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Hi everyone,

In the process of studying a problem in the Johnson association scheme I came across the following sum: $$\sum_{k\geq 0}(-1)^k\binom{n}{k}\binom{a-k}{a-b}\binom{c+k}{b}.$$ All the variables are non-negative integers. I've tried to no avail to simplify this expression using Gosper's algorithm, as well Wilf-Gosper (but it becomes unwieldy).

Is there perhaps a simpler form for this sum? Is there any connection with Eberlein polynomials?

Thanks in advance

Moshe

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2 Answers 2

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This is a hypergeometric function times a binomial coefficient, isn't it? To see exactly which one, one can do the usual procedure described in books, e.g. in A=B.

Then, perhaps, hypergeometric identities can be applied, but at least having your sum encoded like this might help.

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  • $\begingroup$ The sum, in fact, equals $\binom{a}{b}\binom{c}{b} _{3}F_{2}(c+1,−b,−n;c−b+1,−a;1)$. I've already tried looking up hypergeometric identities but couldn't find one that does the job and reduces it to something nicer. $\endgroup$ Jan 10, 2013 at 17:26
  • $\begingroup$ Well, it might be it - no simpler formula. Often this is the case, e.g. check out recent work of A.Schrijver and his collaborators on improved bounds for codes, such as arxiv.org/abs/1005.4959, homepages.cwi.nl/~lex/files/codes.pdf. However, often sums like yours come up as entries of a certain matrix, and this matrix might have a reasonably nice $LU$ decomposition. E.g. in your case $L$ might be having entries ${n \choose k}$. $\endgroup$ Jan 11, 2013 at 1:48
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Where did it come from? What kind of answer are you looking for and why do you expect that there would be one? In the case $n,a,b,c=9,4,3,6$ your sum has four non-zero terms and comes out to the negative of a prime number: $-3889.$ So there is likely not an expression as a simple product.

MUCH LATER On the other hand, a small search (most cases with $n \le 30,a \le b \le 50, c \le 50$) turned up a few hundred cases where the sum is prime, but all of them had either $b \in \{{2,3,4\}}$ or $a=b$ or $a=b+1$ so perhaps examination of the factorizations would be productive. Two more prime cases are $19,14,4,26$ and $19,14,4,23.$

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