gpt pro 5.4 attempt on Math-747

Erdős Problem 291

1) ROUND-2 OBJECTIVE

Path (C): correct and substantially strengthen the Round 1 understanding of the already-solved "$(a_n,L_n)>1$ infinitely often" half, while keeping the "$(a_n,L_n)=1$ infinitely often" half clearly separated as the unresolved part.

Why this is the right continuation of Round 1: Round 1 proved only the easy infinite family coming from the prime $3$. The real missing opportunity was to exploit the leading-digit criterion much more systematically.

2) Round-1 FOUNDATION USED

I use the following Round 1 result.

• $F1$ If $q_n:=\gcd(a_n,L_n)$ and $p\le n$ is prime, then \[ p\mid q_n \iff p\mid \operatorname{num}(H_k), \] where $k$ is the leading base-$p$ digit of $n$ and $H_k=1+\frac12+\cdots+\frac1k$.

3) NEW INSIGHT / TOOL (ROUND-2)

There are three new ingredients.

• $N1$ An exact set-theoretic description of \[ Q_p:=\{n\ge 1: p\mid q_n\} \] in terms of base-$p$ leading digits.
• $N2$ An exact counting formula for $Q_p\cap [1,p^b]$.
• $N3$ A new unconditional theorem: for any two distinct odd primes $p<q$, there are infinitely many $n$ with \[ pq\mid q_n. \] So the easy "$3\mid q_n$ infinitely often" family from Round 1 can be strengthened to simultaneous divisibility by any prescribed pair of odd primes.

4) ATTACK PLAN (ROUND-2)

The exact gap after Round 1 was that the leading-digit criterion was only used one prime at a time.

I therefore prove the following claims.

• $C1$ For each odd prime $p$, the set $Q_p$ is a union of explicit base-$p$ digit blocks.
• $C2$ For each odd prime $p$, the digit $p-1$ is always bad, so $Q_p$ is infinite.
• $C3$ For any two odd primes $p<q$, infinitely many of the $p$-blocks and $q$-blocks overlap.

Claim $C3$ is the real new theorem of this round.

5) WORK (ROUND-2)

Let $H_n=u_n/v_n$ be reduced, so that $q_n=L_n/v_n=\gcd(a_n,L_n)$. For an odd prime $p$, define \[ E_p:=\{m\in\{1,2,\dots,p-1\}: p\mid \operatorname{num}(H_m)\}, \qquad Q_p:=\{n\ge 1: p\mid q_n\}. \]

Proposition 291.1 (exact structure of $Q_p$).

For every odd prime $p$, \[ Q_p=\bigcup_{m\in E_p}\ \bigcup_{a\ge 1}[mp^a,(m+1)p^a-1]. \]

Proof.

Fix $n\ge 1$ and write its leading base-$p$ digit as $m$, so \[ n=mp^a+r, \qquad 1\le m\le p-1, \qquad 0\le r<p^a. \] Then by the Round 1 criterion $F1$, \[ p\mid q_n \iff p\mid \operatorname{num}(H_m) \iff m\in E_p. \] But the condition that the leading base-$p$ digit of $n$ equals $m$ is exactly \[ mp^a\le n<(m+1)p^a. \] Since $n$ is an integer, this is precisely \[ mp^a\le n\le (m+1)p^a-1. \] Taking the union over all possible $a\ge 1$ and all bad digits $m\in E_p$ gives the formula. $\square$

Corollary 291.2 (exact counting on powers of $p$).

For every integer $b\ge 1$, \[ |Q_p\cap[1,p^b]|=|E_p|\,\frac{p^b-p}{p-1}. \]

Proof.

Inside $[1,p^b]$, the blocks in Proposition 291.1 are exactly those with $a=1,2,\dots,b-1$, because the block $[mp^a,(m+1)p^a-1]$ lies below $p^b$ if and only if $a\le b-1$. For fixed $m$ and fixed $a$, the block length is exactly $p^a$. Hence \[ |Q_p\cap[1,p^b]| =|E_p|\sum_{a=1}^{b-1}p^a =|E_p|\,\frac{p^b-p}{p-1}. \] $\square$

Corollary 291.3 ($Q_p$ is infinite for every odd prime $p$).

For every odd prime $p$, one has $p-1\in E_p$. Hence $Q_p$ is infinite.

Proof.

Write \[ H_{p-1}=\sum_{j=1}^{p-1}\frac1j =\sum_{j=1}^{(p-1)/2}\left(\frac1j+\frac1{p-j}\right) =\sum_{j=1}^{(p-1)/2}\frac{p}{j(p-j)}. \] Each denominator $j(p-j)$ is prime to $p$, so the reduced numerator of $H_{p-1}$ is divisible by $p$. Thus $p-1\in E_p$. Now Proposition 291.1 shows that for every $a\ge 1$, \[ [(p-1)p^a,p^{a+1}-1]\subseteq Q_p, \] so $Q_p$ is infinite. $\square$

Theorem 291.4 (new unconditional two-prime theorem).

Let $p<q$ be distinct odd primes. Then there are infinitely many integers $n$ such that \[ pq\mid q_n=\gcd(a_n,L_n). \]

Proof.

By Corollary 291.3, for every $a\ge 1$ the interval \[ I_a:=[(p-1)p^a,p^{a+1}-1] \] is contained in $Q_p$, and for every $b\ge 1$ the interval \[ J_b:=[(q-1)q^b,q^{b+1}-1] \] is contained in $Q_q$. Thus it suffices to show that $I_a\cap J_b\neq\emptyset$ for infinitely many pairs $(a,b)$.

Set \[ \alpha:=\frac{\log p}{\log q}. \] This number is irrational: otherwise $\alpha=r/s\in\mathbb Q$ would imply \[ p^s=q^r, \] impossible for distinct primes. Hence the set of fractional parts $\{a\alpha\}$ is dense in $[0,1]$.

For each $a\ge 1$, let \[ b:=\lfloor a\alpha\rfloor, \qquad \Delta_a:=a\log p-b\log q=\log q\cdot \{a\alpha\}. \] Then $0\le \Delta_a<\log q$. Consider the open interval \[ \mathcal J:=\left(\log\frac{q-1}{p},\ \log\frac{q}{p-1}\right). \] This interval is nonempty because \[ \frac{q-1}{p}<\frac{q}{p-1} \iff (q-1)(p-1)<pq, \] and the last inequality is equivalent to $p+q-1>0$. Also $\mathcal J\subset(0,\log q)$ because $q-1>p$ (distinct odd primes differ by at least $2$) and $p-1>1$.

By density of $\{a\alpha\}$, there are infinitely many $a$ for which \[ \Delta_a\in\mathcal J. \] For such $a$ (with $b=\lfloor a\alpha\rfloor$) we have \[ \log\frac{q-1}{p}<a\log p-b\log q<\log\frac{q}{p-1}. \] Exponentiating gives \[ (q-1)q^b<p^{a+1}, \qquad (p-1)p^a<q^{b+1}. \] Since all quantities are integers, these strict inequalities are exactly what is needed for the integer intervals $I_a$ and $J_b$ to overlap. Hence for infinitely many pairs $(a,b)$ there exists some integer \[ n\in I_a\cap J_b. \] For such $n$, Proposition 291.1 yields simultaneously \[ p\mid q_n \qquad\text{and}\qquad q\mid q_n, \] so $pq\mid q_n$. Because there are infinitely many such pairs $(a,b)$, there are infinitely many such $n$. $\square$

Example.

Taking $(p,q)=(3,5)$, one explicit overlap occurs at $n=20$. Indeed $20$ begins with digit $2$ in base $3$ and digit $4$ in base $5$, so $15\mid q_{20}$. In fact \[ H_{20}=\frac{55835135}{15519504}, \qquad L_{20}=232792560, \qquad q_{20}=\frac{L_{20}}{15519504}=15. \]

Literature note.

Proposition 291.1 is exactly the block description behind Shiu's Theorem 2, and Shiu also proves the harmonic density formula \[ \delta(Q_p)=\frac{1}{\log p}\sum_{m\in E_p}\log\left(1+\frac1m\right). \] This is stronger than mere infinitude of $Q_p$, but it is not needed for Theorem 291.4.

6) ADVERSARIAL VERIFICATION

• Critical check on irrationality. The proof of Theorem 291.4 only uses the irrationality of $\log p/\log q$, which is unconditional. This is safe.
• Important correction of a tempting overclaim. Shiu's multivariate theorem for finitely many primes requires linear independence of \[ \frac{\log p_1}{\log p_i} \quad (1\le i\le k), \] which is equivalent to linear independence of the reciprocals $1/\log p_i$ over $\mathbb Q$. That hypothesis is not known in general. So one must not upgrade Theorem 291.4 from pairs of primes to arbitrary finite sets without adding an explicit hypothesis.
• Interval-overlap criterion. The inequalities \[ (q-1)q^b<p^{a+1}, \qquad (p-1)p^a<q^{b+1} \] are exactly the right integer conditions for $I_a\cap J_b\neq\emptyset$ because the right endpoints are $p^{a+1}-1$ and $q^{b+1}-1$.
• Dependence on Round 1. The only place Round 1 is used is Proposition 291.1, via the already-proved leading-digit criterion.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 makes genuinely new, rigorous progress:

• it gives an exact structural description of each one-prime divisibility set $Q_p$;
• it derives an exact counting formula for $Q_p\cap[1,p^b]$;
• it proves unconditionally that for any two distinct odd primes $p<q$, there are infinitely many $n$ with $pq\mid \gcd(a_n,L_n)$.

Thus the "$(a_n,L_n)>1$" half is now strengthened far beyond the easy base-$3$ family from Round 1. The remaining hard part is still the infinitude of $n$ with $\gcd(a_n,L_n)=1$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 55%

9) REFERENCES

• [Er291] T. F. Bloom, Erdős Problem #291, website entry, accessed 2026-03-07.
• [Sh24] P. Shiu, The denominators of harmonic numbers (Revised), arXiv:1607.02863v2, 2024.
• [WuYa22] B.-L. Wu and X.-H. Yan, On the denominators of harmonic numbers. IV, C. R. Math. Acad. Sci. Paris 360 $2022$, 53--57.