GPT 6 Astra Ultra attempt on Math-5823

Erdős Problem 690: densities of the $k$-th prime factor

Statement and claim.

Let $d_k(p)$ be the density of integers whose $k$-th smallest distinct prime factor is $p$. The universal unimodality assertion is false: it fails for $k=4$. We also prove it for $k=1,2,3$. This reconstructs results of Cambie, Resolution of Erdős' problems about unimodularity, https://arxiv.org/abs/2501.10333, with a direct monotonicity argument for the first three cases. The existing GPT-5.2 attempt contains the same $k=4$ witness; the rational computations are re-derived here. See https://www.erdosproblems.com/690.

Exact density formula.

Divisibility by distinct primes is independent on a complete residue system modulo their product. Thus the relevant density exists, and \[ d_k(p)=\frac1p[z^{k-1}] \prod_{q<p}\left(1-\frac1q+\frac zq\right) =\frac{C_p}{p}e_{k-1}(p), \] where \[ C_p=\prod_{q<p}(1-1/q),\qquad \prod_{q<p}\left(1+\frac z{q-1}\right) =\sum_{r\geq0}e_r(p)z^r. \] All products here are finite. If $p<q$ are consecutive primes, then \[ C_q=C_p\frac{p-1}{p},\qquad e_r(q)=e_r(p)+\frac{e_{r-1}(p)}{p-1}. \] Whenever $e_{k-1}(p)>0$, it follows that \[ \frac{d_k(q)}{d_k(p)} =\frac{p-1+e_{k-2}(p)/e_{k-1}(p)}q. \tag{1} \]

The first three cases.

For $k=1$, directly $d_1(q)/d_1(p)=(p-1)/q<1$. For $k=2$, $d_2(2)=0$, and for $p\geq3$ we have $e_1(p)=\sum_{\ell<p}1/(\ell-1)\geq1$. Equation $1$ is then at most $p/q<1$, so the unique peak occurs at $p=3$.

For $k=3$, the initial nonzero term is at $p=5$. At that point $e_1=3/2$, $e_2=1/2$, so $3e_2=e_1$. Adding another variable $a=1/(\ell-1)>0$ changes these to $e'_1=e_1+a$, $e'_2=e_2+ae_1$, and \[ 3e'_2-e'_1=(3e_2-e_1)+a(3e_1-1)\geq0. \] Hence $e_1(p)/e_2(p)\leq3$ for every $p\geq5$. Consecutive odd primes have $q\geq p+2$, so $1$ is at most $(p+2)/q\leq1$. The two equal peaks at $5,7$ have value $1/30$; all subsequent terms decrease. This proves unimodality for $k=1,2,3$.

A complete counterexample for $k=4$.

Expanding the finite products gives \[ d_4(13)=\frac{31}{5005},\qquad d_4(17)=\frac{206}{36465},\qquad d_4(19)=\frac{1308}{230945}. \] These can also be computed without fractions until the final step: the coefficient of $z^3$ in $\prod_{q<p}(q-1+z)$, divided by $p\prod_{q<p}q$, is exactly $d_4(p)$. The strict comparisons reduce to \[ 31\cdot36465=1130415>1031030=206\cdot5005, \] \[ 206\cdot230945=47574670<47696220=1308\cdot36465. \] Thus the sequence decreases from $13$ to $17$ and increases from $17$ to $19$, which no unimodal sequence can do.

Scope.

This completely disproves the assertion for arbitrary fixed $k$ and independently proves the positive cases $1\leq k\leq3$. It does not purport to classify every individual $k\geq5$; the additional computational cases in Cambie's paper are not needed for the disproof.

Completion estimate.

COMPLETION: 100%