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While the (common) value of the two determinants is symmetric in v and w, and in x and y, unlike the Jacobi-Trudi and N¨agelsbach-Kostka determinants, it is not symmetric in all four variables, much less in v, w, x, y, and z. The purpose of this article is to introduce p-Stirling numbers of the first and second kinds. | Turk J Math 24 (2000) , 379 – 399. ¨ ITAK ˙ c TUB The p-Stirling Numbers Russell Merris Dedicated to the memory of Henry C. Diehl. Abstract The purpose of this article is to introduce p-Stirling numbers of the first and second kinds. Key Words: Binomial coefficient; Character; General linear group; Partittion; Representation; Stirling number; Symmetric function. 1. Introduction Pronounced “m-choose-n”, C(m, n) is the number of n-element subsets of {1, 2, . . . , m}. Let Cm be the m-by-m matrix whose (i, j)-entry is C(i, j). Then, for example, 1 0 0 2 1 0 C5 = 3 3 1 4 6 4 5 10 10 0 0 0 0 0 0 . 1 0 5 1 −1 Because det(Cm ) = 1, not only is Cm invertible, but Cm is an integer matrix. Indeed, −1 can be among the many wonderful properties of binomial coefficients is the fact that Cm −1 is obtained from Cm by inserting a few well chosen minus signs: The (i, j) -entry of Cm (−1)i+j C(i, j). Thus AMS Numbers: Primary 05A05; Secondary, 15A69 379 MERRIS C5−1 1 0 −2 1 = 3 −3 −4 6 5 −10 0 0 0 0 1 0 −4 1 10 −5 0 0 0 . 0 1 One way to insert this “checkerboard array” of minus signs is by means of matrix multiplication: −1 Cm = Dm Cm Dm , (1) −1 = diag(−1, 1, −1, . . . , (−1)m ) is the m − by − m diagonal matrix of where Dm = Dm ∓1’s. The Stirling number of the second kind, S(m, n), is the number of ways to partition {1, 2, . . . , m} into a disjoint union of n nonempty subsets caled the parts of the partition. The 2-part partitions of {1, 2, 3, 4} are {1} ∪ {2, 3, 4}, {2} ∪ {1, 3, 4}, {1, 2} ∪ {3, 4}, {3} ∪ {1, 2, 4}, {1, 3} ∪ {2, 4}, {4} ∪ {1, 2, 3}, and {1, 4} ∪ {2, 3} Thus, S(4, 2) = 7. More colorfully, S(m, n) is the number of ways to distribute m distinguishable (labeled) cows among n identical (unlabeled) pastures, with each pasture containing at least one cow. So, S(m, m) = 1 = S(m, 1), m ≥ 1, and S(m, n) = 0 if m ≥ 1 > n or n > m. Moreover, S(m + 1, n) = S(m, n − 1) +
