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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' ...


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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: ...


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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 ...


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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 ...


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(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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...



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