GPT 6 Astra Ultra attempt on Math-5826
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Contents
Erdős Problem 696: prime chains and divisor chains
Statement and claim.
Let $h(n)$ be the longest increasing chain of prime divisors $p_1,\ldots,p_h$ of $n$ satisfying $p_i\mid p_{i+1}-1$. Define $H(n)$ in the same way with arbitrary positive divisors. We prove, for almost all $n$, \[ h(n)\asymp H(n)\asymp\log_* n,\qquad H(n)/h(n)\leq18. \tag{1} \] In particular the proposed almost-everywhere divergence of the ratio is false. This is a reconstruction of the coarse resolution reported at https://www.erdosproblems.com/696 and its forum, using the prime-successor method in the manuscript hosted at https://github.com/davidturturean/erdos-696. That manuscript attributes its stronger estimates to work with GPT-5.5. No new resolution or sharp normal-order constant is claimed here.External analytic inputs.
We use the classical Siegel--Walfisz theorem in the following fixed-exponent form: for coprime $a,q$, uniformly for $q\leq\log t$, \[ \pi(t;q,a)=\frac{\operatorname{Li}(t)}{\varphi(q)} +O(t e^{-c\sqrt{\log t}}). \tag{2} \] See A. Walfisz, Zur additiven Zahlentheorie. II, Math. Z. 40 $1936$, 592--607; the cited manuscript states and applies the same uniform input. We also use the elementary classical bounds $\sum_{p\leq z}\log p=O(z)$ and $\sum_p1/p=\infty$. All other probability, counting, and tower estimates are given below.Few short successor steps in a typical divisor chain.
Call a pair bad if $d\geq Y$, $d<e<\exp(\sqrt d)$, and $e\equiv1\pmod d$. Then $\gcd(d,e)=1$. For fixed $d$, writing $e=1+jd$ gives \[ \sum_{\substack{d<e<\exp(\sqrt d)\\e\equiv1\ (d)}}\frac1{de} \leq\frac1{d^2} \sum_{1\leq j\leq \exp(\sqrt d)/d}\frac1j \ll d^{-3/2}. \] The number of $n\leq x$ divisible by both members of any bad pair is therefore, by a union bound, \[ \leq x\sum_{d\geq Y}\sum_e\frac1{de}\ll xY^{-1/2}. \tag{3} \] For every other $n$, consecutive chain terms above $Y$ satisfy $d_{i+1}\geq f(d_i)$, where $f(t)=\exp(\sqrt t)$. For all sufficiently large $t$, \[ f(f(t))=\exp\left(\exp(\sqrt t/2)\right)\geq\exp(t). \] Define $T_0=e$, $T_{j+1}=\exp(T_j)$, and let $L=\min\{j\geq0:x\leq T_j\}$. This differs from any usual convention for $\log_*x$ by at most a fixed constant. Every two chain steps above $Y$ raise the tower level by at least one. There are at most $2L+O(1)$ such terms, and at most $Y$ terms below $Y$. Taking $Y=\lfloor\sqrt L\rfloor$, we obtain \[ H(n)\leq3L \quad\hbox{for all but }o(x)\hbox{ integers }n\leq x. \tag{4} \]Enough prime successors in a very large window.
For large $y$, put \[ Z(y)=\exp(\exp y),\qquad F(y)=\exp(\exp(y^2)). \] For every prime $p\leq y$, partial summation in $2$ gives, uniformly in $p$, \[ \sum_{\substack{Z(y)<q\leq F(y)\\q\ {\rm prime}\\q\equiv1\ (p)}} \frac1q =\frac{y^2-y}{p-1}+O\left(e^{-c'\sqrt{\log Z(y)}}\right) \geq y/3. \tag{5} \] Indeed, the modulus condition holds throughout the interval. The integrated error is bounded by a constant times $\int_{\sqrt{\log Z(y)}}^\infty u e^{-cu}\,du$, and the main integral is $(\log\log F(y)-\log\log Z(y))/(p-1)$. Thus no constant in $5$ depends on the chosen prime $p$.Independent prime choices and transfer to integers.
Let $y_0=L$, $y_{j+1}=F(y_j)$, and $K=\lfloor L/5\rfloor$. Consider independent Bernoulli variables indexed by primes $q\leq y_K$, with success probability $1/q$. With probability tending to one, there is a successful prime $p_0\leq L$, since the probability of none is $\prod_{p\leq L}(1-1/p)\longrightarrow0$. Having chosen $p_j\leq y_j$, seek a successful prime \[ Z(y_j)<p_{j+1}\leq y_{j+1},\qquad p_{j+1}\equiv1\pmod{p_j}. \] The window is disjoint from every window previously inspected. Conditional on all previous choices, $5$ shows that the probability of failure is at most \[ \prod_q(1-1/q)\leq\exp\left(-\sum_q1/q\right) \leq e^{-y_j/3}\leq e^{-L/3}. \] A union bound over $K$ stages now gives a prime chain of length $K+1\geq L/6$, with probability $1-o(1)$.The size of the prime cutoff matters for transferring this model to uniform integers $n\leq x$. Let $r_0=\min\{r:L\leq T_r\}=o(L)$. For large $y$, $y^2\leq e^y$, so $F(y)\leq\exp(\exp(\exp y))$. Inductively, \[ y_K\leq T_{r_0+3K}\leq T_{L-3}<\log\log x \] for all sufficiently large $x$; the last inequality uses $x>T_{L-1}$. Set $Q=\prod_{p\leq y_K}p$. The classical bound quoted above gives \[ Q\leq\exp(O(y_K))\leq(\log x)^{O(1)}=o(x). \] On a uniformly random residue modulo $Q$, the divisibility events for its primes are exactly the independent model, by the Chinese remainder theorem. For every event depending on that residue, its probability on $\{1,\ldots,\lfloor x\rfloor\}$ differs by $O(Q/x)=o(1)$. Consequently \[ h(n)\geq L/6 \quad\hbox{for all but }o(x)\hbox{ integers }n\leq x. \tag{6} \]