GPT 6 Astra Ultra attempt on Math-5650
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Erdős Problem #440: small adjacent least common multiples
Partial reconstruction with a sharp universal upper coefficient, 24 September 2026.1) Statement
Let $A=\{a_1<a_2<\cdots\}\subseteq\mathbb N$ be infinite, and write \[B(x)=\#\{i:\operatorname{lcm}(a_i,a_{i+1})\le x\}.\] The problem calls this count $A(x)$; the letter $B$ avoids confusing it with the counting function of the underlying set. Is $B(x)\ll\sqrt x$? How large can $\liminf_{x\to\infty}B(x)/\sqrt x$ be?We prove the first assertion, including the known upper coefficient \[c=\sum_{h\ge1}\frac1{\sqrt h(h+1)}.\] We also obtain the example with lower limit one. The missing direction is that every set has lower limit at most one.
2) Telescoping the good edges
Call an index good when its least common multiple is at most $x$. If $a=a_i$, $b=a_{i+1}$ and $d=b-a$, then $\gcd(a,b)\mid d$, so \[\frac1x\le\frac1{\operatorname{lcm}(a,b)} =\frac{\gcd(a,b)}{ab}\le\frac d{ab}=\frac1a-\frac1b.\tag{1}\] For any $T>0$, sum $1$ over good indices with $a_i>T$. All summands $1/a_i-1/a_{i+1}$ are nonnegative, and the sum over the entire tail is at most $1/T$. Consequently there are at most $x/T$ such good indices. There are at most $\lfloor T\rfloor$ indices starting at $a_i\le T$. Taking $T=\sqrt x$ proves \[B(x)\le2\sqrt x.\tag{2}\] All counts here are finite, since a good edge has both endpoints at most $x$.3) Refine by the integer spacing
Fix a positive integer $H$. For $1\le h\le H$, consider good edges with starting point in \[\big(\sqrt{(h-1)x},\sqrt{hx}\big].\] Their lengths obey $d\ge ab/x>a^2/x>h-1$, and hence $d\ge h$. Two consecutive good starting points in this interval are therefore separated by at least $h$: the later start is at least the endpoint of the earlier edge. The number of starts in this interval is at most \[\frac{\sqrt x(\sqrt h-\sqrt{h-1})}{h}+1.\] The tail with starts greater than $\sqrt{Hx}$ contributes at most $\sqrt x/\sqrt H$ by $1$. Thus \[\frac{B(x)}{\sqrt x}\le \sum_{h=1}^H\frac{\sqrt h-\sqrt{h-1}}h+\frac H{\sqrt x}+\frac1{\sqrt H}.\] First take the upper limit in $x$, with $H$ fixed, and then let $H\to\infty$. Summation by parts gives \[ \sum_{h=1}^H\frac{\sqrt h-\sqrt{h-1}}h =\sum_{h=1}^{H-1}\frac1{\sqrt h(h+1)}+\frac1{\sqrt H}, \] and the series converges by comparison with $\sum h^{-3/2}$. Therefore \[\limsup_{x\to\infty}\frac{B(x)}{\sqrt x}\le c, \qquad B(x)\le(c+o(1))\sqrt x.\tag{3}\] The proof is uniform in the choice of the underlying sequence.4) Lower limit one is attainable
For $A=\mathbb N$, adjacent terms are coprime, so \[B(x)=\#\{i:i(i+1)\le x\} =\left\lfloor\frac{\sqrt{1+4x}-1}{2}\right\rfloor.\] It follows that $B(x)/\sqrt x\to1$. Thus the supremum of the possible lower limits is at least one. Estimate $3$ bounds it above by $c$, but does not by itself give the required upper bound one: an upper-limit estimate and a sharp lower-limit estimate are different assertions.5) Outcome and attribution
UNRESOLVED in this attempt. PARTIAL PROGRESS on the two-part problem. The $O(\sqrt x)$ bound and the refined universal coefficient are fully proved. Erdős and Szemerédi proved that the largest lower limit is exactly one and that $c$ is optimal for the universal upper coefficient. Neither the general lower-limit bound nor a construction attaining the upper coefficient is reproduced here.Sources: https://www.erdosproblems.com/440 and the primary mathematical comments by T. Tao and W. van Doorn at https://www.erdosproblems.com/forum/discuss/440, checked 24 September 2026. The interval-spacing argument is a reconstruction of the method discussed there, not a claim of a new result. The telescoping estimate makes the treatment of the infinite tail explicit.
The estimate concerns the two-part problem, with its missing lower-limit direction taken into account.