Bailey-Crandall's mutual-nondivisibility question for Stoneham numbers: $\alpha_{6,4}$ is not $6$ -normal, and $\alpha_{b,c}$ is $b$ -normal iff some prime of $c$ misses $b$

For integers $b, c \ge 2$ let $\alpha_{b,c} = \sum_{k \ge 1} \dfrac{1}{c^k\, b^{c^k}}$ . Bailey and Crandall proved in 2002 that $\alpha_{b,c}$ is normal in base $b$ whenever $b$ and $c$ are coprime, and asked whether coprimality can be weakened to the condition that neither of $b$ , $c$ divides the other: is $\alpha_{b,c}$ then still $b$ -normal?

Result: Disproved(see note)
Status: Resolved
AI contribution: AI-discovered
Method: Argument
Field: Normal numbers and digit expansions
Posed by: David H. Bailey and Richard Crandall, Random Generators and Normal Numbers, Experimental Mathematics 11 (2002), Section 7 "Conclusions and open problems", p. 544
Year posed: 2002
Years open: 24y
Solved: 2026-09-12
Model: GPT-6 Astra
Vendor: OpenAI
Collaborators: —
Verification: Lean-verified
Publication: Announced
Significance: 18 / 100
Disclosed cost: —
Wikipedia: No dedicated article

What was actually shown

Mutual nondivisibility is not enough: $b = 6$ , $c = 4$ is a counterexample, since every prime dividing 4 divides 6, and then the zero digit of $\alpha_{6,4}$ in base 6 has limiting frequency one. The exact criterion is that $\alpha_{b,c}$ is $b$ -normal if and only if some prime dividing $c$ does not divide $b$ . Sufficiency applies Bailey and Crandall's own exponential-sum estimate to the part of $c$ coprime to $b$ ; the repository's prior-art note records that Kano's 1993 theorem on Korobov-type constructions already reaches some non-coprime cases, $\alpha_{4,6}$ among them, so the classification rather than the method is what is new. Reversing the parameters, $\alpha_{4,6}$ is 4-normal, is also formalised.

What the AI did

Per the submitter, the model found the question in Bailey and Crandall's list of open problems, produced the counterexample and the exact criterion, and wrote the Lean 4 development; the paper carries no byline and the repository names no human author. The repository's own prior-art note, dated 12 September 2026, is unusually careful: it records that Kano's 1993 theorem already reaches some non-coprime cases, that the sufficiency direction is derived from Bailey and Crandall's existing exponential-sum estimate, and that the search is not a certification of novelty.

Verification

The posed question was read at source. Bailey and Crandall, Section 7, p. 544: "Instead of coprimality of b, c, just specify that neither integer divides the other. Can a result on b-normality then be effected?" The submission quotes it faithfully.

The Lean development was read at commit 5919ad4 on 15 September 2026. `normal_iff` states the criterion for `stonehamConstant b c`, defined as $\sum_{k \ge 0} 1/(c^{k+1} b^{c^{k+1}})$ , the standard series; `counterexample` states $6 \nmid 4$ , $4 \nmid 6$ and non-normality of $\alpha_{6,4}$ in base 6; `NormalInBase` counts every finite word at every prefix length through a radix orbit and is the same definition audited for this author's two-sum entry. The audit file guards the transitive axioms of every headline theorem to exactly propext, Classical.choice and Quot.sound; no sorry, no native_decide. CI rebuilt the project on GitHub twice on 12 September, both green. Not read here: the analytic sufficiency proof line by line. The failing direction is elementary: when every prime of $c$ divides $b$ , $1/c^k$ terminates in base $b$ within $O(k)$ digits and the terms sit at positions $c^k$ , so the zero digit has limiting frequency one.

Sources

Submitted by nufrogcaca on 13 Sep 2026

Bailey-Crandall's mutual-nondivisibility question for Stoneham numbers: α6,4\alpha_{6,4}α6,4​ is not 666-normal, and αb,c\alpha_{b,c}αb,c​ is bbb-normal iff some prime of ccc misses bbb

What was actually shown

What the AI did

Verification

Sources

Discussion

Bailey-Crandall's mutual-nondivisibility question for Stoneham numbers: $\alpha_{6,4}$ is not $6$ -normal, and $\alpha_{b,c}$ is $b$ -normal iff some prime of $c$ misses $b$