188 ON THE NUMBER OF DISTINCT TERMS IN A SYMMETRICAL OR [580 
and using the first of these to eliminate the term <£(m —3, 1) and resulting terms 
(f) ( m _ 4 ( i) 5 &c. which present themselves in the second equation, this, after a succession 
of reductions, becomes 
(f> (m, 0) = 1, 0) 
+ (m — 1) </> (m — 2, 0) 
m .ni—1 . , . n 
H 2 10 ( m — 
+ (ra — 3) (f) (m — 4, 0) 
+ (m — 3) ... 3.2<£ (1, 0) 
+ (m-3)... 3.2.1 }; 
or, observing that the last term (m — 3) ... 3.2.1 is, in fact, = (ra — 3) ... 3.2.1(f) (0, 0), 
this may be written: 
2(f) (ra, 0) — (f> (ra — 1, 0) — (ra — 1) </> (ra — 2, 0) = 0 (m — 1, 0) 
+ ('m — 1) cf> (ra — 2, 0) 
+ (ra — 1) (m — 2) (f> (m — 3, 0) 
+ (ra — 1) .. 3.2.1<£ ( 0,0). 
And hence assuming 
rr> rp 2 /vW 
u = <f)(0, 0) + |«/>(l, 0) + r -2^(2, 0)+ ■..+ 1 ; 2::: ^<A(m, 0)+..., 
we find at once 
2-t u — xu = , 
a# 1 — x 
that is, 
2 — = cfoc (1 + # + 
1 -x ’ 
or integrating and determining the constant so that u shall become =1 for x = 0, we 
have 
u = 
gte+i*“ 
Vl — X ’ 
wherefore we have 
glz+iz' 2 
cf) (ra, 0) = 1.2 ... m coefft. x m in 
vl —a