A397711: from my conjecture to Roberto Blanco Gómez's proof

In July 2026 I found a pattern in some counts and listed it in the OEIS as a conjecture, and in October Roberto Blanco Gómez posted a proof of it. Claude (Anthropic) was involved at every step, on both sides; see the section AI at every step, below. The details are on the earlier pages, linked at the end, each dated to when its events happened.

July: the counts and the pattern#

With Claude, I counted dependency charts: numbered tasks in which each task waits on at most two others, with no circular waiting. I call that limit of two the cap; it can be set to other numbers, and I write r for it (the July page calls it k). The same charts are the possible structures of a Bayesian network in which each variable has at most two parents (a Bayesian network is a probability model drawn as a chart of arrows; the arrows say how the model is built, each variable modeled given its parents, the variables with arrows into it, and are often read causally, as direct influence). The counts went into the OEIS as A397711↗, submitted under my name on July 5, 2026 and approved on July 12.

Dividing each count by the number of tasks gave a pattern in the remainders. For prime sizes the remainder is 1, and I could prove it. For the sizes that argument does not reach, odd sizes that are not prime and all even sizes from 4, I saw a rule: remainder 1 for the odd ones, and one more than half the size for even sizes from 4 on (443 charts on four tasks leave 3). I checked it for every size up to 100 tasks, could not prove it, and wrote it into the entry as a conjecture. I also counted charts with a cap of three, and there my counts showed the odd-size rule failing: with nine tasks the remainder is 7, not 1.

October 7: a proof is posted#

Roberto Blanco Gómez, an independent researcher in Spain, found the conjecture while going through recent OEIS entries with open conjectures, and posted a proof for every size in a short preprint on Zenodo↗, which has not yet been refereed. The preprint also gives a proof of a general version for any cap r: n divides lcm(1, 2, ..., r) × (the count − 1), where lcm(1, 2, ..., r), the least common multiple, is the smallest number that 1, 2, ..., r all divide. With a cap of two the multiplier is 2, and an odd n shares no factor with 2, so n divides the count minus 1 itself: the remainder is 1, which is the odd half of my rule (the even half needs a further step, on the proof page). With a cap of three the multiplier is 6, and for nine tasks the general version forces only that the count leaves remainder 1 when divided by 3; the remainder 7 I saw on division by 9 does leave 1 when divided by 3, so the general version allows it. His preprint's authorship note says that he chose the target and designed the method of checking, and that the computations, the search for the argument and the drafting were done with Claude working under his direction. He added a comment to the OEIS entry pointing to the proof, and an OEIS editor approved the comment on October 12.

Later in October: the re-check of the proof#

On October 24, 2026, Claude agents, working under my direction as part of a research project, re-checked his result. New programs, written from the paper's statements without his code, confirmed the pattern up to 400 tasks. The counting formula his proof starts from, a recurrence published by Eggeling and co-authors in 2019, reproduces all 101 values of my table (0 to 100 tasks), which I had computed by a different method, a formula that sorts the tasks into levels; that checks one method against the other, not the remainder argument. The argument itself was read step by step by a fresh Claude session, which found no gap. Apart from its author, no person, me included, has yet read the core of the argument line by line.

AI at every step#

My July counts were computed with Claude. He found the conjecture in a sweep of the 314 OEIS entries numbered A395000 or higher that contain conjectures, carried out with Claude, and the argument was found and drafted with Claude under his direction, as his authorship note says (above). A separate Claude session, working under my direction, checked the proof. None of this is independent in the way a human reader would be, since checks run by the same kind of system can share its mistakes. The parts of my check that do not depend on an AI reading the paper correctly are my July table, computed from the definition by a different method, and direct counts of every chart for small sizes.

Where it stands now#

• The live A397711 entry, as I last read it on October 25, 2026, carries both my conjecture line and his comment that it holds for every size, with a link to the preprint.
• The proof is a preprint and has not been refereed.
• The re-check by Claude agents found nothing wrong; it is an AI check, not a referee's.

@misc{orden2026a397711fromconjecturetoproof,
  author       = {Tyler Satchel Orden},
  title        = {A397711: from my conjecture to Roberto Blanco Gómez's proof},
  year         = {2026},
  howpublished = {Cyborg Mathematics},
  url    = {https://cyborgmathematics.com/writing/a397711-from-conjecture-to-proof}
}