (Problem u) give 
P(10, 7 ) 3 = 0 , because — 3 ^ is no integer. 
