436 
whence S(k, n) is divisible by k/ and in fact S(n, n) = (—1)" n! Also, since 
S(1, n)<0, it follows that for fixed k, S(k, n) preserves a constant sign (or 
vanishes) for all values of n; and this sign is the same as that of C05 
These numbers possess a recursion formula 
(4) S(k, n) = k[S(k, n— 1) —S(k—1, n—1)] n, k = 0, 1, eee 
by means of which may be constructed, 
A TABLE OF VALUES OF S(k, n) 
| 
| r=ok=11 k=2| h=8 | k=4 | k=5 | k=6 | k=7 | B= 
| | _ | | 
20) | | 
n=1') 0|-1| 
neo a | 2 
n=3 | O|-1] 6 | -6 
n=4 | O|-1|] 14 | —36| 24 
n=5 | O|-1| 30 | —150| 240 | —120| 
n=6 | O|-1| 62 | —S540/ 1560 | —1800| 720 | 
n=7 | O}-1| 126 | —1806 | 8400 16800 | 15120 | —5040 
| n=8 | 0|-1) 254 | —5796 | 40824 126000 191520 —141120) 40320 
Subtract any entry from the one on its right, multiply by the value of k above the latter. 
n n 
(5) pe S(k, n) = (—1)" p> S(k, n) = 1+ cos nx 
k=0 k=2 
(6) vy Sth, n) _ 9 n= 23. 
k=1 
(7) > [" Z z S(k,i) = (kK+1)2 | S(k, i) 
i=k * i=k * 
(8) > ("| sc, i) = S&, n) — SE +1, 2) 
i=k : 
Setting m = k + 1 in (7) we obtain 
(k 
Ly ree 
3 | Si, k) 
(9) S(k,k +1) = 
and similarly we can express S(k, k + 2), S(k, k + 3), ete. in terms of S(k, k). 
From (4) 
Rikon = Si. i n) — 1 Skt oes ken = 0,4 
