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The reason this is called the "exp-log" correspondence is that it can be written in terms of formal power series as follows: write $G(z) = \sum g_n z^n, F(z) = \sum f_n z^n$. We need the additional hy …

$\newcommand\Schur{\mathrm{Schur}}\newcommand\op{^\text{op}}\newcommand\FinVect{\mathrm{FinVect}}\DeclareMathOperator\Rep{Rep}\newcommand\Vect{\mathrm{Vect}}\DeclareMathOperator\GL{GL}\DeclareMathOper …

This is the Hankel determinant associated to the sequence $m_n = \mathbb{E}(X^n) = n!$ of moments of an exponential distribution with mean $1$. Some general results can be used to show that the sequen …

The Bell numbers satisfy $\frac{\ln B_n}{n} \sim \ln n$ which is faster than exponential, so the ordinary generating function $\sum B_n x^n$ has zero radius of convergence. As a more elementary argume …

I'll rename all three of your variables; you are asking for the number of partitions of $k$ that fit into an $m \times n$ box. This is famously known to be the coefficient of $q^k$ in the $q$-binomial …

Bjorn Poonen addresses this question for commutative (associative, unital) algebras in The moduli space of commutative algebras of finite rank; asymptotically we have $$q^{\frac{2}{27} n^3 + O(n^{8/3} …

Expanding on Richard's comment: let me rename your graph to $S$ and consider the adjacency matrix $A$ abstractly as a linear operator acting on the free vector space $\mathbb{C}[S]$ on (the vertices o …

Yes. Take the companion matrix $M$ of the characteristic polynomial of your original recurrence. Then the squared recurrence satisfies a recurrence with the characteristic polynomial of the symmetric …

We have the identity $$\frac{1}{1 - x^k} = \prod_{i \ge 0} (1 + x^{k \cdot 2^i})$$ which is equivalent to the uniqueness of binary representations, and is also straightforward to prove using a teles …

Equivalently, we'll show that we cannot have $$\frac{1}{P(x)} = \frac{1}{\prod_{j=1}^{\infty} (1 - x^{s_j})}$$ as formal power series. The idea is that the LHS has a pole of finite order at $x = 1$ …

Here is a more informative version of this identity. Let $Z_n$ denote the cycle index polynomial of the symmetric group $S_n$, namely $$Z_n = \frac{1}{n!} \sum_{\sigma \in S_n} z_1^{c_1(\sigma)} z_2^{ …

The standard permutation representation $\mathbb{C}^n$ of $S_n$ breaks up as a direct sum of a trivial representation $1$ and an irreducible representation $V$ of dimension $n - 1$. Hence the natural …

Yes. $C_n$ counts the number of paths from $(0, 0)$ to $(2n, 0)$ which involve moving by either the vector $(1, 1)$ or $(1, -1)$. Each such path passes through $(n, k)$ for a unique $0 \le k \le n$, a …

Here is Eytan's proof in more detail. First, there is a canonical way to write the cycle decomposition of a permutation. You order the cycles in descending order based on the largest member they conta …

Everything that's been written so far about the classification of finite groupoids reducing to the classification of finite groups is true but, I think, misleading. In order to actually produce a list …