gpt pro 5.4 attempt on Math-5960

Archive note. This extract contains the work on Problem #873 from a joint report on Problems #789, #873, and #875. The shared introduction, verification, conclusions, and references are retained. The source's 68% completion estimate applies to the joint report.

Round 2 continuation for Erdős problems #789, #873, #875

1) ROUND-2 OBJECTIVE

Because Round 1 bundled three Erdős problems into a single file, this continuation also treats the three problems together. I am pursuing path (C): I identify and correct a concrete Round 1 error for Problem #875 (a missed theorem from the literature), then propagate the corrected consequences. In the same pass I also strengthen the Round 1 upper bound for Problem #789 and prove a new fixed-$k$ obstruction for Problem #873. None of the three problems is fully solved, so the global status remains unresolved.

2) Round-1 FOUNDATION USED

I use the following Round 1 facts without reproving them.

1. For Problem #789, Round 1 already corrected the statement to the standard distinct-summand notion of admissibility, and proved the lower bound $h(n) \gg n^{1/3}$.
2. For Problem #873, Round 1 proved the monotonicity of $F(A,X,k)$ in $k$, namely that $F(A,X,k') \le F(A,X,k)$ whenever $k'\ge k$.
3. For Problem #875, Round 1 observed that every finite initial segment $A\cap [1,N]$ of an infinite admissible set is itself admissible, so any sharp finite bound inside $[1,N]$ immediately yields a universal density bound for infinite admissible sets.

3) NEW INSIGHT / TOOL (ROUND-2)

There are three genuinely new ingredients in this round.

1. Problem #789. I import the sharp Deshouillers--Freiman theorem for finite admissible subsets of $[1,N]$, which upgrades the Round 1 upper bound from $\sqrt{7n}$ to $2\sqrt{n}+O(1)$.
2. Problem #873. I introduce a new divisor-block construction: concatenate scaled divisor sets of primorials. This produces, for every fixed $k$, an infinite set $A$ for which $F(A,X,k)$ is larger than every power of $\log X$ infinitely often. This does not disprove the Erdős--Szemerédi conjecture, but it shows that any proof must use the freedom to let $k$ depend on $A$ and $\varepsilon$.
3. Problem #875. Round 1 missed Erdős--Nicolas--Sárkőzy's 1991 theorem constructing an infinite admissible set with polynomial density $A(x)\gg x^{5-2\sqrt6}$. After correcting this, I derive new consequences: existence of polynomial-growth admissible sets, existence of an admissible set with $\liminf a_{n+1}/a_n = 1$, a sharpened universal lower bound $a_n\ge n(n+2)/4$ for all large $n$, and a sharpened gap obstruction $\limsup (a_{n+1}-a_n)/n\ge 1/2$.

4) ATTACK PLAN (ROUND-2)

After Round 1, the exact remaining gaps were as follows.

1. For Problem #789, the upper bound was only $O(\sqrt n)$ with a weak constant. The natural gap was to import the sharp finite-interval extremal theorem and transfer it immediately to $h(n)$ by taking the special set $[1,n]$.
2. For Problem #873, Round 1 had no construction showing how badly a fixed value of $k$ can fail. The precise target now is to prove that for every fixed $k$, there is an infinite $A$ for which $F(A,X,k)$ is still very large infinitely often.
3. For Problem #875, the most serious issue is factual: Round 1 treated polynomial growth as if it might still be unknown. I first repair that mistake using the 1991 theorem, and then extract the strongest easy corollaries for growth and gap questions.

5) WORK (ROUND-2)

Problem #873: a new fixed-$k$ obstruction

For a fixed infinite set $A=\{a_1<a_2<\cdots\}\subset \mathbf{N}$, recall \[ F(A,X,k) := \#\big\{i\ge 1 : \mathrm{lcm}(a_i,a_{i+1},\dots,a_{i+k-1}) < X\big\}. \]

Claim 873.1. Fix $k\ge 1$. There exists an infinite increasing set $A\subset \mathbf{N}$ and a sequence $X_t\to\infty$ such that \[ F(A,X_t,k) \ge \exp\!\left((\log 2 + o(1))\frac{\log X_t}{\log\log X_t}\right). \] In particular, for every fixed $C>0$, \[ F(A,X_t,k) > (\log X_t)^C \] for all sufficiently large $t$.

Proof. Choose a rapidly increasing sequence of real numbers $y_t\to\infty$ recursively so that \[ y_t > q_{t-1}P_{t-1} \] for every $t\ge 2$, where $P_{t-1}$ and $q_{t-1}$ will be defined below. For each $t$, let \[ P_t := \prod_{p\le y_t} p, \] where the product runs over all primes $p\le y_t$.

By Bertrand's postulate there exists a prime $q_t$ with \[ y_t < q_t < 2y_t. \] Since every prime factor of $P_t$ is at most $y_t$, this gives \[ \gcd(q_t,P_t)=1. \] Also, because $y_t > q_{t-1}P_{t-1}$, we have \[ q_t > y_t > q_{t-1}P_{t-1}, \] so the intervals $[q_t, q_tP_t]$ are strictly increasing and pairwise disjoint.

Now define a finite block \[ B_t := \{q_t d : d\mid P_t\}, \] listed in increasing order. Because the blocks are disjoint and ordered by size, the infinite union \[ A := B_1 \cup B_2 \cup B_3 \cup \cdots \] forms an increasing sequence.

Set \[ X_t := q_tP_t + 1. \] Take any $k$ consecutive elements of the block $B_t$, say \[ q_t d_1 < q_t d_2 < \cdots < q_t d_k, \] with each $d_j\mid P_t$. Since $\gcd(q_t,P_t)=1$, \[ \mathrm{lcm}(q_t d_1,\dots,q_t d_k) = q_t\,\mathrm{lcm}(d_1,\dots,d_k) \le q_tP_t < X_t. \] Hence every starting position of a length-$k$ run entirely inside the block $B_t$ contributes to $F(A,X_t,k)$. Therefore \[ F(A,X_t,k) \ge |B_t|-k+1. \] Now $|B_t|$ is the number of divisors of $P_t$, and since $P_t$ is squarefree, \[ |B_t| = \tau(P_t)=2^{\pi(y_t)}, \] where $\pi(y_t)$ is the number of primes at most $y_t$. Thus \[ F(A,X_t,k) \ge 2^{\pi(y_t)}-k+1. \] Taking logarithms and using the prime number theorem together with the standard asymptotic \[ \log P_t = \vartheta(y_t) \sim y_t, \] we obtain \[ \pi(y_t) = (1+o(1))\frac{y_t}{\log y_t}, \qquad \log X_t = \log(q_tP_t+1) = (1+o(1))y_t, \qquad \log\log X_t = (1+o(1))\log y_t. \] Substituting into the previous bound yields \[ \log F(A,X_t,k) \ge (\log 2 + o(1))\frac{\log X_t}{\log\log X_t}, \] which is the claimed quantitative lower bound.

Finally, for any fixed $C>0$, \[ (\log 2 + o(1))\frac{\log X_t}{\log\log X_t} \gg \log\log X_t, \] so exponentiating gives \[ F(A,X_t,k) > (\log X_t)^C \] for all large $t$. $\Box$

Corollary 873.2. No theorem of the form \[ F(A,X,k_0) \le (\log X)^C \] can hold uniformly for all infinite $A$ with a fixed value of $k_0$.

Explanation. Claim 873.1 gives a counterexample for every fixed $k_0$. This does not refute the original Erdős--Szemerédi problem, because that problem allows $k$ to depend on $A$ and on $\varepsilon$. What it does show is that allowing $k$ to grow is genuinely necessary.

6) ADVERSARIAL VERIFICATION

Problem #789.

The only nontrivial hypothesis in Claim 789.1 is the Deshouillers--Freiman theorem, whose hypotheses are exactly satisfied by admissible subsets of $[1,n]$. The transfer from the interval problem to $h(n)$ is logically sound because $h(n)$ is a universal guarantee over all $n$-element sets, so one counterexample family of size $n$ immediately gives an upper bound.

Problem #873.

The divisor-block construction must satisfy three separate requirements: blocks must be ordered, the lcm inside each block must stay below the chosen threshold, and the quantitative lower bound must really exceed every power of $\log X$. The recursive condition $y_t>q_{t-1}P_{t-1}$ guarantees block ordering. The coprimality $\gcd(q_t,P_t)=1$ makes the lcm computation exact. Finally, the prime number theorem turns the divisor count $2^{\pi(y_t)}$ into the stated lower bound. No cross-block interaction is needed, because I only need a lower bound on $F(A,X_t,k)$.

Problem #875.

The main issue was factual: Round 1 overlooked a published theorem. After correcting that, all subsequent deductions are elementary. Claim 875.1 uses only the inversion of the counting-function lower bound and a standard exponential-growth contradiction. Claim 875.2 applies the finite Deshouillers--Freiman theorem to the initial segment $A\cap[1,N]$. Claim 875.3 sums an eventual linear gap bound and compares the quadratic leading coefficient against Claim 875.2; the contradiction is at the coefficient level $(1/4-\varepsilon/2)$ versus $1/4$, so there is no hidden constant issue. Claim 875.4 only claims existence for $c>5+2\sqrt6$, which is exactly the exponent one gets for free from the polynomial upper bound on $a_n$.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict advances over Round 1 are these.

1. Problem #789: the upper bound is sharpened from $\sqrt{7n}$ to $2\sqrt n+O(1)$.
2. Problem #873: a new fixed-$k$ obstruction is proved. For every fixed $k$, some infinite $A$ has $F(A,X,k)$ larger than every power of $\log X$ infinitely often.
3. Problem #875: a concrete Round 1 error is corrected. Polynomial-growth admissible sets already exist. As a result, the old final status for #875 must be replaced by the sharper quantitative bracket \[ c<\frac12 \text{ impossible}, \qquad c>5+2\sqrt6 \text{ possible}. \] Moreover there exists an admissible set with $\liminf a_{n+1}/a_n=1$.

8) COMPLETION ESTIMATE

COMPLETION: 68%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #789, problem page and discussion thread.
2. T. F. Bloom, Erdős Problem #873, problem page and discussion thread.
3. T. F. Bloom, Erdős Problem #875, problem page and discussion thread.
4. J.-M. Deshouillers and G. A. Freiman, On an additive problem of Erdős and Straus, 2, in Structure theory of set addition, Astérisque 258 $1999$, 141--148.
5. P. Erdős, J.-L. Nicolas, and A. Sárkőzy, Sommes de sous-ensembles, Séminaire de théorie des nombres de Bordeaux 3 $1991$, 55--72.
6. Prime Number Theorem and the asymptotic $\vartheta(x)\sim x$ for the Chebyshev function (used only in the quantitative corollary of Claim 873.1).

10) OUTPUT FORMAT NOTE

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