gpt pro 5.4 attempt on Math-5960, version 2

Round 2 continuation for Erdős Problem #873

1) ROUND-2 OBJECTIVE

I pursue path (C). The most promising Round-2 advance is to identify a genuine obstruction to any fixed-$k$ strategy. Round 1 already explained that the known Erdős--Szemerédi bound for $k=3$ settles the problem only for exponents $\varepsilon>1/3$. What was still missing was a rigorous reason that one cannot hope to settle the full conjecture with one universal, fixed value of $k$. I will prove exactly that.

2) Round-1 FOUNDATION USED

I use the following Round 1 results without reproving them.

1. Round 1 defined \[ F(A,X,k)=\#\Big\{i\ge 1:\operatorname{lcm}(a_i,a_{i+1},\dots,a_{i+k-1})<X\Big\} \] for $A=\{a_1<a_2<\cdots\}\subset \mathbf{N}$.
2. Round 1 proved monotonicity in the block length: \[ k'\ge k \implies F(A,X,k')\le F(A,X,k). \]
3. Round 1 noted that the known Erdős--Szemerédi theorem for $k=3$ gives \[ F(A,X,3)\ll X^{1/3}\log X, \] so the original problem is only difficult for $\varepsilon\le 1/3$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is a divisor-block construction. For each large parameter $y$, take the primorial \[ P(y):=\prod_{p\le y}p, \] pick a prime $q>y$ coprime to $P(y)$, and form the block \[ B=\{qd:d\mid P(y)\}. \] Inside such a block, every $k$ consecutive terms have least common multiple at most $qP(y)$, while the block length is exactly the divisor count $\tau(P(y))=2^{\pi(y)}$. By concatenating many such blocks far apart, one gets an infinite sequence for which $F(A,X,k)$ is very large infinitely often for any fixed $k$.

4) ATTACK PLAN (ROUND-2)

The precise Round 1 gap was the absence of a theorem showing that $k$ really must depend on the set $A$. I will now prove the following claim.

Target claim. For every fixed integer $k\ge 1$, there exists an infinite set $A\subset \mathbf{N}$ and a sequence $X_t\to\infty$ such that \[ F(A,X_t,k)\ge \exp\!\left((\log 2+o(1))\frac{\log X_t}{\log\log X_t}\right). \]

This implies that no upper bound of polylogarithmic size can hold uniformly for all sets $A$ with a single fixed $k$. Hence any eventual proof of the original conjecture must use the freedom to choose $k=k(A,\varepsilon)$.

5) WORK (ROUND-2)

Let $k\ge 1$ be fixed once and for all.

Claim 873.1. 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 constant $C>0$, \[ F(A,X_t,k)>(\log X_t)^C \] for all sufficiently large $t$.

Proof. Choose a sequence $y_t\to\infty$ recursively so rapidly that, for every $t\ge 2$, \[ y_t>q_{t-1}P_{t-1}, \] where $P_{t-1}$ and $q_{t-1}$ are objects defined at the previous stage.

For each $t$, define \[ P_t:=\prod_{p\le y_t}p. \] 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 $y_t$, we have \[ \gcd(q_t,P_t)=1. \]

Now define the finite block \[ B_t:=\{q_t d:d\mid P_t\}, \] listed in increasing order. Because $q_t>q_{t-1}P_{t-1}$, every element of $B_t$ is larger than every element of $B_{t-1}$. Hence the disjoint union \[ A:=B_1\cup B_2\cup B_3\cup \cdots \] is an infinite increasing sequence.

Set \[ X_t:=q_tP_t+1. \] Consider any run of $k$ consecutive elements lying entirely inside 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 $q_t$ is coprime to $P_t$, \[ \operatorname{lcm}(q_t d_1,\dots,q_t d_k)=q_t\,\operatorname{lcm}(d_1,\dots,d_k)\le q_tP_t<X_t. \] Therefore every starting position of a length-$k$ window wholly contained in $B_t$ contributes to $F(A,X_t,k)$. So \[ F(A,X_t,k)\ge |B_t|-k+1. \]

Because $P_t$ is squarefree, the number of divisors of $P_t$ is \[ |B_t|=\tau(P_t)=2^{\pi(y_t)}, \] where $\pi(y_t)$ is the number of primes at most $y_t$. Hence \[ F(A,X_t,k)\ge 2^{\pi(y_t)}-k+1. \]

Now use the prime number theorem and the standard asymptotic $\vartheta(y)\sim y$ for the Chebyshev function. Since \[ \log P_t=\vartheta(y_t)\sim y_t, \qquad \pi(y_t)\sim \frac{y_t}{\log y_t}, \] and $q_t<2y_t$, we get \[ \log X_t=\log(q_tP_t+1)=(1+o(1))y_t, \qquad \log\log X_t=(1+o(1))\log y_t. \] Substituting into the lower bound for $F(A,X_t,k)$ yields \[ \log F(A,X_t,k)\ge (\log 2+o(1))\frac{\log X_t}{\log\log X_t}, \] which is equivalent to \[ F(A,X_t,k)\ge \exp\!\left((\log 2+o(1))\frac{\log X_t}{\log\log X_t}\right). \]

Finally, since \[ \frac{\log X_t}{\log\log X_t}\gg \log\log X_t, \] the above lower bound dominates every power of $\log X_t$. Thus for any fixed $C>0$, \[ F(A,X_t,k)>(\log X_t)^C \] for all sufficiently large $t$. $\Box$

Corollary 873.2. Fix $k_0\ge 1$. There is no bound of the form \[ F(A,X,k_0)\le (\log X)^C \] that can hold uniformly for all infinite $A\subset \mathbf{N}$ and all large $X$.

Proof. Apply Claim 873.1 with $k=k_0$. The resulting set $A$ violates any such bound along the sequence $X_t$. $\Box$

Interpretation. This does not disprove the Erdős--Szemerédi problem, because the original statement allows $k$ to depend on $A$ and on $\varepsilon$. What it proves is a genuine obstruction: a proof with one fixed universal $k$ cannot work.

6) ADVERSARIAL VERIFICATION

I check the potential failure points one by one.

1. Are the blocks really increasing? Yes. The recursive choice $y_t>q_{t-1}P_{t-1}$ implies $q_t>y_t>q_{t-1}P_{t-1}$, so every element of $B_t$ exceeds every element of $B_{t-1}$.
2. Do the lcm computations use hidden assumptions? No. Since $\gcd(q_t,P_t)=1$, one has \[ \operatorname{lcm}(q_t d_1,\dots,q_t d_k)=q_t\operatorname{lcm}(d_1,\dots,d_k). \]
3. Do cross-block windows matter? No. I only need a lower bound for $F(A,X_t,k)$, so it suffices to count windows lying entirely inside $B_t$.
4. Is the quantitative asymptotic consistent? Yes. The divisor count of a primorial is $2^{\pi(y_t)}$, and the prime number theorem gives exactly the stated exponential-in-$\log X/\log\log X$ size.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 proves a new obstruction absent from Round 1: for every fixed $k$, there exists an infinite set $A$ for which $F(A,X,k)$ is larger than every power of $\log X$ infinitely often. Thus any proof of the original conjecture must use the freedom to let $k$ depend on $A$ and $\varepsilon$. The full Erdős--Szemerédi question nevertheless remains open.

8) COMPLETION ESTIMATE

COMPLETION: 38%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #873, problem page.

10) OUTPUT FORMAT NOTE

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