Reused writeup on Math-5972 (gpt pro 5.2)

Erdős Problem 890 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/890

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/890.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/890.tex.

1) FORMAL RESTATEMENT

Let $\omega(m)$ be the number of distinct prime divisors of $m$, and let $\pi(k)$ be the number of primes $\le k$. For a fixed integer $k\ge 1$, define \[ S_k(n) := \sum_{i=0}^{k-1} \omega(n+i). \] Question (A): Is it true that \[ \liminf_{n\to\infty} S_k(n) \le k + \pi(k) ? \] Question (B): Is it true that \[ \limsup_{n\to\infty} S_k(n)\,\frac{\log\log n}{\log n} = 1? \]

2) QUICK LITERATURE/CONTEXT CHECK

The problem text states that Erdős and Selfridge observed \[ \liminf_{n\to\infty} S_k(n) \ge k+\pi(k)-1, \] and sketches a justification referencing a theorem of Pólya about $k$-smooth integers having unbounded gaps, together with the simple fact that $n(n+1)\cdots(n+k-1)$ is divisible by every prime $\le k$. The problem also states (as a known fact) that \[ \limsup_{n\to\infty} \omega(n)\,\frac{\log\log n}{\log n} = 1. \] I do not use any external results beyond what is explicitly stated.

3) ATTACK PLAN

For (A):

• The lower bound suggests $S_k(n)$ is forced to be at least $k+\pi(k)-1$ for all large $n$ (depending on $k$). Proving an upper bound $\le k+\pi(k)$ for the liminf would require producing infinitely many $n$ where the block $[n,n+k)$ has "as few prime factors as possible".
• Attempt to construct $n$ where one of the $k$ numbers is $k$-smooth (contributing many small primes at once), while the other $k-1$ numbers are each a product of a single prime $>k$ (so each contributes exactly $1$).

For (B):

• Since $S_k(n)\ge \omega(n)$, the limsup is at least $1$. To show it equals $1$, one would need an upper bound preventing several of the consecutive integers from simultaneously having near-maximal $\omega$.
• A disproof would exhibit infinitely many $n$ for which several of $n,\dots,n+k-1$ are extremely "smooth" and squarefree, pushing $S_k(n)$ above the single-term extremal behavior.

4) WORK

Lemma 890.1 (Deriving the Erdős--Selfridge lower bound from a strong "gap" property). Assume the following interpretation of the Pólya statement in the problem text:

(*) For each fixed $k\ge 1$ there exists $N_k$ such that among any $k$ consecutive integers all exceeding $N_k$, at most one is $k$-smooth (i.e. has no prime factor $>k$).

Under (*), for every fixed $k$ we have \[ S_k(n) \ge k+\pi(k)-1\qquad\text{for all }n\ge N_k. \] Consequently $\liminf_{n\to\infty} S_k(n)\ge k+\pi(k)-1$.

Proof. Fix $k$ and take $n\ge N_k$. By (*), among the integers $n,n+1,\dots,n+k-1$ there are at least $k-1$ integers that are not $k$-smooth. For each such integer $m$, there exists a prime $p>k$ dividing $m$, and therefore $\omega(m)\ge 1$. Summing over these $k-1$ integers yields \[ \sum_{i=0}^{k-1} \omega(n+i) \ge k-1 + \sum_{\substack{0\le i<k\\ n+i\ \text{$k$-smooth}}} \omega(n+i). \] Now we lower bound the contribution coming from primes $\le k$. For each prime $p\le k$, among the $k$ consecutive integers $n,n+1,\dots,n+k-1$ there exists an index $0\le i<p\le k$ such that $p\mid(n+i)$. Whenever $p\mid(n+i)$, that prime contributes $1$ to $\omega(n+i)$. Summing over all primes $p\le k$ therefore guarantees at least $\pi(k)$ prime-divisor contributions to the total sum $S_k(n)$. Formally, write \[ \sum_{i=0}^{k-1} \omega(n+i) = \sum_{i=0}^{k-1} \sum_{p\mid(n+i)} 1, \] where the inner sum is over distinct primes dividing $n+i$. For each prime $p\le k$, at least one of the inner sums contains a $1$ coming from $p$, so the whole double sum is at least $\pi(k)$. Combining this with the $k-1$ lower bound coming from the $k-1$ non-$k$-smooth integers gives \[ S_k(n)\ge (k-1)+\pi(k)=k+\pi(k)-1. \] This holds for all $n\ge N_k$, hence the liminf is also at least $k+\pi(k)-1$. $\square$

Lemma 890.2 (Trivial bounds for the limsup in (B)). Assume the fact stated in the problem text that \[ \limsup_{n\to\infty} \omega(n)\,\frac{\log\log n}{\log n}=1. \] Then for every fixed $k\ge 1$, \[ 1 \le \limsup_{n\to\infty} S_k(n)\,\frac{\log\log n}{\log n} \le k. \]

Proof. For the lower bound, note that $S_k(n)\ge \omega(n)$ for all $n$, hence \[ \limsup_{n\to\infty} S_k(n)\,\frac{\log\log n}{\log n} \ge \limsup_{n\to\infty} \omega(n)\,\frac{\log\log n}{\log n}=1. \]

For the upper bound, for each $n$ we have the pointwise inequality \[ S_k(n)=\sum_{i=0}^{k-1}\omega(n+i) \le k\cdot \max_{0\le i<k}\omega(n+i). \] Multiply by $(\log\log n)/(\log n)$ and take limsup: \[ \limsup_{n\to\infty} S_k(n)\,\frac{\log\log n}{\log n} \le k\cdot \limsup_{n\to\infty} \Big(\max_{0\le i<k}\omega(n+i)\Big)\frac{\log\log n}{\log n}. \] For each fixed $i$, the classical limsup identity for $\omega(n)$ also applies to the shifted sequence $\omega(n+i)$ (because $\log(n+i)\sim \log n$ and $\log\log(n+i)\sim \log\log n$ as $n\to\infty$). Hence \[ \limsup_{n\to\infty} \omega(n+i)\,\frac{\log\log n}{\log n}=1\quad\text{for each fixed }i. \] The limsup of the maximum of finitely many sequences equals the maximum of their limsups, so \[ \limsup_{n\to\infty} \Big(\max_{0\le i<k}\omega(n+i)\Big)\frac{\log\log n}{\log n} = 1. \] Combining with the previous inequality gives the upper bound $\le k$. $\square$

Fast reality check (computation). Using a sieve for $\omega(m)$, I computed $S_k(n)$ for $n\le 200{,}000$ and also restricted-range minima on $n\in[100{,}000,200{,}000]$. Selected exact outputs:

• Over $n\in[100{,}000,200{,}000]$, the minimum values found were: \[ \min S_2(n)=2\ \text{at }n=131071,\qquad \min S_3(n)=4\ \text{at }n=131071. \]
• Over $n\le 200{,}000$, the maxima found were: \[ \max S_1(n)=6\ \text{at }n=30030,\qquad \max S_2(n)=10\ \text{at }n=38570,\qquad \max S_3(n)=11\ \text{at }n=30029. \]
• Over $n\le 200{,}000$, the maxima of the scaled quantity $S_k(n)\,\frac{\log\log n}{\log n}$ were: \[ \max_{n\le 200{,}000} S_1(n)\frac{\log\log n}{\log n}\approx 1.363\ \text{at }n=30030, \] \[ \max_{n\le 200{,}000} S_2(n)\frac{\log\log n}{\log n}\approx 2.043\ \text{at }n=30029, \] \[ \max_{n\le 200{,}000} S_3(n)\frac{\log\log n}{\log n}\approx 2.498\ \text{at }n=30029. \]

These are small-$n$ phenomena; they are consistent with the fact that the asymptotic limsup for a single $\omega(n)$ equals $1$.

5) VERIFICATION

Lemma 890.1 cleanly separates the elementary divisibility fact (every prime $\le k$ divides some number among $k$ consecutive integers) from the nontrivial smooth-gap input (*). The counting step uses only that each such prime contributes at least $1$ to some $\omega(n+i)$.

Lemma 890.2 uses only inequalities and the stated classical limsup for $\omega(n)$. The step "limsup of max of finitely many sequences" is valid because for any finite family $(a^{(i)}_n)$, $\limsup_n \max_i a^{(i)}_n = \max_i \limsup_n a^{(i)}_n$.

Computation is a finite-range check and does not address the asymptotic questions directly.

6) FINAL

UNRESOLVED

1. (i) Strongest proved partial result here. Assuming the strong gap property (*) attributed in the problem text to Pólya, one obtains the universal lower bound $S_k(n)\ge k+\pi(k)-1$ for all sufficiently large $n$, hence $\liminf_{n\to\infty} S_k(n)\ge k+\pi(k)-1$. Also, using only the stated classical limsup for $\omega(n)$, we have the unconditional bounds $1\le \limsup S_k(n)\frac{\log\log n}{\log n}\le k$.
2. (ii) First gap (crisp). Prove either (A) $\liminf_{n\to\infty} S_k(n)\le k+\pi(k)$ for each $k$, and (B) $\limsup_{n\to\infty} S_k(n)\frac{\log\log n}{\log n}=1$ (i.e. show that having $k$ terms does not increase the extremal order beyond the single-term behavior).
3. (iii) Top 3 next moves.
   • For (A), attempt to explicitly construct infinitely many $n$ where exactly one of $n,\dots,n+k-1$ collects most primes $\le k$ (a $k$-smooth number) and the other $k-1$ terms are each the product of a single prime $>k$.
   • For (B), attempt an upper bound of the form $S_k(n)\le (1+o(1))\frac{\log n}{\log\log n}$, perhaps by showing that at most one of $n,\dots,n+k-1$ can be composed mostly of the smallest primes.
   • Extend computations to larger $n$ for fixed $k$ to estimate the growth of $\min_{m\ge n} S_k(m)$ and to search for candidate configurations that might achieve the conjectured liminf upper bound.
4. (iv) Minimal counterexample structure. A counterexample to (B) would require infinitely many $n$ for which at least two (or more) of the consecutive integers $n,\dots,n+k-1$ each have close to the maximal possible number of distinct prime factors (comparable to $(\log n)/(\log\log n)$). Such a configuration would force many small primes to divide multiple nearby integers, which is highly constrained by coprimality of consecutive integers.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.