436 



whence S(k, n) is divisible by k! and in fact S(n, n) = ( — \) n 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 ( — 1) . 

 These numbers possess a recursion formula 



(4) S{k, n) = k[S{k, n— 1) — S(k — 1, n — 1)] n, k = 0, 1, 2, . . . . 



by means of which may be constructed, 



A TABLE OF VALUES OF S(k, n) 





fc = 



Jb=l 



k = 2 



k = 3 



k = 4 



k = 5 



k = Q 



k = 7 



k = 8 



n = 



1 



















71= 1 







— 1 

















n = 2 







—1 



2 















n = 3 







— 1 



6 



—6 













n = 4 







— 1 



14 



—36 



24 











n = 5 







— 1 



30 



—150 



240 



—120 









n = 6 







— 1 



62 



—540 



1560 



—1800 



720 







n = 7 







— 1 



126 



—1806 



8400 



—16800 



15120 



—5040 





71 = 8 









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. 



(5) 



(6) 

 (7) 

 (8) 



2 S(k, n) = (-1)" 



v Sjk,n) = Q 

 k = i k 



2 S(k, n) = 1 + cos n% 



k = 2 



n = 2, 3, 4, . . . 



V ^J 1 ] S(Jb, »)-(* + 1)2 ?).50b,t) 



i = A 



2 ["j S(fc, i) = Sik, n) - S(k + 1, n) 



S(k,k+ 1) = \ k 2 1 ) S(k,k) 



Setting n = k 4- 1 in (7) we obtain 



(9) 



and similarly we can express S(k, k + 2), £(&, fc + 3), etc. in terms of S(k, k). 



From (4) 



Sik, n) = Sik + 1, n) — =-^ S(* + 1, n + 1) *, n = 0, 1, 2, 3, . 



k + 1 



