The board / number theory / #1134
Erdős Problem #1134
DISPROVED LEANdatabase updated 2026-06-21
The statement as posed is false. A solution has also been checked by machine, in Lean.
What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).
The smallest set of natural numbers which contains `1` and is closed under the operations `x ↦ 2x + 1`, `x ↦ 3x + 1` and `x ↦ 6x + 1`.
Let \(A\subseteq \mathbb{N}\) be the smallest set which contains \(1\) and is closed under the operations \[x\mapsto 2x+1,\quad x\mapsto 3x+1,\quad x\mapsto 6x+1.\] Does \(A\) have positive lower density? Lagarias [La16] reports that Erdős asked this in 1972, offering £10 for a solution. (Although Hilton told Lagarias that this problem may have been formulated by Klarner, and that Erdős liked it and offered a prize for its solution.) Erdős had earlier proved (as reported in [KlRa74]) that if \(A\) is the smallest set which contains \(1\) and is closed under the operations \(x\mapsto m_ix+b_i\) for some (possibly infinite) collection of \(m_i\geq 1\) and \(b_i\geq 0\) then, if \(\sigma>0\) is such that \(\sum \frac{1}{m_i^\sigma}=1\) then, for all large \(X\), \(\lvert A\cap [1,X]\rvert \ll X^{\sigma+o(1)}\). This result does not help with the given problem since \(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1\). This was answered in the negative soon afterwards by Crampin and Hilton (as reported in [Kl82]), who proved that in fact, for all large \(X\), \(\lvert A\cap [1,X]\rvert \ll X^{\tau+o(1)}\) where \(\tau\approx 0.900626\) is the unique positive root of \[6^{-\tau}+\sum_{k\geq 0}(3\cdot 2^k)^{-\tau}=1.\] Their proo
Statement as formalized in Formal Conjectures (Google DeepMind, Apache-2.0) · their label: research solved · 2 further variants / known partial results in the Lean file. The authoritative wording is on erdosproblems.com/1134.Prize£10
Topicsnumber theory
Statement in LeanFormal Conjectures
Solution in Leanyes
OEISA185661
Original pageerdosproblems.com/1134
Attempts
submit one →If you take a run at this one — with your own AI subscription or without — the guide says how, and the submission form gets you an honest read of what you found.