ON GENERALIZED STIRLING NUMBERS AND POLYNOMIALS

( )n∑ n(10) S (α) (n + 1, k) − αS (α) (n, k) = S (α) (i, k − 1),ii=0(n∑ j(11)(α + j) n = i!Si)(a) (n, k)i=0and related polynomialsA n (x, α) =with generating function (see [6]):n∑S (α) (n, k)x kk=0(12)∞∑n=0t nn! A n(x, α) = e αt−x(1−et) .Starting with the recurrence relation for numbers S (α) (n, k) (8), and usingsupstitution α → α r, we getr n S ( α r ) (n, k) = rr n−1 S ( α r ) (n − 1, k − 1) + (α + rk)r n−1 S ( α r ) (n − 1, k)We see that(13)S (α) (n, k, r) = r n S ( α r ) (n, k)because the recurrence relation for numbers S (α) (n, k, r) isS (α) (n, k, r) = rS (α) (n − 1, k − 1, r) + (α + rk)S (α) (n − 1, k, r)In the same way from (10) and (11), we have other recurrences which are provedby Sinha and Dhawan [17] and Shrivastava [13].Now, we consider the generating function (9). Using supstitution α → α rand relation (13), we get∞∑n=0S (α) (n, k, r) tnn! = ∞ ∑n=0S ( α r ) (n, k) (rt)nn!= (−1)k e αt (1 − e rt ) k ,k!i.e the result from [17] and [13].For the polynomials T n (α) (x, r, −p), using supstitution x → px r , α → α r andt → rt, we have from (12)∞∑ (rt) nA n (px r , α n! r ) = (1−e rt )eαt−pxrn=03