gpt pro 5.4 attempt on Math-5893

Problem #783

1. ROUND-2 OBJECTIVE

Here I do pursue path (C): the literal "take the consecutive largest primes" prescription is too rigid as an exact finite-$N$ statement, but the correct asymptotic minimum-value problem can now be solved completely. So the Round-2 objective is:

1. exhibit a concrete obstruction to the literal finite formulation;
2. state the minimal corrected asymptotic statement;
3. prove that corrected statement rigorously.

2. Round-1 FOUNDATION USED

I use the following Round-1 ingredients.

1. The notation \[ U_A(N):=\{1\leq n\leq N:\ a\nmid n\ \mbox{for all }a\in A\}, \] for an admissible pairwise coprime set $A\subseteq\{2,\dots,N\}$ with \[ \mu(A):=\sum_{a\in A}\frac{1}{a}\leq C. \]
2. The Round-1 heuristic that the prime-tail construction should leave essentially the $y$-smooth numbers uncovered, suggesting the density $\rho(e^C)$.
3. The special-regime argument from Round 1 showing asymptotic optimality of large primes when $C<\log 2$.

3. NEW INSIGHT / TOOL (ROUND 2)

There are two genuinely new ingredients.

1. A concrete finite obstruction to the literal greedy prime-tail prescription: it is not always the exact best finite set.
2. Tao's 2026 theorem [Tao26]: if $A$ is pairwise coprime, does not contain $1$, and $\mu(A)=O(1)$, then \[ \sigma_N(A):=\frac{|U_A(N)|}{N}\geq \rho(e^{\mu(A)})+o(1) \qquad (N\to\infty). \] Combined with the prime-tail upper bound, this settles the asymptotic minimum uncovered density.

This completely upgrades the Round-1 heuristic for the weak/asymptotic form into a theorem.

4. ATTACK PLAN (ROUND 2)

The exact gap left by Round 1 was the absence of a matching lower bound for all pairwise coprime $A$. The prime-tail construction gave the candidate upper bound $\rho(e^C)$, but there was no proof that arbitrary pairwise-coprime composite sets could not do better.

Round 2 overcomes that gap in two stages:

1. first, show by explicit example that the literal "consecutive largest primes" rule is not the right finite-$N$ statement;
2. then prove the corrected asymptotic theorem \[ \min_A |U_A(N)| = \left(\rho(e^C)+o(1)\right)N, \] where the minimum is over all admissible pairwise coprime $A\subseteq\{2,\dots,N\}$.

5. WORK (ROUND 2)

5.1 A concrete obstruction to the literal greedy prime-tail rule.

Take \[ N=30,\qquad C=1. \] The literal greedy prescription from the problem statement chooses the largest primes in descending order until the reciprocal budget is exhausted: \[ A_0=\{29,23,19,17,13,11,7,5\}, \] because \[ \frac1{29}+\frac1{23}+\frac1{19}+\frac1{17}+\frac1{13}+\frac1{11}+\frac17+\frac15<1, \] but adding $1/3$ would exceed $1$.

The integers up to $30$ not divisible by any element of $A_0$ are \[ U_{A_0}(30)=\{1,2,3,4,6,8,9,12,16,18,24,27\}, \] so \[ |U_{A_0}(30)|=12. \]

Now consider instead the prime set \[ A_1=\{23,19,17,13,11,7,5,3\}. \] Its reciprocal sum is \[ \frac1{23}+\frac1{19}+\frac1{17}+\frac1{13}+\frac1{11}+\frac17+\frac15+\frac13 \approx 0.998956<1, \] so $A_1$ is also admissible. But now \[ U_{A_1}(30)=\{1,2,4,8,16,29\}, \] hence \[ |U_{A_1}(30)|=6<12=|U_{A_0}(30)|. \]

So the literal finite-$N$ rule \[ \mbox{``take the consecutive largest primes''} \] is not an exact extremal prescription.

This does not refute the asymptotic minimum-value prediction; it only shows that the statement needs correction.

5.2 Corrected asymptotic theorem.

Define \[ F_N(C):=\min \left\{ |U_A(N)|:\ A\subseteq\{2,\dots,N\}\ \mbox{pairwise coprime},\ \sum_{a\in A}\frac1a\leq C \right\}. \]

The corrected statement is:

Theorem. For every fixed $C>0$, \[ F_N(C)=\left(\rho(e^C)+o(1)\right)N \qquad (N\to\infty), \] where $\rho$ is the Dickman function.

I now prove this.

5.3 Lower bound via Tao's theorem.

Let $A\subseteq\{2,\dots,N\}$ be admissible. Then:

1. $A$ is finite;
2. $A$ is pairwise coprime;
3. $1\notin A$ automatically;
4. $\mu(A)=\sum_{a\in A}1/a\leq C=O(1)$.

Therefore Tao's Theorem 1.1 applies and gives \[ \sigma_N(A):=\frac{|U_A(N)|}{N}\geq \rho(e^{\mu(A)})+o(1). \] Since $\mu(A)\leq C$ and the Dickman function $\rho(u)$ is decreasing for $u\geq 1$, we obtain \[ \sigma_N(A)\geq \rho(e^C)+o(1). \] Multiplying by $N$ gives \[ |U_A(N)|\geq \left(\rho(e^C)+o(1)\right)N. \] As this holds for every admissible $A$, \[ F_N(C)\geq \left(\rho(e^C)+o(1)\right)N. \]

5.4 Upper bound via a tail of primes.

Now choose $y=y(N)$ so that \[ \sum_{y<p\leq N}\frac1p = C+o(1), \] and let \[ A_y:=\{p\ \mbox{prime}: y<p\leq N\}. \] By Mertens' theorem, this choice corresponds to \[ y=N^{e^{-C}+o(1)}. \]

For this set, an integer $n\leq N$ is not divisible by any element of $A_y$ if and only if all prime factors of $n$ are at most $y$; in other words, $n$ is $y$-smooth. Therefore \[ |U_{A_y}(N)|=\Psi(N,y), \] where $\Psi(N,y)$ is the counting function for $y$-smooth integers up to $N$.

Tao explicitly records on the first page of [Tao26] that for this prime-tail example, \[ \sigma_N(A_y)=\rho(e^C)+o(1), \] equivalently \[ |U_{A_y}(N)|=\left(\rho(e^C)+o(1)\right)N. \]

If one wants the budget condition $\mu(A)\leq C$ rather than $C+o(1)$ exactly, remove from $A_y$ a set of smallest primes in the tail whose total reciprocal mass is $o(1)$. By the Lipschitz lemma in [Tao26], \[ |\sigma_N(A')-\sigma_N(A_y)|\leq \mu(A_y\triangle A')=o(1), \] so this trimming changes the uncovered density by only $o(1)$.

Hence there exists an admissible set $A$ with \[ |U_A(N)|\leq \left(\rho(e^C)+o(1)\right)N. \] Therefore \[ F_N(C)\leq \left(\rho(e^C)+o(1)\right)N. \]

5.5 Conclusion.

Combining the lower and upper bounds yields \[ F_N(C)=\left(\rho(e^C)+o(1)\right)N. \]

This is the corrected asymptotic answer to Erdős's question.

5.6 How Tao's theorem closes the Round-1 obstruction.

Round 1 had identified the main obstruction correctly: a naive Buchstab recursion closes cleanly for a tail of primes but not for general composite moduli. Tao's note [Tao26] overcomes this as follows.

1. Large composite elements of a pairwise coprime set are sparse (Lemma 2.1), so they contribute only $o(1)$ total reciprocal mass and may be deleted with only $o(1)$ change in the sieve density by the Lipschitz bound (Lemma 2.2).
2. After this deletion, one splits \[ A=A_1\cup A_2, \] where $A_1$ contains the small moduli and $A_2$ contains only large primes, separated by a prime-free gap. Lemma 2.4 shows that the truncated inclusion--exclusion sum approximately factorizes: \[ \sigma_{N,r}(A)\approx \sigma_{N,r}(A_1)\sigma_{N,r}(A_2). \]
3. The small-modulus piece $A_1$ is replaced by its limiting Euler product using the pure Brun sieve (Lemma 2.5), while the large-prime piece $A_2$ is handled by the prime case (Theorem 1.2, due originally to Hildebrand).
4. Finally one combines the two pieces using the multiplicative structure of $\sigma_\infty(A_1)$ and the Dickman inequality \[ \rho(u_1)\rho(u_2)\geq \rho(u_1u_2), \] which is exactly what was missing from the Round-1 heuristic.

So the Round-2 theorem is not just a black box citation: it directly resolves the precise structural obstacle left open in Round 1.

6. ADVERSARIAL VERIFICATION

I checked the main possible weak points.

1. Does the explicit $N=30$ example only show a finite glitch? Yes, exactly. It shows that the literal finite-$N$ greedy prescription is too rigid. It does not contradict the corrected asymptotic theorem.

2. Did the lower bound use Tao's theorem within its hypotheses? Yes. Our admissible sets are finite, pairwise coprime, avoid $1$, and have $\mu(A)\leq C=O(1)$, exactly as required.

3. Does the upper bound really satisfy the budget $\leq C$ rather than $C+o(1)$? Yes, after trimming $o(1)$ reciprocal mass from the prime tail; the Lipschitz lemma guarantees only an $o(1)$ change in the density.

4. Could the corrected theorem be stronger than what Tao proved? No. Tao's theorem gives the universal lower bound \[ \sigma_N(A)\geq \rho(e^{\mu(A)})+o(1), \] and he already records that the prime-tail construction attains this asymptotically. I only combine these two statements.

5. What remains open after this correction? The exact identity of the finite-$N$ extremal set remains open in general. The theorem proved here determines the asymptotic minimum value, not the unique exact optimizer.

7. FINAL

Partial Known SOLUTION -- OBSTRUCTION/CORRECTION.

The literal finite-$N$ prescription "take the consecutive largest primes" is not correct as an exact statement; an explicit counterexample is provided at $(N,C)=(30,1)$.

The minimal corrected asymptotic statement is:

\[ \min |U_A(N)|=\left(\rho(e^C)+o(1)\right)N, \] where the minimum is taken over all pairwise coprime \[ A\subseteq\{2,\dots,N\} \qquad\mbox{with}\qquad \sum_{a\in A}\frac1a\leq C. \]

That corrected statement is now fully proved, by combining the prime-tail upper bound with Tao's 2026 lower bound for arbitrary pairwise coprime sets.

8. COMPLETION ESTIMATE

COMPLETION: 92%

Note: This write-up largely follows the discussion https://www.erdosproblems.com/forum/thread/783

9. References used in this section

1. [EP783 ] Erdős Problems and conjectures, Problem #783, erdosproblems.com/783. Checked March 7, 2026.
2. [Tao26 ] T. Tao, Sieving by coprime numbers, note dated February 21, 2026.
3. [Choj26 ] P. Chojecki, A Buchstab--Dickman note and an attempted port to an extremal coprime sieve problem, 2026 note posted at ulam.ai/research/erdos783.pdf. Used here only for historical/contextual comparison and the $C\leq \log 2$ regime.