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4 votes
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For a combinatorial proof of a symmetric identity

In my paper Supercongruences involving dual sequences [Finite Fields Appl. 46(2017), 179-216], I gave a new symmetric identity which states that if $x+y=-1$ then $$\sum_{k=0}^n(-1)^k\binom xk^2\binom{...
Zhi-Wei Sun's user avatar
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2 votes
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Bijective proof of a combinatorial identity: $\sum\limits_{k=0}^n\binom nk^2 \binom k{n-m}=\binom nm \binom{n+m}m$

Identity \begin{equation} \sum_{k=0}^n\binom nk^2 \binom k{n-m}=\binom nm \binom{n+m}m \tag{1} \end{equation} was used in an answer here. As shown in that answer, (1) easily reduces to \begin{...
Iosif Pinelis's user avatar