From: "Burcin Erocal" <[EMAIL PROTECTED]>
>
> it doesn't give an answer. This means that your expression doesn't have
> a hypergeometric closed form in the sense of A=B, p. 143 [1]:
>
> http://www.cis.upenn.edu/~wilf/AeqB.html

Is this a joke?

After converting binomial coefficients to Pochhammer symbols, it is exactly 
the definition of Hypergeometric1F1[-n,-2n,2x] for positive integer n, or 1 
for n=0.

> Sage will have symbolic summation capabilities soon, but I doubt if it
> will ever be able to simplify this expression.

I hope it will. Eric Weisstein wrote to me that the current development 
version of Mathematica gives the answer Hypergeometric1F1[-n,-2n,2x].

> What are your expectations from a "correct answer"? How did you come
> across this example? If you know/conjecture a simpler form for this
> expression, you can prove that they are equal by showing that they both
> satisfy the above recurrence, and agree on the inital values.

I came across this example in a recent thread in Maple newsgroup.

The correct answer is 1 for n=0 and Hypergeometric1F1[-n, -2 n, 2 x] for
integer n>0, which would be equal to the expression given by Mathematica if
n was not a positive integer.

Another form of the correct answer is

(2 x)^(n+1/2) E^x BesselK[n+1/2,x] n!/(2 n)!/Sqrt[Pi]

Alec


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