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Definitions: An upper arch system of order $n$ is a subset of the plane consisting of $n$ non-intersecting closed semicircles in the upper half-plane whose endpoints belong to the set $\{(k,0)\mid 1\ …

Here is one possible expression. I have not tried to simplify it, maybe there actually is much simpler formula, I don't know. $$ P_k(q)=1-(1-q)\frac{H(q^{k+2})}{H(q^{k+3})} $$ with $$ H(z)={}_1\phi_1 …

Probably this is not very helpful, but it is an explicit expression after all. I get $$ S(a,b)=\frac{(-1)^{a+1}}{b!}{}_2F_1(-a+1,b+1;2b-a+1;2)\binom b{2b-a}2^{2b-a}. $$ This follows from $$ S(a,b)=\su …

For brevity, rewrite the recursion as$$f_k(n)=\sum_{1\leqslant i\leqslant\frac n2}\binom nif_k(i)f_k(n-i).$$Now divide it by $n!k^n$ and rewrite like this:$$\frac{f_k(n)}{n!k^n}=\frac1{n!}\sum_{1\leqs …

Trying to count certain combinatorial structures, I arrived at a construction of their generating function through a very inconvenient procedure. I realize that anybody who will read this has right t …

(Initially I had this in my other answer, then I followed a suggestion by S. Carnahan and made it a separate one) There are several impressive combinatorial proofs of some mysterious identities involv …

The Lie algebra $\mathfrak{so}_3$ has a basis $x_1,x_2,x_3$ with the multiplication table $[x_1,x_2]=x_3$, $[x_2,x_3]=x_1$, $[x_3,x_1]=x_2$. Moreover there is an isomorphism $\mathfrak{so}_3(\mathbb C …

One more version - analog of $\sum_{i=0}^n(-1)^i\binom ni=0$: $$ \sum_{i=0}^n(-1)^i\binom ni_q=\begin{cases}0,&n=2k-1\\ \prod_{j=1}^k(1-q^{2j-1}),&n=2k\end{cases} $$

Not sure if this fits, but I find a proof of the Jacobi triple product formula in the form $$ \prod_{n>0}(1+q^{n-\frac{1}{2}}z)(1+q^{n-\frac{1}{2}}z^{-1})=\left(\sum_{l\in\mathbb{Z}}q^{l^2/2}z^l\right …

Just an illustration - for $m$ up to 6000, $r$ up to 120. Quite mysterious looking, I would say. Version 2: in the coordinate system suggested by the answer of Lucia it looks somehow more regular, …

It is easy to see that $$ YS=\sum_{\substack{i+j+k=p-3\\i,j,k\geqslant0}}\binom{i+j+k+2}{i+j+2}\binom{i+j+1}{i+1}\theta^i(Y)\theta^j(Y)\theta^k(Y) $$ It follows that the coefficient at $\theta^i(Y)\th …

This is definitely very far from being an answer, but still. Let $f(t)=\sum_nF(n)t^n$ be the ogf for the numbers in question. As in the comment by Timothy Chow, call a pair $\langle\boldsymbol a,\bo …

This is a spinoff, see Collecting stones in n buckets. Frankly speaking my only motivation is that I became curious: what happens if one redistributes the stones into the same buckets? More rigorousl …

I've got ten (projective) planes in projective 3-space: \begin{align} &x=0\\ &z=0\\ &t=0\\ &x+y=0\\ &x-y=0\\ &z+t=0\\ &x-y-z=0\\ &x+y+z=0\\ &x-y+t=0\\ &x+y-t=0 \end{align} If I did not make a mistake …

Encouraged by the proof in the answer by Fedor Petrov, I started random search and found such subsets for each $n\leqslant200$ starting from $n=7$, which in addition satisfy the condition imposed by F …