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It's easy to prove that for $r=3+2\sqrt{2}$ and for $n=1$ or $n=3$ the sequences of iterates diverge (as they satisfy some simple recursive relation).

Define $\text{rad}_{23}(2^m3^nr)=2^{\text{sign}(m)}3^{\text{sign}(n)}r$, where $m,n\ge0$ and $2,3\nmid r\in\mathbb{N}$. For a triple $a+b=c$ define the quality $q_{23}(a,b,c)=\frac{\log(c)}{\log(\tex …

If $\{p_i\}$ is the sequence of all primes, is it possible that there exist a non constant $P\in \mathbb{Z}[x_1,\dots x_n]$ such that $P(p_i,p_{i+1},\dots p_{i+n-1})$ is bounded in $i$? More precisely …

Equivalently, we want to know if $\mathrm{sqceiling}(n^2/3) > n^4/(3n^2-5) = n^2/3 + 5/9 + 25/(27n^2) + ...$ where $\mathrm{sqceiling}()$ is the function taking a real to the next exact square. Thi …

The answer is no. Start with $\alpha$ between 3.25 and 3.75; take squares of those 2 to see that one can restrict $\alpha^2$ to being between 11.25 and 11.75; take 3/2 powers of those to see that one …

The first two equalities imply $x>m$ and $y>n$ so one can substitute $x=m+X$, $y=n+Y$ and $k=X+Y$, with still $X,Y \in \mathbb N$: ${X \choose 2}=nX+nY-mX\tag{1}$ ${Y \choose 2}=mX+mY-nY\tag{2}$ Fr …

Here is a completely elementary proof, inspired by Pasten's comments. Let $P(n)=an^2+bn+c$. Take $n=a^5x^4+2a^3(ab+2a+1)x^3+a(2a^3c+a^2b^2+6a^2b+3ab+6a^2+5a+1)x^2+(ab+2a+1)(2a^2c+2ab+b+2a+1)x+a^3c^2+2 …

Given a set $P$ of real numbers $\ge 1$, define the gap among different products in $P$ as $$g(P) = \inf \big\{\prod_{i=1}^n p_i^{a_i} - \prod_{i=1}^n p_i^{b_i} \mid p_i\in P;\,\, p_i\ne p_j \,\text{ …

If the Collatz conjecture is strongly false, in the sense that there is an infinite orbit, let $S_n$ be the set of natural numbers $\le n$ whose orbit goes to infinity. If $c=\liminf _{n\rightarrow\in …

Can one color the positive integers with finitely many colors, so that no two different numbers of the same color add to a square? Some easy to prove remarks: at least 4 colors are needed, since the …

I think $\big\{-1, 0, \frac{1+\sqrt{5}}{2}\big\}$ is a counterexample. THIS IS WRONG, see comments, but I'll leave it up as a warning.

Modulo 6 the squares are 0,1,4,3,4,1 and the squares+7 (or -5) can only be 1,2,5,4,5,2, of which 3/6 can at all be prime. The squares-7 (or +5) are 5,0,3,2,3,0 of which only 1/6 can be prime. Obviousl …

$u_n=s_na + t_nb + s_{n+1}c$ satisfies the same recurrence relation as $s_n$ and $t_n$: $u_n = u_{n-1} +2u_{n-2} + 4u_{n-3}$. The question is whether $2^{n+1}\mid u_n$. Since $v_n=u_n/2^{n+1}$ satisfi …

I suspect the answer is no. First rewrite $x_n=y_n+1$, then the recursion becomes $(n+1)y_{n+1}=ny_n(y_n+2)=(y_n+2)(y_{n-1}+2)\cdots (y_2+2) (y_1+2)y_1$ and for the integrality of $y_{n+1}$ it is su …

Define the size, possibly $\infty$, of a set $S\subseteq \mathbb{N}$ as $|S|=\sum\limits_{n\in S} \frac{1}{n}$. Then the Erdős-Turan conjecture states that if $|S|=\infty$, S must contain arbitrarily …