gpt pro 5.2 attempt on Math-988

1) FORMAL RESTATEMENT

Let $n\ge 1$ be an integer and factor \[ n(n+1)=2^k3^l m, \] where $k:=v_2(n(n+1))$, $l:=v_3(n(n+1))$, and $m$ is an integer coprime to $6$ (so $(m,6)=1$). (Here $v_p$ is the usual $p$-adic valuation.) Let $\log$ denote the natural logarithm. The question is whether \[ \limsup_{n\to\infty} \frac{2^k3^l}{n\log n}=\infty. \]

2) QUICK LITERATURE/CONTEXT CHECK

From the provided problem statement: Mahler proved a more general result implying $2^k3^l<n^{1+o(1)}$. Erdős claimed it is easy to see that $2^k3^l>n\log n$ for infinitely many $n$, and Steinerberger noted a simple proof via $n=2^{3^r}$.

3) ATTACK PLAN

• Lower bound constructions: choose special $n$ so that $v_2(n(n+1))$ and/or $v_3(n(n+1))$ are large and computable via LTE (lifting-the-exponent).
• Try to amplify the ratio: look for $n$ where one of $n$ and $n+1$ is close to a large power of $2$ and the other is close to a large power of $3$.
• Computation: scan $n$ up to a moderate bound to see how large the ratio actually gets and what $n$ achieves it.

4) WORK

Lemma 933.1 (an infinite family with ratio $>1$ via $n=2^{3^r}$). For every integer $r\ge 1$, let $n=2^{3^r}$. Then \[ v_2(n(n+1))=3^r,\qquad v_3(n(n+1))=r+1, \] and hence \[ \frac{2^k3^l}{n\log n}=\frac{3}{\log 2}>1. \] In particular, $2^k3^l>n\log n$ for infinitely many $n$.

Proof. Let $n=2^{3^r}$. Then $n$ is even and $n+1$ is odd, so \[ v_2(n(n+1))=v_2(n)=3^r. \] Next, $2\equiv -1\pmod 3$, so $2^{3^r}\equiv -1\pmod 3$, hence $3\mid (2^{3^r}+1)=n+1$. We compute the exact 3-adic valuation using the standard LTE identity for odd primes $p$: if $p\mid (a+b)$ and $p\nmid ab$ and $t$ is odd, then \[ v_p(a^t+b^t)=v_p(a+b)+v_p(t). \] Apply this with $p=3$, $a=2$, $b=1$, and $t=3^r$ (odd). Since $3\mid (2+1)$ and $3\nmid 2$, LTE gives \[ v_3(2^{3^r}+1)=v_3(2+1)+v_3(3^r)=1+r. \] Because $n$ itself is a power of $2$, $v_3(n)=0$, so \[ v_3(n(n+1))=v_3(n+1)=r+1. \] Therefore $2^k3^l=2^{3^r}3^{r+1}=n\cdot 3^{r+1}$. Also $\log n = \log(2^{3^r})=3^r\log 2$. Hence \[ \frac{2^k3^l}{n\log n}=\frac{n\cdot 3^{r+1}}{n\cdot 3^r\log 2}=\frac{3}{\log 2}. \] $\square$

Lemma 933.2 (another infinite family via $n=3^{2^r}-1$). For every integer $r\ge 1$, let $n=3^{2^r}-1$. Then \[ v_3(n(n+1))=2^r,\qquad v_2(n(n+1))=r+2, \] and \[ \lim_{r\to\infty}\frac{2^k3^l}{n\log n}=\frac{4}{\log 3}>1. \] In particular, for all sufficiently large $r$ we have $2^k3^l>n\log n$.

Proof. Let $n=3^{2^r}-1$. Then $n+1=3^{2^r}$, so $v_3(n+1)=2^r$ and $v_3(n)=0$ (since $n\equiv -1\pmod 3$). Hence \[ v_3(n(n+1))=v_3(n+1)=2^r. \] Next, $n$ is even and $n+1$ is odd, so $v_2(n(n+1))=v_2(n)=v_2(3^{2^r}-1)$. We use the $2$-adic LTE formula for odd $a$ and even exponent $t$: \[ v_2(a^t-1)=v_2(a-1)+v_2(a+1)+v_2(t)-1. \] Apply this with $a=3$ and $t=2^r$. We have $v_2(3-1)=v_2(2)=1$, $v_2(3+1)=v_2(4)=2$, and $v_2(2^r)=r$. Therefore \[ v_2(3^{2^r}-1)=1+2+r-1=r+2. \] So $2^k3^l=2^{r+2}\cdot 3^{2^r}$. Since $n=3^{2^r}-1\sim 3^{2^r}$ and $\log n\sim 2^r\log 3$ as $r\to\infty$, we obtain \[ \frac{2^k3^l}{n\log n} =\frac{2^{r+2}3^{2^r}}{(3^{2^r}-1)\log(3^{2^r}-1)} \longrightarrow \frac{2^{r+2}}{2^r\log 3}=\frac{4}{\log 3}. \] The limit is $>1$, so for all sufficiently large $r$ the ratio exceeds $1$, giving $2^k3^l>n\log n$ infinitely often. $\square$

FAST REALITY CHECK (local computation up to $2{,}000{,}000$). I scanned all $2\le n\le 2{,}000{,}000$ and computed $R(n):=2^{v_2(n(n+1))}3^{v_3(n(n+1))}/(n\log n)$. The maximum value observed in this range was \[ \max_{2\le n\le 2{,}000{,}000} R(n)\approx 4.328085\,=\frac{3}{\log 2}, \] achieved for example at $n=2$ and $n=8$. (This is evidence only; it does not bound the true limsup.)

5) VERIFICATION

• Lemma 933.1: checked the hypotheses of LTE for $v_3(2^{3^r}+1)$: $3\mid 2+1$ and $3\nmid 2\cdot 1$ and exponent $3^r$ is odd.
• Lemma 933.2: checked the $2$-adic LTE conditions: base $3$ is odd and exponent $2^r$ is even.
• Computation: the scan used exact $v_2$ via trailing-zero count and exact division for $v_3$.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result here. There are infinitely many $n$ with $2^k3^l>n\log n$ (Lemma 933.1 gives an exact constant ratio $3/\log 2$ along $n=2^{3^r}$; Lemma 933.2 gives another constant $4/\log 3$ along $n=3^{2^r}-1$). Hence \[ \limsup_{n\to\infty}\frac{2^k3^l}{n\log n}\ \ge\ \frac{3}{\log 2}. \]

(ii) First gap (crisp). Either prove that the ratio $\frac{2^k3^l}{n\log n}$ is unbounded (limsup $=\infty$) or produce an explicit upper bound $\frac{2^k3^l}{n\log n}\le C$ for all sufficiently large $n$.

(iii) Top 3 next moves.

1. Search for $n$ such that one of $n,n+1$ is divisible by an unusually large power of $2$ while the other is divisible by an unusually large power of $3$, in a way that makes $2^k3^l$ close to $n^2$; this reduces to studying near-collisions of the form $2^a u \pm 1=3^b v$ with small $u,v$.
2. Extend the brute-force scan to larger $n$ using log-scale comparisons to see whether any $R(n)$ exceed $3/\log 2$ and to guess growth.
3. Try to bound $v_2(n)$ and $v_3(n\pm 1)$ simultaneously using $p$-adic methods (e.g. lifting-exponent plus constraints from exponential Diophantine equations) to either force boundedness or construct sequences with growing $v_2$ and $v_3$.

(iv) What a minimal counterexample would look like. If the conjectured limsup $=\infty$ is false, there would exist a constant $C$ and $n_0$ such that for all $n\ge n_0$, the $\{2,3\}$-part of $n(n+1)$ satisfies $2^k3^l\le C\,n\log n$. Any extremal $n$ should then avoid having simultaneously large $2$-adic valuation on one of $n,n+1$ and large $3$-adic valuation on the other.