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By Marko Petkovsek, Herbert S. Wilf

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The term ratio is tk+1 = tk n k+1 [x, d]k+1 [y, d]n−k−1 n k [x, d]k [y, d]n−k n−k−2 (n − k) kj=0 (x − jd) j=0 (y − jd) = k−1 n−k−1 (k + 1) j=0 (x − jd) j=0 (y − jd) (k − n)(k − xd ) (n − k)(x − kd) = . 1), so we have identified our series as the hypergeometric series −n − xd 1 . 4. Suppose we are wondering if the sum n n (−1)k k! k k=0 can be evaluated in some simple form. 2 The next step would then be to look up that hypergeometric series in the database to see if anything is known about it. Let’s do the first step here.

That’s a lot easier. ” All right. We’re trying to construct a collection of identities that will be equipped to discover if somebody’s sum can or cannot be expressed in a much simpler form. We’re two-thirds of the way there. We have a (slightly mythical) collection of data, and a single rather precise query. What we are missing is the algorithm. If some user asks the system whether or not a certain sum can be expressed in some simple form, exactly what steps shall the system take in order to answer the question?

2n)! n! Our answer now has become f (n) = 4n (−1)n (2n)! (−2n − 12 )! (n − 12 )! 2(−n − 12 )! (− 12 )! A similar argument shows that (−2n − 12 )! (−1)n (n − 12 )! = , (−n − 12 )! (2n − 12 )! 2 2n . n (2n − 12 )! (− 12 )! But for every positive integer m, 1 1 3 1 1 (m − )! = (m − )(m − ) · · · ( )(− )! 2 2 2 2 2 (2m − 1)(2m − 3) · · · 1 1 = (− )! 2m 2 (2m)! 1 = m (− )!. 4 m! 6 Using the database 47 So we can simplify our answer all the way down to f(n) = labor is that we have found the identity (−1)k k 2n k 2k k 2n 2 .

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