gpt pro 5.2 attempt on Math-5999

1) FORMAL RESTATEMENT

Let k$\ge$ 2 be an integer. For x$\ge$ 2, define the property (*)_{k,x}(n): \[(\*)_{k,x}(n)\quad\text{means: for each } i\in\{1,2,\dots,k\},\; n+i \text{ is divisible by at least one prime } p\le x.\] Equivalently, each of the k consecutive integers n+1,…,n+k has a prime factor $\le$ x.

Let A_{k,x}:={n$\in\mathbb N$: (*)_{k,x}(n) holds}. Define S(k) to be the minimal real x such that A_{k,x} has positive (natural) density, i.e. \[ \limsup_{N\to\infty} \frac{|A_{k,x}\cap[1,N]|}{N} > 0. \] (For fixed k, the intended meaning of “positive density set of n” in the problem statement is: the asymptotic density exists and is >0, or at least that the upper density is >0; for our purposes we will show that once A_{k,x} is nonempty it already contains an arithmetic progression and hence has positive density.)

Question: Estimate S(k) as k$\to\infty$. In particular, is S(k)$\ge$ k^{1-o$1$}?

2) QUICK LITERATURE/CONTEXT CHECK

The problem statement itself records: - A lower bound from Rosser’s sieve: S(k) > k^{1/2-o$1$}. - A trivial upper bound: $S(k)\le k+1$ (e.g. take $n\equiv 1 \pmod{(k+1)!}$). - An improved upper bound from large prime gaps (Ford–Green–Konyagin–Maynard–Tao): $S(k) \ll k \frac{\log\log\log k}{\log\log k\,\log\log\log\log k}.$

I do not use or assert any additional external results beyond what is stated in the problem file.

3) ATTACK PLAN

Proof-track ideas (lower bounds on S(k)): - Translate the condition into a covering problem modulo the primorial $P(x)=\prod_{p\le x} p$ and apply sieve lower bounds to show one cannot cover an interval of length k using only primes $\le$ x when x is too small.

Construction-track ideas (upper bounds on S(k)): - Use Chinese remainder / covering congruences to force each of n+1,…,n+k to be divisible by a chosen small prime, yielding explicit arithmetic progressions.

Best path here: I can prove exact equivalences and the trivial upper bound rigorously, and I can compute small values of S(k) exactly for small k by exhaustive search modulo primorials. The asymptotic lower bound S(k)$\ge$ k^{1-o$1$} remains open.

4) WORK

Lemma 929.1 (nonempty $\Rightarrow$ positive density). Fix k$\ge$ 2 and x$\ge$ 2. Let M(x):=$\prod_{p\le x}$ p (the product of all primes $\le$ x). If there exists n_0 such that (*)_{k,x}$n_0$ holds, then A_{k,x} contains the entire arithmetic progression {n_0 + t M(x): t$\in\mathbb Z_{\ge 0}$}. In particular, A_{k,x} has natural density at least 1/M(x), hence positive.

Proof. Assume (*)_{k,x}$n_0$ holds. Fix i$\in${1,…,k}. By definition, there exists a prime p_i$\le$ x such that p_i divides n_0+i. Since p_i$\le$ x, the prime p_i is one of the factors of M(x), so p_i divides M(x). Therefore for any t$\ge$ 0, \[ n_0 + tM(x) + i \equiv n_0 + i \pmod{p_i}, \] so p_i divides n_0+tM(x)+i. Thus for each i, the integer n_0+tM(x)+i is divisible by a prime $\le$ x. Hence (*)_{k,x}(n_0+tM(x)) holds for every t$\ge$ 0. So the full arithmetic progression is contained in A_{k,x}. The progression has density exactly 1/M(x), so A_{k,x} has density at least 1/M(x)>0. $\square$

Corollary 929.1a. For fixed k, S(k) equals the minimal x such that there exists at least one integer n with (*)_{k,x}(n).

Proof. If A_{k,x} is nonempty then by Lemma 929.1 it has positive density, so such x is admissible for S(k). Conversely, if A_{k,x} has positive density then it is certainly nonempty. Thus “positive density” and “nonempty” are equivalent for A_{k,x}, and taking minima over x gives the corollary. $\square$

Lemma 929.2 (trivial upper bound S(k)$\le$ k+1). For every integer k$\ge$ 2, we have S(k)$\le$ k+1.

Proof. Let x:=k+1 and take n satisfying \[ n \equiv 1 \pmod{(k+1)!}. \] Then for each i$\in${1,…,k}, we have \[ n+i = 1 + i + t (k+1)! = (i+1) + t (k+1)! \] for some integer t, hence $i+1$ divides n+i because $i+1$ divides $k+1$!. Therefore n+i is divisible by a prime factor of $i+1$, and every prime factor of $i+1$ is $\le$ i+1$\le$ k+1=x. So (*)_{k,x}(n) holds. By Corollary 929.1a, this implies S(k)$\le$ x=k+1. $\square$

FAST REALITY CHECK (exact small-k computation). Using Corollary 929.1a, I computed S(k) for k$\le$ 30 by brute force search modulo primorials. I also computed, for each x, the maximum length L(x) of a run of consecutive integers all having a prime factor $\le$ x. Exact values found: - L$2$=1, L$3$=3, L$5$=5, L$7$=9, L$11$=13, L$13$=21, L$17$=25, L$19$=33. From these, the exact S(k) for k$\le$ 30 is: S$1$=2, S$2$=3, S$3$=3, S$4$=5, S$5$=5, S$6$=7, S$7$=7, S$8$=7, S$9$=7, S$10$=11, S$11$=11, S$12$=11, S$13$=11, S$14$=13, S$15$=13, S$16$=13, S$17$=13, S$18$=13, S$19$=13, S$20$=13, S$21$=13, S$22$=17, S$23$=17, S$24$=17, S$25$=17, S$26$=19, S$27$=19, S$28$=19, S$29$=19, S$30$=19.

An explicit witness for k=33 and x=19 is n=60043, since each of 60044,60045,…,60076 has a prime factor $\le$ 19; by Lemma 929.1 this yields a positive-density arithmetic progression of solutions.

5) VERIFICATION

- Lemma 929.1: Checked that using M(x)=$\prod_{p\le x}$p ensures every prime p$\le$ x divides M(x), so divisibility persists under translation by M(x). - Lemma 929.2: Checked that for each i, i+1 divides $k+1$!, hence divides n+i. - Computational check: Verified the run length L$19$=33 by direct gcd computation against the primorial 2$\cdot$3$\cdot$5$\cdot$7$\cdot$11$\cdot$13$\cdot$17$\cdot$19=9699690.

6) FINAL

UNRESOLVED (i) Strongest proved partial result: For each fixed (k,x), existence of a single n with the property implies a positive-density infinite arithmetic progression of such n (Lemma 929.1), and in particular S(k) can be characterized as the minimal x for which at least one such n exists (Corollary 929.1a). Also S(k)$\le$ k+1 holds by an explicit congruence construction (Lemma 929.2). Exact values of S(k) were computed for k$\le$ 30. (ii) First gap (crisp): Prove a near-linear lower bound $S(k)\ge k^{1-o(1)}$ (or exhibit a family of coverings showing $S(k)\le k^{1-\delta}$ for some fixed $\delta>0$ infinitely often). (iii) Top 3 next moves: 1. Reformulate the problem as a covering congruence problem: choose for each prime $p\le x$ one residue class a_p (mod p) so that {1,…,k} is covered by $\bigcup_{p\le x}\{i: i\equiv a_p\ (\mathrm{mod}\ p)\}$. Use this to search algorithmically for better constructions and to attempt lower bounds. 2. Attempt to prove a sieve lower bound on the Jacobsthal-type function for primorials: show the maximum run of integers not coprime to $\prod_{p\le x} p$ is $\le x^{1+o(1)}$ (or similar), then translate to S(k) lower bounds. 3. Extend brute force computations of S(k) to larger k (within feasible primorial sizes) to guess the true order of growth and test the plausibility of $S(k)\ge k^{1-o(1)}$. (iv) Minimal counterexample structure: A counterexample to $S(k)\ge k^{1-o(1)}$ would provide a sequence $k_j\to\infty$ and primes bounded by $x_j\le k_j^{1-\delta}$ together with residue classes mod those primes that cover all indices 1..k_j; equivalently, runs of length k_j in the sequence of integers all having a prime factor $\le$ x_j with x_j substantially smaller than k_j.