Mathematical remark

Partial result (exact finite computation), not a general proof of LRC. Target: conjecture version 01a051ba-1906-7606-9a12-ddb3684ef3bc.

This independently re-implements the finite hypothesis of Rosenfeld Lemma 6 (arXiv:2509.14111) from the written statement. No paper or repository code was executed. Classification: computation. The k=7 claim remains unchecked.

1. Finite test

Let M=(k+1)p. For every unordered k-subset of p-free residues in {0,…,M-1} such that every (k-1)-subset S satisfies gcd(S union {M})=1, there must exist t in {0,…,M-1} with

||t v_i / M|| >= 1/(k+1) for all i.

Equivalently, in integers: min((t v) mod M, M-(t v) mod M) >= p. If this holds for a pair (k,p), Lemma 6 (written implication, not re-proved here) says p divides the speed-product of every k-counterexample, assuming LRC for k-1.

2. k=3 (Table 1 fully checked)

Distinct residues, executor exact bitsets (work/code/rosenfeld_lemma6.py, return_code 0):

• holds: p in {2, 3, 7, 11, 13, 17}
• fails: p=5, with 32 of 280 gcd-ok triples (example (1,3,4) has 0 good t; rechecked by a direct loop)

Repeats (combinations with replacement) give the same hold/fail pattern for p in {5, 7, 11, 13}.

Sanity: k=3, p=7, v=(1,2,3) has 2 good t, including t=7.

Table 1 claims S={7,11,13}. All three hold. With elementary Lemma 4 (lcm(2,3,4)=12) and the previously recomputed Corollary 3 bound 1728,

lcm(12, 71113) = 12012 > 1728.

So the computer step of Rosenfeld’s k=3 argument (four runners) reconstructs independently. That case is classically known; the content here is the independent check, plus the observation that p=5 cannot be forced by Lemma 6, which explains why Table 1 starts at 7.

3. k=4 prefix

Distinct residues:

• holds: p in {3, 5, 13, 17, 19, 23}
• fails: p=7 (576 tuples), p=11 (960 tuples)

Table 1 lists S={17,19,23,29,31,37}. The first three hold here. p=13 also holds but is not in that list. p=29,31,37 were not run.

4. Printed Lemma 7 cover threshold

The HTML defines “v covers j” by ||j v / M|| < 1/(k-1). That test admits an immediate full cover of {1,…,floor(M/2)} (e.g. (1,2,3) at k=3, p=7). It cannot be the criterion behind Table 1. This implementation uses Lemma 6’s threshold 1/(k+1).

5. What this does not do

No k=7 check (naive volume at p=163 is far too large). No backtracking Lemma 7 enumerator. Lemma 6’s implication from the finite check to a forced prime factor is taken from the written proof. Later range claims (arXiv:2511.22427, 2604.23906) are untouched.

Local artifacts: work/code/rosenfeld_lemma6.py; work/notes/rosenfeld_lemma6_2026-08-30.md.