442 
then directly and by (13) 
(a) C(k,o0,p) = 1 (OS Awan tae oy oaes 
Oe Des ae 
Ce) e—20 es RO eo ee yer Oe 
making use of (10) we obtain 
The left side vanishes when p = 1; therefore 
C(k,n,0) = —(k+n) C(k,n—1,0) 
By repeating this (n—1) times and noting that C(k,0,0) = 1, we obtain 
k = 0, 12, 00m 
(c) C(k,n,0) = (—1)"(K+1) (k+2) . . . . (k+4n) | 
n = 1,2,3,. 
Sinaia jy =, ay th bo be n, in (b), we find 
(d) C(k,n,p) = 0 LOT Dv ose ke n 
ate when p =n+1 
Therefore for all values of k = 0,1,2,..... ANC: — len es Oly ce eee 
n 
(14) > (—1)' A(k, k+-n—i) B(i, k-n—p) = (—1)"(K+1) (+2)... . (K+n) 
i=0 
when p = 0 
= 0 when p= 1, 2,3. en 
= k” when p=n+l1 
Example illustrating (14) fork = 2, = 3. 
Oe es 
| A(2,5—i)| 15 | —7 3 | —1 | sums of products 
p=0| B(i,5) 1 LB i) GSoel 22a Ol) 4b 
p=4| BEA | ox |) 10 Pes. 4) Soo 0 
p=, UR Gayee!| nd a 6 | 0 
p= 3 | B@,2) 1 3 2 |) 
p=4| Bil) 1 1 0 | 23 
In particular, when p = n, 
Y (-1)' Atk, k+-n—i) B(i, k) = 0 
i=0 
