Plectis

Cross-problem paper

Reading Eight Erdős Problems Together

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

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.

In this paper

The positive direction and the arithmetic cost

Suppose c>1 and the support gaps are eventually at most d=⌈c⌉−1. Discard a finite prefix and write

rj=nj!nj−1!,Fj=njc,wj=1nj!.

Then rj≥nj→∞, while rj≤njd and therefore Fj/rj≥njc−d→∞. Define

hj=infk≥jmin{rk2,Fk2rk}.

This lower envelope is nondecreasing and tends to infinity. Begin sufficiently late that h1≥1, set M0=1, and let Mj be the largest factorial at most hj. These moduli are nested and eventually divisible by every fixed integer. Moreover,

2Mj≤rj≤rjMj−1,2rjMj−1≤Fj.

The second inequality uses Mj−1≤hj, also valid for j=1. Finally, Mj/nj!≤rj/(2nj!)≤1/(2nj!)→0. Lemma 3.4 fills [1/n0!,2/n0!] using the positions nj, j≥1. Put zero digits elsewhere. Once q∣MJ, all later digits and cumulative sums at the original integer indices have the divisibilities in (4). This proves the positive direction, including common cutoffs.

For comparison, remove condition (2) and call the resulting attainable set E0(S,c).

Proposition 3.5. For an infinite support and c≥1, E0(S,c) contains an interval exactly when its successive gaps are eventually at most ⌊c⌋. Otherwise it is null and meagre. For 0<c<1 it is always null and meagre.

Proof. For necessity repeat Section 3.3 with r=⌊c⌋+1. The bound in (5) now tends to zero. With lattice spacing 1/N!, it is eventually bounded by 1/(2N!), so the same compact-set argument applies with q=1, B=1/2. For sufficiency, rj≤njc along the support. The ordinary mixed-radix expansion using digits 0≤bj<rj fills [0,1/n0!]. To see this directly, the maximum tail after nj is 1/nj!, since (rk−1)/nk!=1/nk−1!−1/nk!. The adjacent digit intervals therefore meet, and their lengths tend to zero. ◻

Thus the difference between the two gap thresholds occurs exactly at integer c. At c=m≥2, imposing eventual congruences raises the least support density from 1/m to 1/(m−1). At c=1 it destroys interval filling entirely. With quadratic allowances, deleting only the positions 2k destroys interval filling, although the permitted support still has density one. This is the simplest example of why support density loses decisive information. These statements concern intervals of values on a common permitted support; they are not lower bounds for representing one specially chosen value.

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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