448 



changing the order of summation and replacing g( — l,x,k,n) by its value, 



we have 



k j 



2 x ; 2 (— 1) J_1 B(k— j+i,k) S(k,n— i) . 



(14) /(— l,x,k,n) = j=1 i=1 



x(x+l) (x+2) (x+k) 



n > k = 0, 1, 2 



the numerator being a polynomial arranged according to ascending powers 

 of x; on arranging this in descending powers of x, taking account of (14) §3. 



(15) f(—l,x,k,n) 



k—l j 



2 x h ~ j 2 (— 1)* B(j— i,k) S(k,n+i) 



x(x+l) (x+2) (x+k) 



n > k = 0, 1, 2, 3 



It is obvious that (14) does not hold for n < k, since in that case 



S(k,n — i) vanishes, i = 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 n~<k, the coefficient of every 

 power of x vanishes except that of .t and this turns out to be 



(_!)*-« B(0,k)S(k,k) = (—l) n k> which agrees with (12). 



Therefore, 



A— 1 j 



k 1 x~ ] 2 (— 1)* B(j—i,k) S(k,n+i 



2 (— 1) 



y i i\» i k ) i>l _ i=o 



x + l x(x+l) (x+2) (x+k) 



k,n = 1. 2, 3 



but for the case where n ~< k, (12) is simpler. 



Setting in = 2 in (11) 



(17) f(—2,x,k,n) = (co + an) (—x) n n = 0, 1, 2, k—l. 



Put n — 0, n = 1, and determine 



Co = /(— 2,x,k,0) 



ci = — - /(— 2,a;,jfc,l) — f (— 2,x,k,0), which by (7) 

 x 



= -/(— 2,x+l,k— 1,0) — /(— 2,a;,A-,0) 



