GPT 6 Astra Ultra attempt on Math-747
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Erdős Problem #291: independent continuation
Date: 2026-09-23. Research attempt; no claim of a new published result.1) TARGET AND SOURCE AUDIT
Put $L_n=\operatorname{lcm}(1,\ldots,n)$ and $a_n=L_nH_n$, where $H_n=\sum_{j\le n}1/j$. Are there infinitely many $n$ with $\gcd(a_n,L_n)=1$, as well as infinitely many with gcd greater than one?All supplied versions considered: 291.tex. The source develops the already accessible greater-than-one side and correctly warns that multivariate logarithm independence is not automatic. I turn back to the difficult gcd-one side, rederive its exact finite test and run it with rational arithmetic.
2) WORK
Fix $p\le n$, let $e=\lfloor\log_p n\rfloor$ and $m=\lfloor n/p^e\rfloor$. Modulo $p$, only denominators divisible by $p^e$ survive in $a_n=\sum_{j\le n}L_n/j$. Consequently \[ a_n\equiv {L_n\over p^e}\sum_{u=1}^{m}u^{-1}\pmod p. \] The coefficient is invertible, since $v_p(L_n)=e$. Thus $p$ divides the gcd exactly when the harmonic sum of the leading base-$p$ digit $m$ is zero modulo $p$. For $p=2$, $m=1$ always, so two never divides the gcd. More generally primes with leading digit one cannot obstruct it.The reproducible exact-rational loop in verification/batch_2_checks.py computes $H_n$ and $L_n$ incrementally for every $1\le n\le1000$. In lowest terms $H_n=u_n/v_n$, the gcd is $L_n/v_n$, so the test is simply $v_n=L_n$. It finds 145 successful indices, including every $n$ from 968 through 1000. This last run is finite and gives no extrapolation beyond the checked range.