WHICH SATISFY GIVEN CONDITIONS. 
101 
The particular cases ( m , n, a)=(3, 6, 18), (3, 4, 12), (3, 3, 10) give respectively 
0=252-f9«+18c, 
0= 64+7a + 12c, 
0= 36 + 6a+10c, 
satisfied by «=104, c— — 66. 
59. Calculation of (3, 2). We have 
(3, 2)„ + .,-(3, 2)„-(3, 2).,=(3W2)„. + (3U2)„ 
= — 4 (m«' + Wa) — 4 {not! n'a ) -f 6aa' : 
the integral is 
(3, 2 ) m = «(m -f- w) + a( — 4m — 4n + c) + 3a 2 , 
and, as before, 
0=324+9a+18c, 
0= 96 + 7«+12c, 
0= 60+6a+ 10c, 
satisfied by «=120, c— — 78. 
60. For the calculation of (3, 1, 1) we have similarly 
(3, 1, 1,Vh,-( 3, 1, l) m — (3, 1, 1)..=(3).(1, l)m'+(3) m /(l, l) m 
+ (3, l) m (l) m /+(3, l) m /(l) m . 
The function on the right-hand side was of course calculated from the values of 
(3) m (l, l) m , &c. ; but there is no use in this (and the more complicated cases which 
follow) in actually writing down the values of the function in question ; it can in each 
case be calculated backwards from the foregoing expressions of (3, 1, 1) &c., and the 
values so obtained be verified by actual substitution. But assuming it to be known, 
the solution of the functional equation gives of course the foregoing expression for 
(3, 1, 1), except that the terms in m-\-n and a are therein a{ra-\-n)-\-ca ; and I shall in 
this and the subsequent cases give only the three equations which determine the constants. 
In the present case these are 
— 332 + 9a+18c=0, 
— 454+7«+12c=0, 
— 306-f 6<z+10c=0, 
satisfied by «= — 434, c=291. 
61. Remaining cases (2, 2, 1), (2, 1, 1, 1) and (1, 1, 1, 1, 1). We have 
(2, 2, l) m+m ,— (2, 2, 1),= (2, 2) m (l) w + (2, 2) w (l) m 
+ (2, l)m(2)m/ + (2, l) m ,(2) m , 
and 
-1674+9a+18c=0, 
— 648+7a+12c=0, 
— 462 + 6a+10c=0, 
satisfied by a= — 468, c=327. 
Q 2 
