Combinatorial Identities for Stirling Numbers: The by Jocelyn Quaintance, H. W. Gould

By Jocelyn Quaintance, H. W. Gould

This e-book is a distinct paintings which gives an in-depth exploration into the mathematical services, philosophy, and information of H W Gould. it truly is written in a mode that's available to the reader with uncomplicated mathematical wisdom, and but comprises fabric that would be of curiosity to the expert in enumerative combinatorics. This booklet starts with exposition at the combinatorial and algebraic thoughts that Professor Gould makes use of for proving binomial identities. those ideas are then utilized to boost formulation which relate Stirling numbers of the second one type to Stirling numbers of the 1st variety. Professor Gould's thoughts additionally offer connections among either different types of Stirling numbers and Bernoulli numbers. Professor Gould believes his study good fortune comes from his instinct on the right way to observe combinatorial identities.

This ebook will entice a large viewers and will be used both as lecture notes for a starting graduate point combinatorics type, or as a examine complement for the expert in enumerative combinatorics.

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22) n − 2i = 0 if n > 2i. 22), we conclude that n (−1)i i=0 m i 2n − m n−i n (−1)i = (−1)n i=0 ⌊n 2⌋ (−1)i = i=0 m i 2n − m i 2n − i i 2 n−i 2n − 2m . 23) Through a collection of hand calculations Professor Gould conjectured that n (−1)j j=0 m j 2n − m n−j = (−1)n 22n m−1 2 n . 24) page 41 September 15, 2015 12:0 ws-book9x6 42 Combinatorial Identities for Stirling Nu... master Combinatorial Identities for Stirling Numbers He then proved his conjecture by applying the fundamental theorem of algebra.

We obtain n−b n−k n−b i ai,n−k = an−i,k+n−i . 18) to obtain n−b i n−b n−b an−i,k+n−i = i=0 k=0 n n−b an−i,k+n−i = k=0 i=k n n an−i,k−b+n−i k=b i=k−b an−i+b,k+n−i . 19) are just three of the many variations of the standard interchange formula. 20) i=0 k=0 k=b i=0 whose proof is as follows: n an−i,k ai,k = ai,k = i=0 k=n−i i=b k=i k=b i=b n n−b n n k n−b 0 = n−b i an−i,n+k = i=0 k=−i an−i,n−k . 20) has the symmetrical form n n k i ai,k = an−i,n−k . i=0 k=0 k=0 i=0 We end this section with an example of how to use the standard interchange formula in the derivation of a combinatorial identity.

We now ask a related question of how to multiply two series. Suppose we want to calculate ni=1 ai nj=1 bj . By defining ai,j = ai bj , we convert ni=1 ai nj=1 bj into ni=1 nj=1 ai,j and apply the techniques used to derive the iterative series identities of the previous section. In particular, given the iterative series ni=1 nj=1 ai,j we form the two-dimensional array a1,1 +a2,1 +a3,1 +... +an−1,1 +an,1 +a1,2 +a2,2 +a3,2 .... +an−1,2 +an,2 +a1,3 +a2,3 +... +an−1,3 +an,3 +... +... +a3,n−2 .... +...

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