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Math-975-A1 · revision r1 · on Math-975

Reused writeup on Math-975 (gpt pro 5.2)

No claim recorded · LLM-generated · gpt pro 5.2 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem 893 --- collection record
  2. PROVENANCE AND STATUS
  3. REUSED SOURCE 1
  4. COMPLETION ESTIMATE

Erdős Problem 893 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/893

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_5.2/893.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/893.tex.

1) FORMAL RESTATEMENT

Let $\tau(n)$ denote the number of positive divisors of the positive integer $n$. Define \[ f(n):=\sum_{1\le k\le n}\tau(2^k-1) \qquad (n\ge 1). \] Question: does the ratio $f(2n)/f(n)$ converge as $n\to\infty$?

Ambiguity. “Tend to a limit” could mean a finite real limit, or could allow the extended real value $+\infty$. I will treat both:

  • $Q1$ Does $\lim_{n\to\infty} f(2n)/f(n)$ exist as a finite real number?
  • $Q2$ Does $\lim_{n\to\infty} f(2n)/f(n)=+\infty$?

2) QUICK LITERATURE/CONTEXT CHECK

The problem statement records that Kovač and Luca $2025$ proved $\limsup_{n\to\infty} f(2n)/f(n)=\infty$, hence $Q1$ has a negative answer. It also records numerical/theoretical evidence suggesting $Q2$ may hold.

3) ATTACK PLAN

  • Prove elementary structural lower bounds on $\tau(2^k-1)$ using divisibility properties of Mersenne numbers.
  • Do a computational sanity check for $n\le 30$ (or similar) to see whether ratios are stable or exhibit spikes.
  • Identify a crisp next-step lemma that would force $f(2n)/f(n)\to\infty$.

4) WORK

PHASE 1 — FAST REALITY CHECK (computation).

I computed $\tau(2^k-1)$ for $1\le k\le 60$, then $f(n)$ and the ratios $f(2n)/f(n)$ for $1\le n\le 30$.

Selected values of $\tau(2^k-1)$ for $k\le 20$: \[ \begin{array}{c|cccccccccc} k &1&2&3&4&5&6&7&8&9&10\\\hline \tau(2^k-1) &1&2&2&4&2&6&2&8&4&8 \end{array} \qquad \begin{array}{c|cccccccccc} k &11&12&13&14&15&16&17&18&19&20\\\hline \tau(2^k-1) &4&24&2&8&8&16&2&32&2&48 \end{array} \]

Ratios $f(2n)/f(n)$ for $1\le n\le 30$: \[ \begin{array}{c|cccccccccc} n&1&2&3&4&5&6&7&8&9&10\\\hline f(2n)/f(n)&3.00&3.00&3.40&3.00&3.545&3.941&4.053&3.741&4.355&4.744 \end{array} \] \[ \begin{array}{c|cccccccccc} n&11&12&13&14&15&16&17&18&19&20\\\hline f(2n)/f(n)&4.953&4.672&4.768&5.208&5.941&5.337&5.466&8.081&8.051&7.086 \end{array} \] \[ \begin{array}{c|cccccccccc} n&21&22&23&24&25&26&27&28&29&30\\\hline f(2n)/f(n)&7.406&7.488&7.719&7.831&8.047&8.337&8.638&8.057&8.095&15.689 \end{array} \] The maximum among $n\le 30$ in this computation is $f(60)/f(30)\approx 15.689$.

Problem-specific lemmas (elementary lower bounds).

Lemma 893.1 (Divisibility along divisors). If $d\mid k$ then $2^d-1\mid 2^k-1$.

Proof. Write $k=dr$. Then \[ 2^k-1 = 2^{dr}-1 = (2^d)^r-1 = (2^d-1)\left((2^d)^{r-1}+(2^d)^{r-2}+\cdots+1\right), \] so $2^d-1$ divides $2^k-1$. $\square$

Lemma 893.2 (A uniform divisor-count lower bound). For every integer $k\ge 1$, \[ \tau(2^k-1)\ \ge\ \tau(k). \] Consequently, \[ f(n)=\sum_{k\le n}\tau(2^k-1)\ \ge\ \sum_{k\le n}\tau(k). \]

Proof. For each divisor $d\mid k$, Lemma 893.1 gives a divisor $2^d-1$ of $2^k-1$. If $d_1\ne d_2$ then $2^{d_1}-1\ne 2^{d_2}-1$, so this produces at least $\tau(k)$ distinct divisors of $2^k-1$. Therefore $\tau(2^k-1)\ge \tau(k)$. Summing over $1\le k\le n$ gives the displayed inequality for $f(n)$. $\square$

Lemma 893.3 (Crude growth of $f(n)$). For $n\ge 2$, \[ n\log n - n\ \le\ \sum_{k\le n}\tau(k)\ \le\ n(1+\log n). \] In particular, $f(n)\ge n\log n - n$.

Proof. The identity \[ \sum_{k\le n}\tau(k) = \sum_{k\le n}\sum_{d\mid k} 1 = \sum_{d\le n}\#\{k\le n: d\mid k\} = \sum_{d\le n} \lfloor\frac{n}{d}\rfloor \] is standard by swapping the order of summation. For the upper bound, $\lfloor n/d\rfloor\le n/d$ gives \[ \sum_{d\le n}\lfloor\frac{n}{d}\rfloor \le n\sum_{d\le n}\frac{1}{d} \le n(1+\log n), \] using $\sum_{d\le n}1/d \le 1+\int_1^n \frac{dt}{t}=1+\log n$. For the lower bound, $\lfloor n/d\rfloor\ge n/d -1$ gives \[ \sum_{d\le n}\lfloor\frac{n}{d}\rfloor \ge n\sum_{d\le n}\frac{1}{d}-n \ge n\int_1^n \frac{dt}{t} - n = n\log n - n. \] Finally, combine with Lemma 893.2 to get $f(n)\ge \sum_{k\le n}\tau(k)$. $\square$

5) VERIFICATION

  • Lemma 893.1 is the standard factorization $x^r-1=(x-1)(x^{r-1}+\cdots+1)$.
  • Lemma 893.2: injectivity of $d\mapsto 2^d-1$ is immediate since $2^d-1$ is strictly increasing in $d$.
  • Lemma 893.3: the bounds on harmonic sums are correct and fully explicit.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result. An elementary lower bound is $f(n)\ge n\log n - n$ (Lemma 893.3). Computationally, the ratios $f(2n)/f(n)$ for $n\le 30$ range from $3$ up to approximately $15.689$.

(ii) First gap (crisp). Prove or disprove that $f(2n)/f(n)\to +\infty$ as $n\to\infty$. Equivalently, show that the contribution of the block $\sum_{n<k\le 2n}\tau(2^k-1)$ dominates $f(n)$.

(iii) Top 3 next moves.

  1. Prove a quantitative lower bound of the form $\sum_{n<k\le 2n}\tau(2^k-1) \ge c(n)\, f(n)$ with $c(n)\to\infty$.
  2. Identify a subsequence $n_j$ for which many $k\in(n_j,2n_j]$ have unusually large $\tau(2^k-1)$ (e.g. by forcing $2^k-1$ to have many small prime factors).
  3. Computational: extend the exact factorization/ divisor-count computation to larger $k$ (beyond 60) using ECM or specialized Mersenne factorization routines, and empirically locate the growth of $\max_{n\le X} f(2n)/f(n)$.

(iv) Minimal counterexample structure (if $f(2n)/f(n)\not\to\infty$). A minimal counterexample to divergence would require that for infinitely many scales $n$, the block $(n,2n]$ contributes only a bounded multiple of the earlier sum. This would mean that “typical” values of $\tau(2^k-1)$ in $(n,2n]$ are not significantly larger than those for $k\le n$, and that large spikes (like the one visible at $k=60$ in the sanity check) occur too sparsely to force divergence.

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.