论文标题
Bernoulli时钟：通过循环卷积对Bernoulli多项式的概率和组合解释
The Bernoulli clock: probabilistic and combinatorial interpretations of the Bernoulli polynomials by circular convolution
论文作者
论文摘要
众所周知，以$ b_0（x）= 1 $和$ b_n（x）$ for $ n> 0 $为特征的众所周知，b_n（x）/n！$的分解标准化的bernoulli polyenmials $ b_n（x）= b_n（x）/n！$是特征的。我们提供相关的特征：$ b_1（x）= x-1/2 $和$（1）^{n-1} b_n（x）$ for $ n> 0 $是$ n $ - fold圆形卷积$ b_1（x）$。同等地，$ 1-2^n b_n（x）$是$ x \ in（0,1）$的概率密度，$ n $ n $独立的随机变量的总和，每个变量的总和都带有beta $（1,2）$概率密度$ 2（1 -x）$ 2（1 -x）$ $ x \ in（0,1,1）$。该结果具有一个新颖的组合类似物，{\ em bernoulli clock}：标记按$ 2 n $小时的时钟的时间按均匀的随机排列$ \ {1,1，2,2,2,2，\ ldots，n，n，n \} $，这意味着要从$ 2 N $ 2 $ 2均匀地选择两次不同的时间，然后选择两次$ 1 $ 1，然后将两次均匀地标记为$ 1 $ 1，然后将两次相同的零售价为$ 1 $ 1 n -2 $小时，标记它们$ 2 $，依此类推。从小时$ 0 = 2n $开始，顺时针移动到标记为$ 1 $的第一个小时，以$ 2 $标记的第一小时继续前进，等等，依次在伯努利时钟围绕时钟顺时针延伸，直到两个小时的第一个标记为$ n $的第一个，以随机的小时$ i_n $ 1 $ 1 $ $ 1 $ $ 1 $和2n $。我们表明，对于每个正整数$ n $，事件$（i_n = 1）$具有概率$（1-2^n b_n（0））/（2n）$，其中$ n！ b_n（0）= b_n（0）$是$ n $ th bernoulli号码。对于$ 1 \ le K \ le 2 n $，差$δ_n（k）：= 1/（2n） - ¶（i_n = k）$是$ k $的多项式功能，令人惊讶的对称$Δ_n（2 n + 1- k）=（2 n + 1- k）=（-1）^nΔ_n（k） $ b_n（1-x）=（-1）^n b_n（x）$。
The factorially normalized Bernoulli polynomials $b_n(x) = B_n(x)/n!$ are known to be characterized by $b_0(x) = 1$ and $b_n(x)$ for $n >0$ is the antiderivative of $b_{n-1}(x)$ subject to $\int_0^1 b_n(x) dx = 0$. We offer a related characterization: $b_1(x) = x - 1/2$ and $(-1)^{n-1} b_n(x)$ for $n >0$ is the $n$-fold circular convolution of $b_1(x)$ with itself. Equivalently, $1 - 2^n b_n(x)$ is the probability density at $x \in (0,1)$ of the fractional part of a sum of $n$ independent random variables, each with the beta$(1,2)$ probability density $2(1-x)$ at $x \in (0,1)$. This result has a novel combinatorial analog, the {\em Bernoulli clock}: mark the hours of a $2 n$ hour clock by a uniform random permutation of the multiset $\{1,1, 2,2, \ldots, n,n\}$, meaning pick two different hours uniformly at random from the $2 n$ hours and mark them $1$, then pick two different hours uniformly at random from the remaining $2 n - 2$ hours and mark them $2$, and so on. Starting from hour $0 = 2n$, move clockwise to the first hour marked $1$, continue clockwise to the first hour marked $2$, and so on, continuing clockwise around the Bernoulli clock until the first of the two hours marked $n$ is encountered, at a random hour $I_n$ between $1$ and $2n$. We show that for each positive integer $n$, the event $( I_n = 1)$ has probability $(1 - 2^n b_n(0))/(2n)$, where $n! b_n(0) = B_n(0)$ is the $n$th Bernoulli number. For $ 1 \le k \le 2 n$, the difference $δ_n(k):= 1/(2n) - ¶( I_n = k)$ is a polynomial function of $k$ with the surprising symmetry $δ_n( 2 n + 1 - k) = (-1)^n δ_n(k)$, which is a combinatorial analog of the well known symmetry of Bernoulli polynomials $b_n(1-x) = (-1)^n b_n(x)$.
