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Consider an $n \times 1 - $rectangle where the $n$ squares are numbered $1$ to $n$. Cover this rectangle with white squares, black squares, and dominoes. To each covering of the rectangle associate …

The Narayana numbers defined by $N_{n,k}=\frac{1}{k+1}\binom{n}{k} \binom{n+1}{k},$ for $n\geq0,$ $0\leq k \leq n,$ and $N_{n.k}=0$ else, can be constructed recursively via the recurrence $$\binom{n+ …

The idea of Fedor Petrov led me to the following proof of the above formula. In arXiv:1203.5424 R. Duarte and A. G. de Oliveira gave a combinatorial proof of the identity $$S_m(k)=\sum_{i_1+\dots …

Denote by $ {n\brack {k}}$ a $q-$binomial coefficient. Let ${D_{n,k}}(t,q) = \sum\limits_{j = 0}^{n - k} {{q^{{j^2} + kj}}}{n\brack {j}}{n\brack {k+j}}t^j $ and ${R_n}(x,t,q) = \sum\limits_{k = 0 …

In his paper “Ein Beitrag zur additiven Zahlentheorie und zur Theorie der Kettenbrüche” I. Schur has stated that Frobenius has communicated to him a simple direct proof of his finite version of the Ro …

Is the following result about Hankel determinants known or a simple consequence of some known results? Let $f(x) = \frac{\displaystyle 1}{{\displaystyle 1 - \frac{{a x^{m + 2}}}{\displaystyle {1 - \f …

The formula for the measure of the Motzkin numbers as stated by Gjergij follows from the formula for the Catalan numbers if we write the formula for the Catalan numbers in the form $\frac{1}{{2\pi }}\ …

Let $a(n)$ be the number of lattice paths in ${\mathbb{Z}^2}$ of length $n$ which start at the origin $(0,0)$ and end up at $(n,0)$ and have only up-steps $U:(i,j) \to (i + 1,j + 1)$, down-steps $D:( …

I stumbled upon the following (perhaps well-known) identity for a positive integer $k$: $$\sum_{j=0}^n\frac{1}{(k-1)j+1}\binom{kj}{j}\binom{k(n-j)}{n-j}=\frac{1+kn}{1+(k-1)n}\binom{kn}{n}.$$ Could yo …

There are many possibilities, e.g. $\sum_{i=0}^{n}q^i{n \choose i}_{q^2}=(1+q)(1+q^2)...(1+q^n)$ or $\sum_{i=0}^{n}q^{i(i+1)/2}{n \choose i}_{q }=(1+q)(1+q^2)...(1+q^n).$

Let $n$ be a positive integer, $k$ a non-negative integer and $N(n,k)$ be the number of odd elements among the numbers $\binom{n+k}{j}\binom{-n-k}{n-j}$, $0\le{j}\le{n}$, which sum to $0.$ It seems th …

Let $[x]_{q}=\frac{1-q^x}{1-q}$ and $\binom{x}{n}_{q}$ denote a $q$-binomial coefficient. Let $A_n(x,q)$ be the $n\times n $ matrix with entries $$q^\binom{i-j}{2}\binom{i+j+x}{i-j+1}_{q},$$ $0 \le i …

Let $[x]_q=\frac{1-q^x}{1-q}$, $[n]_q!=[1]_q[2]_q\cdots[n]_q$ and ${\binom{x}{n}}_{q}=\frac{[x]_q[x-1]_q\cdots[x-n+1]_q }{[n]_q!}$. Computer experiments suggest that $$\det \left(q^\binom{i-j}{2}\le …

As I said above I have some difficulty to denote specific $q-$analogues as canonical. Consider as example the Catalan numbers $\frac{1}{{n + 1}}{2n\choose n}$ . They have a simple generating function …

Let $$a(n)=a(2^k-n)+k(n-2^{k-1})$$ for $$1 \leqslant {2^{k - 1}} < n \leqslant {2^k}$$ with initial values $a(0)=0, a(1)=0, a(2)=1.$ The first values are $0,0,1,2,4,5,7,9,12,13,15,\dots.$ Compari …