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Take a finite projective plane: for the sake of concreteness, the Fano plane $PG(2, 2)$. It has seven points $P$ and seven lines $L$, where each line goes through three points and each pair of lines i …

The numerator grows faster than the denominator, so we can do better by making a minimal extension of a previous powerset: $2^{[k]} \cup \{[m] : k+1 \le m \le n \} \setminus \emptyset$ gives $$\frac{( …

At first I got the impression that the coefficient is $1$ if $n=m, m+1$ and $2^{n-m-1}$ otherwise. Yes, that's correct: $$\begin{eqnarray} Pf(n+1) &=& f(n+1) + \sum_{i=1}^n Pf(i) \\ &=& f(n+1) + Pf …

Inductively, $a_n$ tells us the index of the ticket selected from the reordered stack $n, 1, 2, \ldots, (n-1)$ to determine $a_{n+1}$. So $$a_n = \begin{cases} 1 & \textrm{if } n = 1 \\ n-1 & \textrm{ …

Wikipedia suggests that no-one has improved on Klarner, D.A.; Rivest, R.L. (1973) A procedure for improving the upper bound for the number of n-ominoes, Canadian J of Math. 25 (3): 585–602, which give …

Hint: it's always worth checking the Online Encyclopedia of Integer Sequences. For $r=3$ the values of $N$ are the sequence A002426. There's a wealth of literature references, a number of comments whi …

0123 1032 2310 3201 1032 2310 3201 0123 2310 3201 0123 1032 3201 0123 1032 2310 ---------------------- 0231 3102 1320 2013 1320 2013 0231 3102 3102 1320 2013 0231 2013 0231 31 …

Wikipedia says "The generalization of the Cantor polynomial in higher dimensions" is $$(x_1,\ldots,x_n) \mapsto x_1+\binom{x_1+x_2+1}{2}+\cdots+\binom{x_1+\cdots +x_n+n-1}{n}$$ Note that this is not e …

The first question may be answered, as you yourself implied in comments, by using a linear programming solver. Applying GNU's lpsolve to the question of which permutations have solutions, I find that …

We can rewrite $$\sum_{i=0}^{|Q|} c_i x^ i = \prod_{q_i \in Q} (1+q_ix)$$ as $$c_i = \sum_{\substack{S \subseteq Q \\ |S| = i}} \prod_{q \in S} q$$ Then $$\sum_{a \in A} (1 - a) \langle C(A_{\neg a}), …

We have a Markov process where the state after $i$ steps is given by the size of the codomain of $g_i$. If at time $i$ we are in a state with $j$ surviving values, we can ignore the other values and c …

For the special case $n_1 = n_2 = n$ and $1 \le m \le 9$ it turns out that the double sum yields a polynomial of the form $P(n)(2n+1)(n+1)n^2$ where $P(n)$ is irreducible of degree $2(m-1)$. Obviously …

First choose your 2-cycles, for a factor of $\binom{n}{2k}(2k-1)!!$. (Note that we require the convention that $(-1)!! = 1$). Then count functions $g: [n-2k] \to [n]$ with no fixed points or 2-cycles. …

Take the LHS: $$\sum_{\substack{\alpha \models i+k} \\ \ell(\alpha) = k \\ \alpha_a \leq \lfloor \frac{l}{2} \rfloor} \prod_{a=1}^k \binom{l}{2(\alpha_a - 1)}$$ Firstly, we can simplifying by rolling …

Let $b_n(t)$ be the Morgan-Voyce polynomial defined by $$\begin{eqnarray}b_0(t) &=& 1 \\ b_1(t) &=& t + 1 \\ b_n(t) &=& (t+2) b_{n-1}(t) - b_{n-2}(t) \end{eqnarray}$$ Then $f_n(t) = (-1)^n b_n(-t)$ fi …