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 $v(n)$

From papers

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.

Progress summary

Open

No public discussion or published progress on this conjecture appears to have been found.

No public discussion or published progress was found.

Current status (as of August 2026): The conjecture appears open, with no recorded public discussion or progress.

Sources & referencesView supporting material

Primary source

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

divisor-sums number-theory

1 week ago

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.

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Shivam Patel · 1 day ago