A problem of Benkoski and Erdős. In other words, this problem asks for an upper bound for the abundancy index of
weird numbers. This could be true with $C=3$. We must have $C>2$ since $\sigma(70)=144$ but $70$ is not the distinct sum of integers from $\{1,2,5,7,10,14,35\}$.
Erdős suggested that as $C\to \infty$ only divisors at most $\epsilon n$ need to be used, where $\epsilon \to 0$.
Weisenberg has observed that if $n$ is a weird number with an abundancy index $\geq 4$ then it is divisible by an odd weird number. In particular, if there are no odd weird numbers (see
[470]) then every weird number has abundancy index $<4$. Indeed, if $l(n)$ is the abundancy index and $n=2^km$ with $m$ odd then $l(n)=l(2^k)l(m)$, and $l(2^k)<2$ so if $l(n)\geq 4$ then $l(m)>2$, and hence $m$ is weird (as a factor of a weird number).
A similar argument shows that either there are infinitely many primitive weird numbers or there is an upper bound for the abundancy index of all weird numbers.
See also
[18] and
[470].
This is part of problem B2 in Guy's collection
[Gu04] (the \$25 is reported by Guy as offered by Erdős for a solution to this question).
This has been solved in the affirmative
by Larsen - in fact, for any $\epsilon>0$ there exists $L$ such that if $n$ has only prime divisors $>L$ and $\sigma(n)>(2+\epsilon)n$ then $n$ is the distinct sum of proper divisors of $n$.
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This page was last edited 01 February 2026. View history
Additional thanks to: Desmond Weisenberg
When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:
T. F. Bloom, Erdős Problem #825, https://www.erdosproblems.com/825, accessed 2026-04-11
