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3 votes
2 answers
244 views

Is there a combinatorial reason for variable-independence of this binomial-coefficient identity?

Consider the following identity $$\sum_{n=0}^{R-t}\binom{n+\ell}n\binom{R-\ell-n}{R-t-n}=\binom{R+1}{t+1}.\tag1$$ It is relatively easy to give an algebraic or mechanical proof of (1). But, I like to ...
6 votes
5 answers
944 views

Combinatorial proof of Catalan's identity

Consider the problem of tiling a board of length $n$ with squares of size $1×1$ and dominoes of size $1×2$, Let's denote $f_n$ to be the number of ways to tile this so-called ($n$)-board.Then $f_n=F_{...
6 votes
2 answers
1k views

Products and sum of cubes in Fibonacci

Consider the familiar sequence of Fibonacci numbers: $F_0=0, F_1=1, F_n=F_{n-1}+F_{n-2}$. Although it is rather easy to furnish an algebraic verification of the below identity, I wish to see a ...
8 votes
2 answers
325 views

A link between hooks and contents: Part II

This is a question in the spirit of an earlier problem. Let $\lambda$ be an integer partition: $\lambda=(\lambda_1\geq\lambda_2\geq\dots\geq0)$. Recall also the notation for the content of a cell $...
20 votes
1 answer
1k views

A proof required for this identity [duplicate]

Experiments support the below identity. Question. Is this true? Combinatorial proof preferred if possible. $$\sum_{m=0}^n\binom{n-\frac13}m\binom{n+\frac13}{n-m}(1+6m-3n)^{2n+1} =\left(\frac43\...