Tuyển tập các báo cáo nghiên cứu khoa học về toán học trên tạp chí toán học quốc tế đề tài: Automated Proofs for Some Stirling Number Identities. | Automated Proofs for Some Stirling Number Identities Manuel Kauers and Carsten Schneidery Research Institute for Symbolic Computation Johannes Kepler University Altenbergerstrafie 69 A4040 Linz Austria Submitted Sep 1 2007 Accepted Dec 4 2007 Published Jan 1 2008 Mathematics Subject Classification 68W30 Abstract We present computer-generated proofs for some summation identities for q- Stir-ling and q- Eulerian numbers that were obtained by combining a recent summation algorithm for Stirling number identities with a recurrence solver for difference fields. 1 Introduction In a recent article 5 summation algorithms for a new class of sequences defined by certain types of triangular recurrence equations are given. With these algorithms it is possible to compute recurrences in n and m for sums of the form n F m n h m n k S n k k 0 where h m n k is a hypergeometric term and S n k are . Stirling numbers or Eule-rian numbers. Recall that these may be defined via S1 n k S1 n 1 k 1 n 1 S1 n 1 k S1 0 k ỏ0 k 1 S2 n k S2 n - 1 k - 1 kS2 n - 1 k S2 0 k ỏo k 2 E1 n k n - k E1 n - 1 k - 1 k 1 E1 n - 1 k E1 0 k ỏ0 k. 3 mkauers@ Partially supported by FWF grants SFB F1305 and P16613-N12 lcschneid@ Partially supported by FWF grants SFB F1305. Part of this work was done while the two authors were attending the Trimestre on methods of proof theory in mathematics at the Max-Planck-Institute for Mathematics Bonn Germany. THE ELECTRONIC JOURNAL OF COMBINATORICS 15 2008 R2 1 The original algorithms exploit hypergeometric creative telescoping 9 . More generally the algorithms can be extended to work for any sequence h m n k that can be rephrased in a difference field in which one can solve creative telescoping problems. Since such problems can be solved in Karr s nS-fields 3 8 we can allow for h m n k any indefinitely nested sum or product expression such as q- hypergeometric terms harmonic numbers Hk E1 1 1 etc. Moreover S n k may satisfy any .
