# Tag Info

2

We give a proof of the strict inequality $$\frac{1 + p(1) + p(2) + \cdots + p(n-1)}{p(n)} < \sqrt{2n}.$$ It is convenient and natural to extend the definition of $p(n)$ to all integers by setting $p(0) = 0$ for all $n < 0$ and $p(0) = 1$. In particular, the numerator $1 + p(1) + p(2) + p(3) + \cdots$ in the fraction of interest is $$s_1 := \sum_{n' ... 1 For small numbers of cells 3,4,5,.. there may be a possibility of finding explicitly the optimal partition (the Y 120 degrees partition in the case n=3, etc). For large numbers of cells in the partitions, analytical results are not known. In order to find candidates for the optimal partitions, some numerical studies were performed: Cox and Fikkema: ... 1 Based on your suggestion, given two partitions P_1 and P_2, you could construct a complete bipartite graph, where P_1 and P_2 are the parts, and assign to every edge the Jaccard similarity weight. Taking the average weight (for example) of the edges in a maximum matching (which you can find using the augmented path algorithm or max-flow) will give ... 1 HINT: Note that$$\binom{\lambda_i'}2=\sum_{k=0}^{\lambda_i'-1}k=\sum_{k=1}^{\lambda_i'}(k-1)\;.\tag{1}$$Now \lambda_i'\ge k if and only if there are at least k parts \lambda_j of \lambda such that \lambda_j\ge i. In terms of the Ferrers diagram of \lambda, the i-th column has \ge k elements if and only if there are at least k rows with ... 1 (Too long for a comment.) Just to give the exact value of (2), we have the identity,$$A = \frac{1}{e^{5\pi/6}\,\eta\big(\tfrac{2\,i}{5}\big)} = \frac{2^{19/8}(-1+5^{1/4})\,\pi^{3/4}}{e^{5\pi/6}\,(-1+\sqrt{5})^{3/2}\,\Gamma\big(\tfrac{1}{4}\big)} = 0.0887758\dots$$which is approximated by,$$B ...

1

Let $x$, $y$ $\in \mathcal{X}$ be such that $P(x)\cap P(y) \neq \emptyset$. Then there is some $c \in P(x)$, $P(y)$. Assume for a contradiction that $P(x) \neq P(y)$ . Then there is $A \in F$ such that $x,c \in A$ but $y \notin A$ (or there is $B \in F$ such that $y,c \in B$ but $x \notin B$. This case follows similarly) Since $F$ is closed under taking ...

1

There are $\binom{n}k$ ways to pick a subset $I$ of $B$ of cardinality $k$ to be the image of a function $f\in S$. There are ${m\brace k}$, or in your notation $S(m,k)$, ways to partition $A$ into $k$ non-empty subsets. And there are then $k!$ ways to pair up these subsets with the $k$ members of $I$. That is, if the $k$ non-empty subsets of $S$ are ...

1

2 examples I like: Fix any natural number $n\geqslant 2$ and define $\sim$ on $\mathbb Z$ by $x\sim y$ if and only if $x$ and $y$ differ by a multiple of $n$ (this means that $x-y$ is a multiple of $n$). Then the equivalence class of $x$ is $[x]=\{\ x+kn\ |\ k\in\mathbb Z\ \}$, so $\mathbb Z$ is partitioned into the subsets of integers that have the same ...

1

Consider the set $A = \{a,b,c,d,e,f,g,h,i \}$, and look at the array: $$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}$$Define $\sim$ on $A$ by saying that $x \sim y \iff x$ and $y$ are in the same column. This way, we have $a \sim d$, $b \sim h$, $f \sim i$, $a \not\sim e$, $h \not\sim c$, for example. I'll leave ...

1

Consider an elementary school. Define an equivalence relation on the children in the school based on $\text{thing}_1 \sim \text{thing}_2$ if and only if they are in the same room. Then every child is assigned to a room and so every child lands in one and only one of the equivalence classes. Thus the set of children is the union of all of the equivalence ...

Only top voted, non community-wiki answers of a minimum length are eligible