448 
changing the order of summation and replacing g(—1,2,k,n) by its value, 
we have 
Ls 
> 2 > C1)! Be Jj+ik) S(k,n—) 
=1 i=1 
ral Ger rol 8 ed ene Bee lee (a+k) 
(14) fC—-L2.km) = ! 
the numerator being a polynomial arranged according to ascending powers 
of zx; on arranging this in descending powers of z, taking account of (14) §3, 
k—1 a : 
> A> C4) BiH, k) S(k,n4+4) 
(15) (‘kn == 
Z(c-+-)) (G2E2) 2 ees ee (c+k) 
a>k=0;1,2,3°:2 .. ae 
It is obvious that (14) does not hold for ee k, since in that case 
S(k,n—i) vanishes, 1 = 1,2,..... n; on the other hand, noting that B(k,n) 
and S(k,n) both vanish if k > n and taking account of (15), §3, it results that 
in the numerator on the right side of (15), when Te k, the coefficient of every 
power of x vanishes except that of x" and this turns out to be 
(—1)*™ B(0,k)S(k,k) = (—1)"k! which agrees with (12). 
Therefore, 
k—1 fj 
i é = tI S (1)! BiH k) S(k,n+1) 
(6) > (41) ;| oS —_— 
i=0 Se roe dee (ed) 2) oo ne oe (a+k) 
kin, =o, 2,00. 2 eee 
but for the case where n = k, (12) is simpler. 
Setting m = 2 in (11) 
(17) f(—2,2, k,n) = (Co == cn) (—z)" nm = 0; A se <  #, B atta ee k—1. 
Put n = 0, n = 1, and determine 
Co = f(—2,2,k,0) 
c= — + f(-22,k,1)—f(22,k,0), whieh by (7) 
- * f(2c+1k—-1,0) — f(—2,2,k,0) 
