439 



(7) 2 [!j] A(k,i) =il(A; + l,n + l) 



(8) 2 A(k, n) S(k — 1, k — 1) = 

 fc=l 



Inversely, since 



.T (r ' +1) = X(X — 1) (X 



2) 



(a; — n ) 



re>£ = 0, 1, 2, . 



n = 2, 3, 4, . 



re = 0, 1, 2, ... . 



if we set 



4- (— if B(k, n)x n ~ k + 



(9) 2 r "+ 1( = x [B(o, n)x n — S(l, «)/ l + . 



+ (—1)" B(n, re)] 



it is evident that B(o, n) = 1, n = 0, 1, 2 B(k, n) = the sum of the 



products of the numbers 1, 2, 3, n. taken k at a time; in particular 



B(k, k) = k! = ( — 1) S{k, k) and B(k, n) = if k>n. For convenience define 

 B(p, n) = 0, if p is a negative integer. 



If we multiply both sides of 



B(l,n — l)x n 2 + 



+ (—1)" 'B(n 



l,n — 1)] 



n— i 



z (M) =i[B(o,b-1K 



by x — n, and equate the coefficients of x" ", we obtain the recursion formula 



(10) B(k, n) = B(k, n — 1) + re £(A~ — 1, n — 1) 



by means of which may be constructed 



A TABLE OF VALUES OF B (k, n) 





£=0 



k = l 



k = 2 



k = 3 



& = 4 



k = 5 



k = 6 



k = 7 



k=8 



n = 



1 



















n = 1 



1 



1 

















re = 2 



1 



3 



2 















n = 3 



1 



6 



11 



6 













n = 4 



1 



10 



35 



50 



24 











re = 5 



1 



15 



85 



225 



274 



120 









re = 6 



1 



21 



175 



735 



1624 



1764 



720 







re = 7 



1 



28 



322 



1960 



6769 



13132 



13068 



5040 





re = 8 



1 



36 



546 



4536 



22449 



67284 



118124 



109584 



40320 



Multiply any entry by the number (n+1) of the next row, and add to the entry on its right, 

 n-l-i- 



(11) B(k,k+n) = 2 \l\ B(k + re — i, k + n — 1) k, re = 0, 1, 2, 3. 



i=k 



