450 
The corresponding results for n = k+2,k+ 3, ete., may be found by 
putting these values successively for n in 
(21) f(2,2,k,n+2) = S(k,n) — 22 f(—2,2,k,n+1) — x? f(—2,2,k,n) 
which results from setting m = 2 in (9). The general result may be put 
into the form 
2-2 kd c 
Sak F S DGji,k) Skin) 
(22) f(—2,2,k,n) = 3=2 ze kan = 1, 23 
7d oes lL) Pe a (a+k)? 
in which the coefficients D, are independent of n: 
OER) — st whens: — 0 
= 0 fe NO a acy ee tae 
io 
D(0j,k) = > B(t, k—1) B(j—t, kK—1) j= 1, 2,300 
t=0 
but I have not been able to determine a general formula for D(i,j,k) by means 
of which to calculate the coefficients of f(—2,z,k,p), p>k+1, without first 
calculating successively those for n = k+2,k+3,..... p—i. 
By making use of (10) § 2, (21) may be reduced to 
2 I 
Digest oS E(i,j,k) S(k,n+7) 
j=0 i=0 
(23) f(—2,2,k,n) = kin = 1,-2) 3a 
with which compare (16) 
Example: 
4 
a?(x-+1)?(2+2)2(2+3)2(x+4)? > (—1)' 
i=0 
{= 
(4 qn 
= 8 8 
li] tai)" S(4,n) x + 
(12 S(4,n) + 8 S(3,n)] 2? + 
[58 S(4,n) + 76 S(3,n) + 36 S(2,n)] 2° + 
(144 S(4,n) + 272 S(3,n) + 288 S(2,n) + 96 S(1,n)] x5 + 
[193 S(4,n) + 460 S(3,n) + 780 S(2,n) + 720 S(1,n)] 24 + 
[132 S(4,n) + 368 S(3,n) + 840 S(2,n) + 1680 S(1,n)] 2? + 
[36 S(4,n) + 112 S(3,n) + 312 S(2,n) + 1200 S(1,n)] x? 
wat Sn ke as 
