449 
In (8) set i = —2,k = 1 
2,1,0) = f(—1,2,1,0) + f(—2,2+1,0,0) 
af eae, 7-9 
whence by (12) and (3) 
1 il 
Tes ey [OY ae Ree 
f( 2,2,1,0) = x 2(x+1) 5\= a(a+1)2 
1! : 
- > 3h . . 1 
See Ey eS Le) tah) 
FT epic aN gee aaa 
Again, setting k = 2 in (8) 
y 
f(—2,2,2,0) = = f(—2,2,1,0) + = f(—2,2+1,1,0) 
2! z 
a. > 
= St BQ-i.2 
PCr CE teen aaa ali 2) 2 
Assume 
k! : 
18 = 2, ae — = a } B(k—i,k 
a I ED) aod (ea ol ea ee (ck)? fe =e eee 
and a complete induction, on taking account of (11) $3, shows that this holds 
for all positive integral values of k. 
Therefore: 
k! £ 
.= : Zl B(k—i,k 
Syaee tas eee ame cee 
k 
k! Ss 
1 bse as. Bepeeoe ik) & 
and 
(19) f(—2,z2,k,n) = (ope S (1t+i—n) B(k—i,k) 2 
ip (a ell) eae oe GEA) =o 
ip = (00 th 2 Pat (Jean BARS ars ete prea k—1 
On computing, by means of (10), the values of f(—2,x,k,k) and 
fC2, z, k ,k +1), we verify that. (19) holds forn=1,2,3....... k+1 
but not for n>k+1, 
Therefore, 
k 
(ke) in (—x)" k! . 
20) Seay - = 1+i—n) B(k—i,k 
eae Gan Per Gah. je 
a0) Slee Deedes etn Sue OSL ph ee Steg os k+1; not n>k+1 
29—4966 
