The integral index-ratio conjecture

About 1 year old · traced to 2025

Let $n$ ) be a positive integer with divisors $d_1<d_2<\cdots<d_{\tau(n)}$ , and let $k$ be its index ratio, meaning that the sum of the even-indexed divisors equals $k$ times the sum of the odd-indexed divisors. In particular, $d_2$ is the second-smallest divisor of $n$ . Integral index-ratio conjecture. If $n$ is a $k$ -index ratio number and $k\in\mathbb{N}$ , then

k=d_2.

Numerical calculations support this claim; the preceding corollaries establish it in some special cases, but the general assertion remains open.

References

Primary source

Brahim Mittou, “A New Classification of Positive Integers Via New Divisor Functions”, arXiv:2509.08844 (2025).

divisor-functions number-theory

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A proposed complete counterexample would disprove the conjecture, but it has not been independently verified, while the published paper gives only partial results.

Brahim Mittou stated the conjecture in a 2025 preprint: whenever the ratio of the sums of even- and odd-indexed divisors is a positive integer $k$ , one must have $k=d_2$ . The paper reports supporting computations but no general proof.

Known results

Even $n$ : integral ratio implies $k=2$ (Mittou, 2025).
Odd $n$ with $\tau(n)\equiv 2\pmod 4$ and $3\mid n$ : integral ratio implies $k=3$ (Mittou, 2025).
Every ratio satisfies $k<d_2+\frac{1}{d_2}$ (Mittou, 2025).
If $k=p$ is prime and $\tau(n)\le 8$ , then $d_{2j}=p d_{2j-1}$ (Mittou, 2025).

Posted attempt

An unverified complete disproof claims $n=2431=11\cdot13\cdot17$ has ratio $k=7$ but $d_2=11$ , and extends this to infinitely many $2431Q^a$ with prime $Q>2431$ . The calculation has not been independently verified.

Current status (as of August 2026): The published conjecture remains unproved, but a complete counterexample has been proposed in reader material and is presently unverified.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

Complete counterexample and infinitely many further counterexamples.

Let $d_1<\cdots<d_{\tau(n)}$ be the positive divisors of $n$ , and put

O(n)=\sum_{j\text{ odd}}d_j,\qquad E(n)=\sum_{j\text{ even}}d_j,\qquad R(n)=E(n)/O(n).

Take $n=2431=11\cdot13\cdot17$ . Its complete ordered divisor list is

(1,11,13,17,143,187,221,2431).

Hence

O(2431)=1+13+143+221=378,

E(2431)=11+17+187+2431=2646=7\cdot378.

Thus $R(2431)=7\in\mathbb N$ , but $d_2=11\neq7$ . This disproves the conjecture.

In fact, for every prime $Q>2431$ and integer $a\ge1$ , the increasing divisors of $N=2431Q^a$ occur in consecutive blocks $Q^j\operatorname{Div}(2431)$ , $0\le j\le a$ . Each block has eight elements, so index parity is preserved and

O(N)=378\sum_{j=0}^aQ^j,\qquad E(N)=2646\sum_{j=0}^aQ^j.

Therefore $R(N)=7$ while $d_2(N)=11$ , giving infinitely many counterexamples.