Stirling1( n, k )
Stirling1 returns the Stirling number of the first kind S_1(n,k) of
the integers n and k. Stirling numbers of the first kind are defined
by S_1(0,0) = 1, S_1(n,0) = S_1(0,k) = 0 if n, k <> 0 and the
recurrence S_1(n,k) = (n-1) S_1(n-1,k) + S_1(n-1,k-1).
S_1(n,k) is the number of permutations of n points with k cycles. Stirling numbers of the first kind appear as coefficients in the series n! {x choose n} = sum_{k=0}^{n}{S_1(n,k) x^k} which is the generating function for Stirling numbers of the first kind. Note the similarity to x^n = sum_{k=0}^{n}{S_2(n,k) k! {x choose k}} (see Stirling2). Also the definition of S_1 implies S_1(n,k) = S_2(-k,-n) if n,k<0. There are many formulae relating Stirling numbers of the first kind to Stirling numbers of the second kind, Bell numbers, and Binomial numbers.
gap> List( [0..4], k->Stirling1( 4, k ) );
[ 0, 6, 11, 6, 1 ] # Knuth calls this the trademark of $S_1$
gap> List( [0..6], n->List( [0..6], k->Stirling1( n, k ) ) );;
gap> PrintArray( last );
[ [ 1, 0, 0, 0, 0, 0, 0 ], # Note the similarity
[ 0, 1, 0, 0, 0, 0, 0 ], # with Pascal\'s
[ 0, 1, 1, 0, 0, 0, 0 ], # triangle for the
[ 0, 2, 3, 1, 0, 0, 0 ], # Binomial numbers
[ 0, 6, 11, 6, 1, 0, 0 ],
[ 0, 24, 50, 35, 10, 1, 0 ],
[ 0, 120, 274, 225, 85, 15, 1 ] ]
gap> Stirling1(50,10);
101623020926367490059043797119309944043405505380503665627365376
GAP 3.4.4