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Source. Stijn Cambie, Resolution of Erdős' problems about unimodularity, arXiv:2501.10333v1 (17 January 2025), Claim 4 and proof, PDF p. 3.

Dependencies. Ford, The distribution of integers with a divisor in a given interval, Theorem 4, printed p. 375, in the existing Fo08 source folder; Mertens' third theorem.

Bears on. #692 and Theorem 3.

Statement

There is a constant $c>0$ such that, for all sufficiently large $n$, if

$$m=\left\lfloor\exp(3n^c)\right\rfloor,$$

then

$$\delta_0(n,m+1)>\delta_1(n,m+1).$$

The same estimates hold for bounded multiplicative perturbations of this choice of $m$, which is the source's $m=\Theta(\exp(3n^c))$ formulation.

Rewritten proof

For large $n$, $m\geq n^2$. Every $1\leq t\leq n$ has a multiple

$$t\left(\left\lfloor\frac nt\right\rfloor+1\right)$$

in $[n+1,n+t]\subseteq[n+1,m]$. Thus the lcm of $1,\ldots,n$ divides the lcm of $n+1,\ldots,m$. Since every number in $[n+1,m]$ also lies in $[1,m]$, the lcm of $n+1,\ldots,m$ divides the lcm of $1,\ldots,m$; hence

$$L=\operatorname{lcm}(n+1,\ldots,m)=\operatorname{lcm}(1,\ldots,m).$$

The interval $(n,m+1)$ contains precisely $n+1,\ldots,m$. A residue $x$ modulo $L$ has no divisor in this set exactly when $\gcd(x,L)\leq n$. To see the converse implication, put $d=\gcd(x,L)>n$. If a prime-power factor of $d$ exceeds $n$, it is at most $m$ and is a divisor in $(n,m]$. Otherwise, multiply the pairwise coprime prime-power factors of $d$ until the first partial product exceeds $n$; the preceding product and the next factor are at most $n$, so this partial product is at most $n^2\leq m$. It is again a divisor in $(n,m]$. The number of residues with $\gcd(x,L)=i$ is $\varphi(L/i)$, so

$$\delta_0(n,m+1)=\frac1L\sum_{i=1}^{n}\varphi(L/i). \tag{1}$$

For $i\mid L$, a prime-by-prime check of Euler's product gives

$$\varphi(L/i)\geq\frac{\varphi(L)}{i}. \tag{2}$$

Indeed, removing a prime power $p^a$ from $L$ contributes a factor $p^{-a}$ to the totient ratio unless the whole $p$-power is removed, in which case the ratio is larger by $p/(p-1)$. Mertens' third theorem gives

$$\frac{\varphi(L)}L=\prod_{p\leq m}\left(1-\frac1p\right) \sim\frac{e^{-\gamma}}{\log m}.$$

Since $e^\gamma<2$, this is greater than $1/(2\log m)$ for all sufficiently large $m$. From (1), (2), and $H_n>\log n$,

$$\delta_0(n,m+1) \geq H_n\frac{\varphi(L)}L >\frac{\log n}{2\log m}. \tag{3}$$

We now use Ford's theorem with a fixed parameter $0<a<1$, say $a=1/2$, independent of the constant $c$ that will be chosen below. Ford defines $H(x,y,z)$ as the number of positive integers at most $x$ having at least one divisor in $(y,z]$, and $H_1(x,y,z)$ as the number having exactly one such divisor. Ford's Theorem 4 states, for fixed $0<a<1$, $y$ sufficiently large, $y+1\leq z\leq x^{5/8}$, and $yz\leq x^{1-a}$, that

$$\frac{H_1(x,y,z)}{H(x,y,z)} \asymp_a \frac{\log\log(z/y+10)}{\log(z/y+10)}. \tag{4}$$

Take $y=n$ and $z=m$. As $x\to\infty$, the hypotheses hold for each fixed $n,m$, and $(y,z]=(n,m+1)$. The limits of $H_1/x$ and $H/x$ are $\delta_1(n,m+1)$ and the density of integers with at least one divisor in the interval. Since the latter density is at most $1$, (4) gives

$$\delta_1(n,m+1) \ll_a\frac{\log\log(m/n+10)}{\log(m/n+10)}. \tag{5}$$

For $m=\exp(3n^c)+O(1)$,

$$\log m=3n^c+O(1),\quad \log(m/n+10)=3n^c+O(\log n),$$

and

$$\log\log(m/n+10)=c\log n+O(1).$$

Thus (3) is asymptotic to $\log n/(6n^c)$, while (5) is at most

$$\left(C_a+o(1)\right)\frac{c\log n}{3n^c}$$

for the implied constant $C_a$. Choose $c>0$ after fixing $a$ so that $2C_ac<1$. The lower bound then exceeds the upper bound for all sufficiently large $n$, proving the claim.

Indexing note. The source writes $\delta_0(n,m)$ while using $L=\operatorname{lcm}(n+1,\ldots,m)$; its displayed calculation is for $\delta_0(n,m+1)$. The statement and proof here use the consistent indexing.