For a positive integer n, define v(n)=∑_d| n, d>1frac1dlog fractau(n)-1d. Let p be the smallest prime factor of n. Lower-bound conjecture for v(n). If n notin1,12,24,30,36,48,60,72,120,180,240,360, then v(n)≥frac1p^2. This would strengthen the proven bound v(n)≥1/(2p^2) for perfect numbers. The source presents the assertion as unresolved and does not explain the exceptional set beyond listing it.

Conjectural lower bound for the weighted divisor statistic $v(n)$

About 1 year old · traced to 2025

For a positive integer $n$ , define

v(n)=\sum_{d\mid n,\ d>1}\frac{1}{d}\log\frac{\tau(n)-1}{d}.

Let $p$ be the smallest prime factor of $n$ . Lower-bound conjecture for $v(n)$ . If

n\notin\{1,12,24,30,36,48,60,72,120,180,240,360\},

then

v(n)\geq\frac{1}{p^2}.

This would strengthen the proven bound $v(n)\geq1/(2p^2)$ for perfect numbers. The source presents the assertion as unresolved and does not explain the exceptional set beyond listing it.

References

Primary source

Joshua Zelinsky and Kyle Zhang, “Kullback-Leibler divergence and primitive non-deficient numbers”, arXiv:2501.04209 (2025).

divisor-sums number-theory

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A posted calculation claims the conjecture is false, but that disproof has not been independently verified and the published source still treats the claim as open.

Joshua Zelinsky and Kyle Zhang (2025) formulate the lower bound outside the listed exceptional set. Their source presents it as unresolved and proves only the weaker bound $v(n)\geq 1/(2p^2)$ for perfect numbers.

Known results

For perfect numbers, Zelinsky and Zhang (2025) prove $v(n)\geq 1/(2p^2)$ ; the conjectured factor $1/p^2$ is not proved.

Posted attempt

A reader claims a complete disproof: $n=6$ is a perfect-number counterexample, and every prime $n=p$ gives $v(p)<0<1/p^2$ . The calculation has not been independently verified.

Current status (as of August 2026): The conjecture is published only as an unresolved assertion with the weaker perfect-number bound known; a posted disproof claim exists but is unverified, so no result is settled.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

Counterexample even among perfect numbers.

Take $n=6$ , whose smallest prime divisor is $p=2$ . This integer is perfect and does not belong to the stated exceptional set. Since $\tau(6)=4$ and the divisors greater than $1$ are $2,3,6$ , the defining statistic equals

\begin{aligned} v(6) &=\frac12\log\frac32+\frac13\log1+\frac16\log\frac12\\ &=\frac16\log\frac{27}{16}. \end{aligned}

Using the strict elementary inequality $\log t<t-1$ for $t>1$ , we obtain

v(6) < \frac16\left(\frac{27}{16}-1\right) = \frac{11}{96} < \frac14 = \frac1{p^2}.

Thus the conjectured lower bound is false even under the additional restriction that $n$ be perfect.

Furthermore, the conjecture as stated has infinitely many counterexamples. For every prime $p$ , take $n=p$ . Then $\tau(p)=2$ , and its only divisor greater than $1$ is $p$ , so

v(p)=\frac1p\log\frac1p =-\frac{\log p}{p} < 0 < \frac1{p^2}.

No prime belongs to the exceptional set. Consequently all primes, as well as the perfect number $6$ , contradict the claimed inequality.