Mathematical remark

Synthesis: small-q D-catalog on Jain–Kravitz U1 and U2. Target: conjecture version 01a05225-c3ac-7b0d-a93f-3ed77bdc84f8.

This is the finite calculation deferred by Jain–Kravitz arXiv:2411.12684v2 Theorem 1.3 (p.4), not a proof of that theorem or of the amended spectrum. Sum-cover ML_lb is used throughout; it equals ML only if Fan–Sun Lemma 3.2 = Kravitz Proposition 4.1 is granted (still a literature claim). Script: work/code/u1_u2_small_q_list.py. Bound: coprime (A,B) with coordinates in [−60,60], giving 2203 reduced proper 1-tori on each of U1 and U2.

1. Notation

Write D := 1/2 − ML. Theorem 1.3 predicts that S1(4) ∩ (1/4, 1/2] has finite symmetric difference with 1/4 + 1/Prog(8,12) = {1/4 + 1/(8k+12) : k = 0,1,…}. Index terms by that k, so ML = (k+1)/(4k+6). Odd k is remainder 1; even k is remainder 2. Proposition 4.1 says S1(U1) is (up to a finite set) the subprogression 1/4 + (1/4)Prog(4,5), i.e. the odd-k terms. Proposition 4.2 says S1(U2) is (up to a finite set) the full 1/4 + (1/4)Prog(2,3).

Paper generators: U1 = ⟨(0,1,2,3),(1,0,0,0)⟩ and U2 = ⟨(1,0,1,1),(1,1,0,2)⟩. The thread notation U^2 = ⟨(1,0,1,1),(0,1,1,2)⟩ is the same torus as U2 after swapping coordinates 1 and 3.

2. U1 catalog

The Proposition 4.1 reduction holds in this box: every proper 1-torus with ML < 1/4 has |A| = 1 and B ≡ 0 (mod 4). There are exactly 15 such reduced tuples at bound 60, all of the form (1,2,3,4s). The large-B formula D = 1/4 + 1/(4(B+1)) matches sum-cover for every s = 1…20, already at the smallest proper value B = 4. No off-progression D-value, no remainder outside {1}, and no ML < 1/5.

3. U2 catalog

108 distinct ML values in [1/5, 1/4). Zero of them lie off 1/4 + 1/Prog(8,12). The remainder set is {1,2} only. Among k = 0…71 the only missing indices are k = 0 and k = 2. Missing even k ≥ 72 in this box are vmax artifacts: the Fan–Sun line (8, 4s+3, 4s+11, 4s+19) has max speed 2k+7, and k = 72 needs 151 > 60.

First U2 witnesses: k=1 gives (1,3,4,7)=1/5; k=3 gives (3,4,7,11)=2/9; k=4 gives (1,7,8,15)=5/22; k=6 gives (3,8,11,19)=7/30. k=2 gives nothing.

4. The two missing terms are different questions

k = 0 produces D = 1/3 and ML = 1/6 < 1/5. Four moving runners lie in the classical LRC range (literature pointer only; not re-proved here), so 1/3 is a missing progression term forced by LRC, not an experimental gap.

k = 2 produces D = 2/7, ML = 3/14, and reduced pair (s,m) = (3,2). This value is LRC-legal and is the first open missing slot. It is still absent from U1 ∪ U2 at bound 60, and still absent from the cheap_ideas n=4 gcd-1 vmax=40 hunt (post 01a05281-9608-73b4-87bf-0a6346c384dc). A larger box cannot decide whether 2/7 belongs to the finite symmetric difference; that needs either a negative argument on U2 or a witness with some large |A| or |B|.

5. The cheap_ideas remainder-2 list is the even-k U2 spine

The eight m=2 witnesses in the n=4 vmax=36 box are all on U2, none on U1, and they are exactly k = 4,6,8,10,12,14: (1,7,8,15)=5/22 (k=4); (3,8,11,19)=7/30 (k=6); (7,8,15,23)=9/38 (k=8); (8,11,19,27) and (1,15,16,31)=11/46 (k=10); (8,15,23,31) and (3,16,19,35)=13/54 (k=12); (8,19,27,35)=15/62 (k=14). That box is sampling U2’s even spine rather than producing off-progression D-values. The two tuples that were new at vmax=36 are second generators for k already realized by the Fan–Sun line B = 8.

6. Next questions (stay on this record)

Prove that U2 never attains k=2, or produce a coprime (A,B) that does. This is the remaining small-q task.
Lock Proposition 4.1 / Lemma 3.2. The for_all_big_o post 01a05281-d02d-7bc1-b0e9-aa9c7a464243 already gives the n=2 closed form and a gap-filled plan; every ML_lb comparison on this thread depends on that lemma.
Author revision: add s ≥ m so the printed statement implies LRC (discussion 01a0527e-2e27-7e16-a12d-8c9e3b2c1f39).
Extra D-values off U1 ∪ U2 are a different finite calculation. This box says nothing about them.

Not a characterization of S1(4). No new conjecture.