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In their article "Stolarsky's conjecture and the sum of digits of polynomial values"( https://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P34_Stolarsky.pdf ), Hare, Laishram and Stoll show in Proposit …

Pyber showed that the number of groups of order $n$ is $\leq n^{\frac{2}{27}\nu(n)^3+C\nu(n)^{3/2}}$, where $\nu$ is the highest power of a prime dividing $n$ and $C$ is an absolute constant. On the o …

In higher dimension things become more complicated. For a finite abelian group $G$ define $\mathfrak{s}(G)$ to be the least integer $N$, such that every sequence $x_1, \ldots, x_N$ of elements of $G$ …

Let $p\neq 3$ be a prime. Then \begin{eqnarray*} \nu_p\left(\frac{(3x)!}{x!^3}\right) & = & \sum_k \left[\frac{3x}{p^k}\right]-3\left[\frac{x}{p^k}\right]\\ & = & \sum_k 3\left\{\frac{x}{p^k}\right\} …

The number of totient divisors of $n$ is $d(n-1)-d((n-1, \varphi(n))$. As $n$ gets large, then almost all $n$ have the property that $\varphi(n)$ is divisble by all small primes. The average number of …

It depends a lot what you exactly you mean by "constructive". You can replace randomization by a greedy algorithm, or by a deterministic strategy similar to the one developed by Beck for combinatorial …

In a quite large range of the parameters we can approximate the fraction by a Taylor series to obtain $$ f(x) = \frac{1}{x!}\frac{1}{\frac{-\log c}{\binom{x+n-1}{n-1}} + \mathcal{O}\left(\frac{\log^2 …

I assume that $k$ is fixed, while $n$ tends to $\infty$. I claim that for $p=2$ the sum in question is asymptotically equal to $k\zeta(2)^{k-1}n^{-2}$. First consider those partitions, which contain p …

The number $N(i,j)$ of paths from $i$ to $j$ is given by the matrix $B=E+A+A^2+\dots$. The number of paths from $i$ to $j$ passing through $k$ is $N(i,k)N(k,j)$, which is the number of times you have …

Put $a=|v_1|$, $b=|v_2|$, $c=|v_1v_2|$. Then we have $$ \sum_{i=0}^a\sum_{j=0}^b\sum_{k=0}^c\binom{a-c}{i-k}\binom{b-c}{j-k}= \sum_{k=0}^c\left(\sum_{i=0}^a\binom{a-c}{i-k}\right)\left(\sum_{j=0}^b\bi …

Let $f$ and $g$ be random mappings. If they commute, then $f(g(1))=g(f(1))$, and this happens with probability $n^{-1}$. Now $f(g(2))=g(f(2))$ also holds with probability $n^{-1}$, but these events ne …

The average score difference does not suffice to predict the probability of the outcome. Suppose all players in team A are of equal strength, while all but one player in team B are somewhat stronger t …

The dihedral groups can be viewed as the set of all functions of the form $x\mapsto\pm x+c$ acting either on $\mathbb{Z}$ or on $\mathbb{Z}/n\mathbb{Z}$. The images of the infinite dihedral group are …

A similar problem, called the Coupon Collector's younger brothers, has been studied by Foata, Han and Lass (Séminaire Lotharingien de Combinatoire, B47a, 20 pages, 2001, obtainable via http://math.uni …

Converging sieves: Let $q_i$ be a sequence of integers with $\sum\frac{1}{q_i}<\infty$, and pick for each $i$ an integer $a_i$ within a finite set $\mathcal{A}$. Then the set of integers $n$ such that …