Almost Perfect Numbers Beyond Powers of 2

Almost Perfect Numbers Other Than Powers of Two

Definitions and mathematical statement

Shared notation.

Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

An almost perfect number is a positive integer $n$ with $\sigma(n)=2n-1$. The powers $2^k$, including $2^0=1$, satisfy this equality. The target is the existence or impossibility of a solution outside $\{2^k:k\geq0\}$. [S1] [S2]

Statement recorded in the supplied source.

Do any almost perfect numbers exist that are not powers of 2? [S1]

Residual task reported by the archive.

The embedded review $2026-08-17$ records: Construct an almost perfect number outside the powers of two or exclude all remaining structural cases. [S1]

Short English statement

Apart from powers of two, can any positive integer have proper divisors whose sum falls short of that integer by exactly one?

Research attempt

Scope and approach.

The equation is $\sigma(n)=2n-1$, so the powers of two, including one, are already solutions. The question is whether there are any others. The Antalan--Dris primary preprint [S2], version 4, was inspected at abstract level on 1 October 2026; its stated result concerns structural restrictions on a hypothetical even example, not a construction. This attempt derives a cofactor identity, eliminates every number with just one distinct odd prime factor, and isolates the remaining problem.

The parity reduction and all known immediate examples.

The right side is odd. For an odd prime $p$, the divisor sum $1+p+\cdots+p^e$ is odd exactly when $e$ is even. Therefore \[ n=2^a m^2 \] with $a\ge0$ and $m$ odd. Conversely this shape merely ensures odd divisor sum; it does not imply the required equality. If $m=1$, then $\sigma(2^a)=2^{a+1}-1=2n-1$, proving all the stated power-of-two examples, including $a=0$. We henceforth take $m>1$.

The exact odd-cofactor identity.

Let $A=2^{a+1}$ and $D(m^2)=\sigma(m^2)-m^2$, the sum of the proper positive divisors of $m^2$. Multiplicativity gives \[ (A-1)\sigma(m^2)=Am^2-1 \quad\Longleftrightarrow\quad (A-1)D(m^2)=m^2-1. \] Thus $A-1$ divides $m^2-1$, and $\gcd(m,A-1)=1$. For a prescribed odd $m>1$, the only possible power-of-two part is determined by \[ A=1+\frac{m^2-1}{D(m^2)}. \] The quotient must be an integer and the resulting $A$ must be a power of two at least two. Conversely, if these tests pass, the original equation follows, so this is a complete criterion for each cofactor. There is at most one exponent $a$ for a fixed odd $m>1$.

The sign matters: the analogous quasiperfect equation would have $m^2+1$ on the right, producing a sum-of-two-squares obstruction. Here $m^2-1=(m-1)(m+1)$ has no such obstruction. Importing that plus-sign argument into this problem would incorrectly eliminate the known powers of two and the unresolved cases.

A complete infinite exclusion: one odd prime factor.

Suppose $m=p^e$ with $p$ odd prime and $e\ge1$. The proper-divisor sum is \[ D(p^{2e})=1+p+\cdots+p^{2e-1} =\frac{p^{2e}-1}{p-1}. \] Substitution into the cofactor identity and cancellation of the positive $p^{2e}-1$ gives $A-1=p-1$, or $p=A=2^{a+1}$. No odd prime is a power of two. Thus no solution outside the powers of two can have only one distinct odd prime factor. This includes all odd prime powers and all integers of the form $2^a p^{2e}$, with no bound on the exponents. Every nontrivial candidate needs at least two distinct odd primes.

A useful size restriction with proof.

Since $m>1$, both $1$ and $m$ are distinct proper divisors of $m^2$. Hence $D(m^2)\ge m+1$. The cofactor identity then implies \[ A-1\le\frac{m^2-1}{m+1}=m-1, \qquad A\le m. \] But $A$ is even and $m$ is odd, so equality is impossible and \[ 2^{a+1}<m. \] This recovers directly the kind of structural inequality stated in [S2], without using its proof. It bounds the even part in terms of the odd square root, not the total size of the candidate. In particular it does not imply that only finitely many $m$ can occur.

There is a residue restriction on factors of $m$ as well. If $3\mid m$, then $3\nmid A-1$ by coprimality, so $a+1$ cannot be even, because $2^{a+1}\equiv1\pmod3$ for even $a+1$. Thus $a$ must be even on that branch. More generally the order of two modulo any odd prime dividing $m$ cannot divide $a+1$. These restrictions discard some exponents but do not exhaust all exponents simultaneously.

Abundancy and the limiting-product issue.

Dividing the defining equation by $n$ gives \[ \frac{\sigma(n)}n=2-\frac1n<2, \qquad \frac{\sigma(m^2)}{m^2} =\frac{A-m^{-2}}{A-1}. \] For fixed $A$, the latter is just below $A/(A-1)$, with a prescribed error $1/((A-1)m^2)$. Having a deficient integer, meaning divisor sum less than twice itself, is much weaker than attaining deficiency exactly one. A sequence of abundancy values approaching two from below does not prove that any one of them has the required exact error.

For a fixed prime support, upper products such as $\prod p/(p-1)$ help bound possible exponents, but the support is not fixed in the original problem. An argument that exhausts a finite list of supports needs a proof that every candidate lies on that list. Neither the square-part reduction nor $A<m$ provides such a list.

Executed finite classification.

The script evaluates $\sigma(n)$ exactly for $1\le n\le200000$. The solutions to $\sigma(n)=2n-1$ are exactly \[ 1,2,4,\ldots,131072, \] namely the powers $2^a$ with $0\le a\le17$. Each integer in the range is tested, rather than sampling only odd squares or a restricted exponent pattern. The script and result are checks.py and results.json in _Local/Erdos_problems/UnsolvedMath/attempt_audit_2026_10_01/number_theory/. The finite classification is independent of the proof of the one-odd-prime exclusion, which covers an infinite family.

Remaining gap and outcome.

For a nontrivial candidate the odd part must be a square with at least two distinct prime factors, its cofactor quotient must give exactly a power of two, and $2^{a+1}<m$. No argument here rules out all such cofactors or produces one that passes. Outcome: UNRESOLVED. The work gives exact reductions and an infinite excluded subclass, without asserting that the finite list of powers of two is globally complete.

Sources

• S1 ULAM.AI, UnsolvedMath, supplied problems.json, record 1251 (NT-044), statement and background. The embedded literature-review date is 2026-08-17; that is the archive’s review date, not a fresh certification. Dataset landing page: https://huggingface.co/datasets/ulamai/UnsolvedMath.

• S2 John Rafael M. Antalan and Jose Arnaldo B. Dris, Some New Results On Even Almost Perfect Numbers Which Are Not Powers Of Two, arXiv:1602.04248. https://arxiv.org/abs/1602.04248. Bibliographic information from the supplied record.

• S3 Encyclopedia of Mathematics, Almost perfect number. https://encyclopediaofmath.org/wiki/Almost_perfect_number. Bibliographic information from the supplied record.

• E1 John Rafael M. Antalan and Jose Arnaldo B. Dris, Some New Results On Even Almost Perfect Numbers Which Are Not Powers Of Two, arXiv:1602.04248v4 $2017$. https://arxiv.org/abs/1602.04248v4. Abstract inspected 1 October 2026. The inequality $2^{a+1}<m$ used above is proved directly in this notebook as well as reported in that source.