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3 votes
1 answer
127 views

Bijective proof of deteminant formula for Hankel matrix of central binomial coefficients

Is there a nice bijective proof of the fact that the determinant of the $(n+1)$-by-$(n+1)$ Hankel matrix whose respective entries are the central binomial coefficients $0 \choose 0$, $2 \choose 1$, $\...
James Propp's user avatar
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2 votes
0 answers
495 views

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
4 votes
0 answers
225 views

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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