Reading Eight Erdős Problems Together

Will Cook

Cross-problem paper

Reading Eight Erdős Problems Together

Eight Erdős problems read together By Will Cook 22 September 2026 Authorship, AI use and citation

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Précis. One account of the mathematics developed across the programmes: an exact tail-capacity criterion under eventual congruences, its sharp factorial support-gap threshold, rational and irrational Lambert subsums across bases, and the limits of finite greedy tests and other irrationality methods. Full arguments, exact counterexamples, unsuccessful routes and attribution are retained together. The principal proofs are ordinary mathematics; cited Lean ingredients have their own stated scope. Historical novelty and independent expert review are not established.

This paper owns the synthesis exposition and ordinary proofs across the covered problems: capacity and congruence constructions, Lambert subsums, rational-point counts, method limits and their research record.

It is not authority for a solution to any original Erdős target, historical novelty, independent expert review, or a full Lean proof of the analytic capacity criterion or Lambert-chain theorem.

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Why a large gap prevents interval filling

The decisive obstruction is local. At a position $N$ , the whole preceding factorial sum lies on a lattice of spacing $1 / N!$ . Eventual divisibility widens that spacing to $q / N!$ , up to a fixed translation. After a sufficiently long gap, all remaining permitted digits reach only a bounded multiple of $1 / N!$ . Choosing one fixed $q$ larger than that bound leaves holes at arbitrarily small scales.

Here are the details, including the target-dependent cutoffs in (1)–(2). Put $r = ⌈ c ⌉$ . Suppose infinitely many successive $S$ -gaps have length at least $r$ , and let $N$ run through the positions immediately before these gaps. For large $k$ ,

\frac{(k + 1)^{c} / (k + 1)!}{k^{c} / k!} = \frac{(1 + 1 / k)^{c}}{k + 1} \leq \frac{1}{2} .

Consequently, for sufficiently large such $N$ ,

\begin{aligned} (1) & N! \sum_{\begin{matrix} k > N \\ k \in S \end{matrix}} \frac{k^{c}}{k!} & \leq 2 N! \frac{(N + r)^{c}}{(N + r)!} \\ (5) & \leq 2^{c + 1} N^{c - r} \leq B, B = 2^{c + 1} . \end{aligned}

In the second inequality we used $N \geq r$ and $(N + 1) \dots (N + r) \geq N^{r}$ .

The bounded capacity along this subsequence invokes the necessity argument of Theorem 3.1, with allowances $F_{n} = ⌊ n^{c} ⌋$ on $S$ and zero elsewhere. It gives nullity and meagreness even with target-dependent congruence and allowance cutoffs.

If $c \leq 1$ , every support gap is at least $r = 1$ , so this proves the first part of Theorem 3.2. If $c > 1$ and (3) fails, it proves the negative part. Notice that infrequent large gaps are enough; average digit counts do not detect this obstruction.

About this paper

Authorship and AI use. Will Cook built and directed the research infrastructure and maintains the public release. He reviewed claims when he could. AI agents did most of the research and drafting. Cook did not independently verify every claim.

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