Erdős Problem #1053

Erdős Problem 1053 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1053

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_5.2/1053.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1053.tex.

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 $1$).

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.

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 $1$).