GPT 6 Astra Ultra attempt on Math-5664
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Erdős Problem 459
Statement.
For an integer $u\geq2$, let $f(u)$ be the largest integer $v>u$ such that no integer in $(u,v)$ has all its prime factors among the prime factors of $uv$. We reconstruct the extremal and typical estimates attributed to Stijn Cambie.An exact alternative definition.
Let $w$ be the smallest integer greater than $u$ composed only of prime factors of $u$. It exists, since $u^2$ is one candidate. Taking $v=w$ introduces no new prime factors, and minimality of $w$ excludes all intermediate integers of the required kind. For every $v>w$, however, $w\in(u,v)$ and its prime factors divide $u$, hence $uv$. Such a $v$ is forbidden. Thus \[ f(u)=\min\{w>u:\text{every prime factor of }w\text{ divides }u\}. \tag{1} \] Since $u+1$ is coprime to $u$ and exceeds one, \[ u+2\leq f(u)\leq u^2. \tag{2} \] For a prime $p$, $1$ gives $f(p)=p^2$. If $u$ is even, the least power of two exceeding $u$ is an admissible upper bound. In particular $f(2^j-2)=2^j$ for $j\geq2$. Both extremes in $2$ therefore occur infinitely often.Two fixed primes give arbitrarily small relative gaps.
Fix distinct primes $p,q$ and $\epsilon>0$. The number $\log q/\log p$ is irrational, by unique factorization. The elementary density theorem for an irrational rotation supplies a finite $B$ such that the residues \[ b\log q\pmod{\log p},\qquad 0\leq b\leq B, \] have every circular gap smaller than $\log(1+\epsilon)$. For each sufficiently large real $x$, choose a residue immediately after $\log x\pmod{\log p}$, using the next distinct residue if equality occurs. Some $b\leq B$ and integer $a$ then satisfy \[ \log x<a\log p+b\log q<\log x+\log(1+\epsilon). \] For $x>q^B$, this $a$ is nonnegative. Hence a number composed only of $p,q$ lies in $(x,(1+\epsilon)x)$. In particular, if $pq\mid n$ and $n$ exceeds a fixed threshold depending on $p,q,\epsilon$, then $f(n)<(1+\epsilon)n$.Almost every integer has two suitable fixed small primes.
For a finite set $\mathcal P$ of primes, the Chinese remainder theorem gives density \[ \prod_{p\in\mathcal P}(1-1/p) \left(1+\sum_{p\in\mathcal P}\frac1{p-1}\right) \tag{3} \] to the integers divisible by at most one member of $\mathcal P$. Take all primes up to $M$, and write $S_M=\sum_{p\leq M}1/p$. The expression in $3$ is at most \[ e^{-S_M}(1+S_M+C), \] where $C$ is absolute because $\sum_p(1/(p-1)-1/p)<\infty$. Euler's divergence $S_M\to\infty$ makes this bound tend to zero.Given $\eta>0$, choose $M$ making $3$ less than $\eta$. There are only finitely many pairs of primes up to $M$, so take the maximum of their thresholds in the preceding paragraph. Outside a set of density at most $\eta$ and finitely many integers, $f(n)<(1+\epsilon)n$. Since $\eta$ was arbitrary, the exceptional set has density zero. This holds for every $\epsilon>0$, proving \[ f(n)/n\longrightarrow1\quad\hbox{in natural density}. \tag{4} \]