GPT 6 Astra Ultra attempt on Math-5458
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FORMAL RESTATEMENT
Erdős #144; partial reconstruction, 2026-09-23. Prove that the positive integers $n$ having divisors $d_1,d_2$ with $d_1<d_2<2d_1$ have natural density one. The stronger version replaces $2$ by any fixed $c>1$.QUICK LITERATURE AND CONTEXT CHECK
Maier--Tenenbaum proved the assertion, including its stronger version. Their primary paper concerns the distribution and concentration of divisors. We give an exact reduction to a set of multiples, a quantitative sufficient condition, and a proved positive-density family. These do not establish density one; the missing analytic step is identified explicitly. No new resolution is claimed.WORK
Fix $c>1$. Define \[ \mathcal M_c=\{ab:1\le a<b<ca,\ (a,b)=1\}. \] Then $n$ has the desired close divisors if and only if it is divisible by a member of $\mathcal M_c$. For the forward implication, write $d_1=ga,d_2=gb$ after cancelling their gcd. Since $\operatorname{lcm}(d_1,d_2)=gab$ divides $n$, so does $ab$. Conversely, if $ab\mid n$ with the stated inequalities, $a,b$ themselves are suitable divisors. This proves the exact identity \[ \{n:\exists d_1\mid n,\ d_2\mid n,\ d_1<d_2<cd_1\} =\bigcup_{m\in\mathcal M_c}m\mathbb N. \tag{1} \]For any finite $F\subset\mathcal M_c$ the union of its multiples is periodic of period $L=\operatorname{lcm}(F)$ and has density \[ \delta_F=\sum_{\varnothing\ne J\subseteq F} \frac{(-1)^{|J|+1}}{\operatorname{lcm}(J)}. \tag{2} \] This is inclusion--exclusion, since the intersection of the sets of multiples for $J$ is exactly the multiples of its least common multiple. Thus the lower density in $1$ is at least every $\delta_F$. In particular, it would suffice to find finite sets $F$ with $\delta_F\to1$. Large overlapping collections cannot be counted by adding their reciprocal densities; $2$ displays the overlaps that must be controlled.
For $c=2$, take the four coprime pairs $(2,3),(3,5),(4,5),(5,7)$, so $F=\{6,15,20,35\}$. Among residues modulo $420$, their individual counts are $70,28,21,12$. Pairwise intersection counts sum to $14+7+2+7+4+3=37$; triple intersection counts sum to $7+2+1+1=11$; the fourfold count is $1$. Consequently \[ \delta_F=(131-37+11-1)/420=26/105. \tag{3} \] This proves a concrete positive lower density. For every $c>1$, choose an integer $a>1/(c-1)$; then $a<a+1<ca$, so all multiples of $a(a+1)$ supply a positive-density family in the stronger version too.
A logarithmic pigeonhole criterion
Suppose $m\mid n$ and list the divisors of $m$ as $1=t_1<t_2<\cdots<t_s=m$. If every consecutive ratio is at least $c$, then multiplying them gives $m\ge c^{s-1}$. Therefore \[ \tau(m)>1+\frac{\log m}{\log c} \quad\Longrightarrow\quad \text{$n$ has two divisors of ratio strictly between $1$ and $c$}. \tag{4} \] More quantitatively, since $\sum_{i=1}^{s-1}\log(t_{i+1}/t_i)=\log m$, at most $\log m/\log c$ of the consecutive ratios can be at least $c$. Thus at least $s-1-\lfloor\log m/\log c\rfloor$ are less than $c$ whenever that number is positive. Taking $m=n$ is allowed, but no argument here proves that $4$ holds for almost all $n$ or for a suitable divisor of almost all $n$.The distinction between $d_2<2d_1$ and $d_2\le2d_1$ matters. Every power of two has no pair of divisors with ratio strictly between $1$ and $2$, so there are infinitely many exceptions. They have density zero, and therefore do not contradict the theorem. The witness pairs used in $3$ satisfy the strict inequality.