By Petkovsek M., Wilf H.S., Zeilberger D.

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Sin nπ)(n − 1)! π (n − 1)! Thus as n approaches a positive integer, we have found that (2n)! (−n)! −→ (−1)n . (−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 .

It’s a myth. Is there a collection of information? The data might be, for instance, a list of all known hypergeometric identities. But there isn’t any such list. If you propose one, somebody will produce a known identity that isn’t on your list. But suppose that problem didn’t exist. Let’s compromise a bit, and settle for a very large collection of many of the most important hypergeometric identities. Fine. Now what are the queries that we would like to address to the database? That’s a lot easier.

N! so we must be doing something wrong. Well, it turns out that if you would really like to simplify ratios of factorials then the thing to do is to read in the package RSolve, because in that package there lives a command FactorialSimplify, which does the simplification that you would like to see. Out[2] = 28 Tightening the Target So let’s start over, this time with In[1] :=<< DiscreteMath‘RSolve‘. ] and we get Out[2] = 1 + n, which is what we wanted. 1). ). The rational function certificate (the key to the lock) for this identity is R(n, k) = − k 2 (3n + 3 − 2k) .

### A=B by Petkovsek M., Wilf H.S., Zeilberger D.

by William

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