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Proposition. $f(n)\ge 2n+\tfrac 13\left(\sqrt{\tfrac{n}3+1}-2\right) $ for any natural $n\ge 13$. Proof. Fix the cake cutting with the minimum number $f=f(n)$ of slices. We shall work with the graph $ …

If $S$ is congruentially equidistributed and contains enough elements .... is it true that $S+S$ contains all the positive integers except a finite number of them? Let $S=\bigcup_{n=1}^\infty \{2^{2 …

There is a huge number of such topologies. Let $G$ be any discrete abelian group. By $\widehat G$ we denote the family (in fact, a group) of its characters, that is of homomorphisms from $G$ to the un …

To mark the question answered, I copied below my accepted answer from Mathematics Stack Exchange. There are no matrices $M_1$ and $M_2$ which you are trying to find because of the following propositio …

For each nonnegative integer $n$ we have $$I_n-J_n=\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)} ( \pi p_k(\ln\pi) - k! )-\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)} ( S_n p_k(\ln\pi) - k! )=$$ $$(\pi-S_n)\sum_ …

For each natural $n$ we have $$f(n)=\left(1-\frac{1}{n}\right)^{g(n)}=\left(\left(1-\frac{1}{n}\right)^{n}\right)^{g(n)/n}.$$ Since $\lim_{n\to\infty} \left(1-\frac{1}{n}\right)^{n}=\frac 1e$, the bou …

GH from MO's answer suggests to study (odd) composite numbers $m$ such that $\sigma_2(m)−1$ is a square. Therefore $\sigma_2(m)$ is not divisible by $4$ and has no (prime) divisors equal $3$ modulo $4 …

There is a special version of this question (for $n=15$ and $m\le 7$) at Mathematics.SE. For each $n\ge 2$ I constructed a set $S$ with the property requiring $m\ge \left\lfloor\tfrac n2\right\rfloor= …

Fedor Petrov's comment shows than for each $k$ there are only finitely many cases for $x_1$ to check, so I wrote a program to do this. The positive answer is already obtained for all natural $k\le 10$ …

Since $p^k<m^2$, we have $p^k<2(m−1)m$, and so we can put $a=m−1$.