gpt pro 5.2 attempt on Math-975

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.