gpt pro 5.2 attempt on Math-6107

FORMAL RESTATEMENT.

Let $\sigma(n)=\sum_{d\mid n} d$ be the sum-of-divisors function for $n\in\mathbb{N}$. Call $n$ $k$-perfect if $\sigma(n)=k n$ for some integer $k\ge 1$. Question: along any sequence of $k$-perfect numbers $n\to\infty$, must the corresponding multipliers satisfy \[ k = o(\log\log n), \] meaning: for every $\varepsilon>0$ there exists $N(\varepsilon)$ such that if $n\ge N(\varepsilon)$ is $k$-perfect then $k\le \varepsilon\log\log n$. Edge cases: $n=1$ is $1$-perfect; $k=2$ are the usual perfect numbers.

QUICK LITERATURE/CONTEXT CHECK.

I did not use external sources. I only use the definitions and context stated in the problem text (e.g. that examples exist up to $k=11$).

ATTACK PLAN.

Proof-track: relate $k=\sigma(n)/n$ to the Euler product over prime powers in $n$; compare to $n/\varphi(n)$ and attempt to control $n/\varphi(n)$ in terms of $\log\log n$. Disproof-track: search for (computational) $k$-perfect numbers with unusually large $k$; attempt constructions where $n$ has many small prime factors. Best path here: derive rigorous elementary bounds on $k$ (too weak to settle $o(\log\log n)$), and record sanity-check computations.

WORK.

Lemma 1 (Euler product and comparison with $n/\varphi(n)$). If $n=\prod_{p^a\parallel n} p^a$, then \[ \frac{\sigma(n)}{n}=\prod_{p^a\parallel n}\frac{p^{a+1}-1}{p^a(p-1)} =\prod_{p^a\parallel n}\frac{1-p^{-(a+1)}}{1-p^{-1}}. \] In particular, \[ \frac{6}{\pi^2}\,\frac{n}{\varphi(n)}\le \frac{\sigma(n)}{n}\le \frac{n}{\varphi(n)}. \]

Proof. The factorization of $\sigma$ over prime powers gives $\sigma(p^a)=(p^{a+1}-1)/(p-1)$, hence $\sigma(n)=\prod\sigma(p^a)$ and the displayed Euler product for $\sigma(n)/n$. For the upper bound, each factor has numerator $\le 1$, so \[ \frac{\sigma(n)}{n}\le \prod_{p\mid n}\frac{1}{1-p^{-1}}=\frac{n}{\varphi(n)}. \] For the lower bound, since $a+1\ge 2$ we have $p^{-(a+1)}\le p^{-2}$, so $1-p^{-(a+1)}\ge 1-p^{-2}$, giving \[ \frac{\sigma(n)}{n}\ge \prod_{p\mid n}\frac{1-p^{-2}}{1-p^{-1}} =\left(\prod_{p\mid n}(1-p^{-2})\right)\frac{n}{\varphi(n)}. \] Because each $(1-p^{-2})\in(0,1)$, the product over the subset $\{p\mid n\}$ is at least the product over all primes: $\prod_{p\mid n}(1-p^{-2})\ge \prod_{p}(1-p^{-2})=1/\zeta(2)=6/\pi^2$. $\square$

Lemma 2 (elementary bound $n/\varphi(n)\le e\,\omega(n)$). Let $\omega(n)$ be the number of distinct prime divisors of $n$. Then \[ \frac{n}{\varphi(n)}=\prod_{p\mid n}\frac{p}{p-1}\le e\,\omega(n)\le e\,\log_2 n. \] Consequently, for any $k$-perfect $n$ one has $k\le e\log_2 n$.

Proof. For each prime $p\mid n$, $\frac{p}{p-1}=1+\frac{1}{p-1}$. Using $\log(1+x)\le x$ for $x>-1$, \[ \log\left(\frac{n}{\varphi(n)}\right)=\sum_{p\mid n}\log\left(1+\frac{1}{p-1}\right) \le \sum_{p\mid n}\frac{1}{p-1}. \] Order the distinct prime divisors as $p_1<p_2<\cdots<p_{\omega(n)}$. The $j$th smallest prime satisfies $p_j\ge j+1$ (there are only $j$ integers in $\{2,3,\dots,j+1\}$), hence $p_j-1\ge j$, so \[ \sum_{p\mid n}\frac{1}{p-1}\le \sum_{j=1}^{\omega(n)}\frac{1}{p_j-1}\le \sum_{j=1}^{\omega(n)}\frac{1}{j} =H_{\omega(n)}\le 1+\log\omega(n). \] Exponentiating yields $n/\varphi(n)\le e\,\omega(n)$. Finally $n\ge 2^{\omega(n)}$ gives $\omega(n)\le \log_2 n$. For $k$-perfect $n$, Lemma 1 gives $k=\sigma(n)/n\le n/\varphi(n)$. $\square$

FAST REALITY CHECK (computation). I searched all $n\le 10^6$ for which $\sigma(n)/n$ is an integer. The results were:

(1,1), (6,2), (28,2), (120,3), (496,2), (672,3), (8128,2),
(30240,4), (32760,4), (523776,3).

Thus among $n\le 10^6$ the largest multiplier found was $k=4$ (attained at $n=30240$ and $n=32760$).

VERIFICATION.

Lemma 1: For $n=12$ one has $\sigma(12)/12 = 28/12 = 7/3$ and $n/\varphi(n)=12/4=3$, so indeed $7/3\le 3$. Also $(6/\pi^2)(n/\varphi(n)) \approx 0.6079\cdot 3\approx 1.8237 \le 7/3$. Lemma 2: For $n=30$, $n/\varphi(n)=(2/1)(3/2)(5/4)=15/4=3.75$ while $e\omega(n)=3e\approx 8.15$.

FINAL: UNRESOLVED.

(i) Strongest proved partial result here. For every $n$, \[ \frac{6}{\pi^2}\frac{n}{\varphi(n)}\le \frac{\sigma(n)}{n}\le \frac{n}{\varphi(n)}\le e\,\omega(n)\le e\log_2 n, \] so in particular every $k$-perfect $n$ satisfies $k\le e\log_2 n$. (ii) First gap. Prove (or disprove) that for $k$-perfect $n$ one necessarily has $n/\varphi(n)=o(\log\log n)$ (or even $O(\log\log n)$), which via Lemma 1 would force $k=o(\log\log n)$. (iii) Top 3 next moves. $1$ Establish an explicit upper bound for $n/\varphi(n)$ in terms of $\log\log n$ using only results allowed in the problem file. $2$ Investigate whether $k$-perfect numbers necessarily have bounded $\omega(n)$ or other restrictions tightening Lemma 2. $3$ Extend computations of known $k$-perfect numbers to record $k$ versus $\log\log n$. (iv) What a minimal counterexample would look like. A sequence of $k$-perfect numbers $n$ with $k$ comparable to $c\log\log n$ (or larger) would force $n/\varphi(n)$ large, hence $n$ with very many distinct small prime divisors and $\sigma(n)/n$ close to $n/\varphi(n)$ (i.e. large exponents on those primes so that the factors $1-p^{-(a+1)}$ are near $1$).