Let p be a prime number, and let n be a p-index ratio number, with divisors d_1<d_2<…<d_tau(n). Prime index-ratio divisor-pair conjecture. Then d_2j=p d_2j-1quadfor all 1≤ j≤ tau(n)/2. Theorem 9 proves this when tau(n)≤ 8, and the source notes that infinitely many examples with larger divisor count satisfy the equalities. The general statement is supported by numerical calculations and remains open.

The prime index-ratio divisor-pair conjecture

About 1 year old · traced to 2025

Let $p$ be a prime number, and let $n$ be a $p$ -index ratio number, with divisors $d_1<d_2<\cdots<d_{\tau(n)}$ . Prime index-ratio divisor-pair conjecture. Then

d_{2j}=p d_{2j-1}\quad\text{for all }1\leq j\leq \tau(n)/2.

Theorem 9 proves this when $\tau(n)\leq 8$ , and the source notes that infinitely many examples with larger divisor count satisfy the equalities. The general statement is supported by numerical calculations and 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

The conjecture has a small-case proof, but an unverified reader-provided claim gives counterexamples that would disprove it, including one in the supposedly settled case.

The conjecture says that, for these numbers and a prime ratio, the increasing divisors must form consecutive pairs with common multiplier $p$ . Mittou’s paper introduces this problem and records the small-divisor theorem, larger examples, and numerical support for the general claim.

Known results

Theorem 9 proves the pairing for $\tau(n)\leq 8$ .
Infinitely many examples with $\tau(n)>8$ satisfy all the asserted pairings.
Numerical calculations support the general conjecture.

Posted attempt

An unverified complete-counterexample claim gives $n=2431=11\cdot13\cdot17$ , with divisor sums ratio $R(n)=7$ but $d_2=11\ne7d_1$ , contradicting the claimed $\tau(n)=8$ theorem. It also gives $n=13113=3^2\cdot31\cdot47$ , where $R(n)=d_2=3$ but $d_4=31\ne3d_3$ ; neither calculation has independent verification here.

Current status (as of August 2026): The general conjecture has no verified proof or disproof; the posted examples, if correct, would overturn even the stated $\tau(n)\leq8$ result.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

Complete counterexamples, including the natural strengthened hypothesis.

For increasing positive divisors $d_1<\cdots<d_{\tau(n)}$ , write

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

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

\operatorname{Div}(n)=(1,11,13,17,143,187,221,2431).

Consequently

O(n)=378,\qquad E(n)=2646=7\cdot378.

Thus $R(n)=7$ is prime, but the first required divisor-pair identity fails:

d_2=11\neq7=7d_1.

The conjecture is therefore false. Since $\tau(n)=8$ , this also contradicts Theorem 9 in the cited paper under its stated hypothesis $\tau(n)\le8$ .

Even imposing the additional condition $R(n)=d_2$ does not repair the claim. Take

n=13113=3^2\cdot31\cdot47,

whose complete divisor list is

(1,3,9,31,47,93,141,279,423,1457,4371,13113).

Here

O(n)=4992,\qquad E(n)=14976=3\cdot4992,

so $R(n)=3=d_2$ , yet

d_4=31\neq27=3d_3.

Hence even the strengthened divisor-pair statement is false.