The integral index-ratio conjecture

From papers

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.

Progress summary

Open

The conjecture remains open: a 2025 paper records supporting computations and several special cases, but no general proof or verified counterexample has appeared.

The conjecture asserts that whenever the ratio of the sums of even- and odd-indexed divisors is a positive integer $k$ , that integer equals the second-smallest divisor $d_2$ . It was recorded as Conjecture 8 in a 2025 preprint.

Known results

For even $n$ with integral index ratio, $k=2$ .
If $n$ is odd, $\tau(n)\equiv 2\pmod 4$ , and $3\mid n$ , then integral $k$ implies $k=3$ .
Every index ratio satisfies $k<d_2+\frac{1}{d_2}$ .
If $n$ has prime index ratio $p$ and $\tau(n)\le 8$ , then $d_{2j}=p d_{2j-1}$ for all $1\le j\le\tau(n)/2$ .

September 2025 preprint

The preprint reports numerical support for the conjecture and the partial results above, but gives no general proof, verified counterexample, or subsequent resolution.

Current status (as of August 2026): The integral index-ratio conjecture remains open; its stated special cases and numerical evidence are known, but the general assertion is unsettled.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

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

divisor-functions number-theory

1 week ago

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.

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Shivam Patel · 1 day ago