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

Dependencies. Claim 4; the Baker--Harman--Pintz theorem that every sufficiently large interval $[x-x^{0.525},x]$ contains a prime, Theorem 1, printed p. 532 of the existing BHP01 source; and the elementary recurrence below.

Bears on. #692.

Statement

For some constant $c>0$, the sequence

$$\bigl(\delta_1(n,m)\bigr)_{m\geq n+2}$$

has $\omega(\exp(n^c))$ local maxima for all sufficiently large $n$.

Rewritten proof

Let $X=\exp(3n^c)$, with $c$ chosen as in Claim 4. If $p$ is a prime of size $\Theta(X)$, adjoining $p$ gives

$$\delta_1(n,p+1) =\frac{p-1}{p}\delta_1(n,p)+\frac1p\delta_0(n,p). \tag{1}$$

Claim 4, with its $m$ parameter shifted by one, gives $\delta_0(n,p)>\delta_1(n,p)$ for the primes in the selected scale. Hence

$$\delta_1(n,p+1)>\delta_1(n,p). \tag{2}$$

There is also a strict fall at every sufficiently large doubled prime. Let $q>n$ be prime and let

$$L_q=\operatorname{lcm}(n+1,\ldots,2q).$$

The residue class $2q\pmod {L_q}$ has exactly one divisor in $(n,2q)$ when $n\geq2$: the divisors of $2q$ are $1,2,q,2q$, and only $q$ lies in that open interval. Every integer in this residue class therefore contributes to $\delta_1(n,2q)$, but after $2q$ is adjoined it has the two divisors $q$ and $2q$. Conversely, no multiple of the new divisor $2q$ can have it as its only divisor because it is also a multiple of $q$. The positive-density residue class proves

$$\delta_1(n,2q+1)<\delta_1(n,2q). \tag{3}$$

It remains to obtain many alternating positions. Put $H=(2X)^{0.525}$. For

$$1\leq i\leq\left\lfloor\frac{X}{16H}\right\rfloor,$$

choose a prime

$$q_i\in\left[\frac X2+4iH,\frac X2+(4i+1)H\right]$$

using BHP at the right endpoint of each interval. Applying BHP at $2q_i$ then gives a prime $p_i\in[2q_i-H,2q_i]$; since $2q_i$ is composite, $p_i<2q_i$. The interval choices give $p_1>X$ and $p_{i+1}>2q_i$, so

$$X<p_1<2q_1<p_2<2q_2<\cdots<p_r<2q_r<2X. \tag{4}$$

The number of selected pairs satisfies

$$r=\left\lfloor\frac{X}{16H}\right\rfloor\asymp X^{0.475}. \tag{5}$$

At each $p_i$ there is a strict rise by (2), and at each $2q_i$ there is a strict fall by (3). The ordering (4) makes these sign changes disjoint, so the maximum of the finite segment from $p_i+1$ through $2q_i$ supplies a local maximum; choose the rightmost occurrence if the maximum has a plateau. These maxima are distinct for different $i$. By (5), their number is

$$\gg\exp(0.475\cdot3n^c),$$

and the ratio of this lower bound to $\exp(n^c)$ tends to infinity. This proves the theorem.

Source correction. The arXiv text has an extra closing parenthesis in the displayed $\omega(\exp(0.475\cdot3n^c))$ count. The paper defines $\omega()$ as a lower bound by a constant multiple (p. 2), so that display says $r\gg\exp(0.475\cdot3n^c)$, which (5) derives explicitly from the prime-gap theorem. Since $0.475\cdot3>1$, the proof also gives the statement's $\omega(\exp(n^c))$ in the usual sense.