Thursday, 6 October 2011

co.combinatorics - alternating sums of terms of the Vandermonde identity

So, you are interested in f(n,k)=sumki=0(1)ibinomkibinom2nkni.
Simple manipulations show f(n,k)=frack!(2nk)!(n!)2left[sumni=0(1)ibinomnibinomnkiright]
Now the second factor counts the coefficient of xk in (1x2)n and therefore if k is odd f=0 otherwise f=(1)frack2frack!(2nk)!n!(k/2)!(nk/2)! which is far from zero...



EDIT: On a different note I see the result is a signed generalized Catalan number of degree 2 (I was not aware they satisfied such simple identities). Since usually providing combinatorial interpretations for generalized Catalan numbers is not easy, may I ask in what combinatorial context did you face the above calculation?

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