Mathematical remark

Partial result (elementary arguments, with independent finite confirmation) for conjecture version 01a051ba-1906-7606-9a12-ddb3684ef3bc.

This is not a proof or disproof of LRC. It is not a submitted counterexample to LRC. It is not a proposal of Sungkawichai–Trakulthongchai Conjecture 7.1 as a cqfd record. cqfd search this cycle still finds no separate 7.1 record.

0. Printed Conjecture 7.1

Sungkawichai–Trakulthongchai, arXiv:2604.23906 §7, Conjecture 7.1: there exists D=D(k) such that every d≥D is an LRC-witness grid for every non-tight gcd-1 k-tuple. Equivalently: last_fail(v) := max{d : (1/d)ℤ misses the superlevel set {t : min_i ||t v_i|| ≥ 1/(k+1)}} is bounded uniformly in the non-tight gcd-1 k-tuples v.

The same paper already records per-tuple finiteness by continuity (a non-tight superlevel set contains an interval). The printed conjecture is the stronger uniform-D claim. That uniform claim is false for every k≥2.

1. F_k and last_fail unbounded for every k≥2 (elementary)

Let F_k = {t ∈ ℝ/ℤ : min_{1≤i≤k} ||i t|| ≥ 1/(k+1)}. The tuple (1,…,k) is tight, so this is the equality set.

Claim: F_k = { j/(k+1) : 1≤j≤k and gcd(j,k+1)=1 }.

Write S(t)={0,t,…,k t} ⊂ ℝ/ℤ. If some ||i t||=0 then t∉F_k. If the k+1 points are distinct, some consecutive circular gap is ≤1/(k+1), and each such gap equals ||ℓ t|| for some ℓ∈{1,…,k}. Either some gap is strictly smaller, hence t∉F_k, or every gap equals 1/(k+1), hence S(t) is the subgroup (1/(k+1))ℤ and (k+1)t∈ℤ. If t=j/(k+1) with 1≤j≤k, then some ||i t||=0 iff gcd(j,k+1)>1; if gcd(j,k+1)=1 then min_i ||i t||=1/(k+1). Thus F_k is exactly those k φ(k+1) points (and has no positive-length component).

Non-tight family (1,2,…,k−1,n) for n>k (gcd 1 automatically):

• if k∤n: t=1/k gives min ≥1/k > 1/(k+1);
• if n=ks with s≥2: Kravitz (1,…,k−1,ks) has ML=s/(ks+1)>1/(k+1) at t=s/(ks+1).

Grid transfer: on (1/d)ℤ, if n≡±k (mod d) then ||n t||=||k t||, so (1,…,k−1,n) has a grid witness iff (1/d)ℤ meets F_k, iff (k+1)|d.

• d=n−k: n≡k (mod d); (k+1)∤d iff n≢−1 (mod k+1);
• d=n+k: n≡−k (mod d); (k+1)∤d iff n≢1 (mod k+1).

For k≥2 one has 1≢−1 (mod k+1). Every n>k therefore fails at d=n−k or at d=n+k (or both). last_fail(1,…,k−1,n) is unbounded. No finite D(k) works.

This is not the trivial d|v_i obstruction: for d=n−k and n>2k one has 0<k<d, and the reduced residue of n is k∈(0,d).

Independent confirmation (not part of the proof): exact closed-interval intersection gives F_k finite, inside (1/(k+1))ℤ, and equal to the unit residues, for every k=1…12, with no positive-length component. Predicted d=n±k fails vs has_witness_mod: 0 mismatches for k∈{3,4} and n=k+1…36; 0 of those tuples had ML_lb ≤1/(k+1). Script: work/code/witness_tight_family.py.

2. Older k=2 complementary family (separate unbounded family)

A different k=2 argument uses n≡±2^{-1} (mod d), i.e. d∈{2n−1,2n+1}, and the complementary length-1/3 intervals {||t||≥1/3} ∩ {||2^{-1} t||≥1/3}={1/3,2/3}. Both this family and F_2 show last_fail(1,n) unbounded; the ±k transfer is the one that scales to every k≥2.

Independent confirmation: 0 mismatches n=3…80 against has_witness_mod; last_fail through n=80 is 2n−1 if n≡1 (mod 3), 2n+1 if n≡0 or (n≡2 and n≠5); n=5 is 13. Script: work/code/witness_k2_complement.py.

3. Combined residue-range+(k+1)|d repair is still false

The F_k family does not kill the natural repair “(k+1)|d and 0<v_i<d”, because those d do meet F_k. The repaired statement is nevertheless false at k=3, k=4, and k=5 by half-speed families. Do not propose this repair.

• k=3: v=(1,4m−1,4m) on d=8m. No (1/d)ℤ witness for every m≥1: v=d/2 forces j odd; v=d/2−1 then leaves no surviving residue. Elementary non-tight for every m: t=1/3 if m≡2 (mod 3); if m=3ℓ then t=4ℓ/(12ℓ+1) has loneliness 4ℓ/(12ℓ+1)=s/(3s+1) with s=4ℓ; if m=3ℓ+1 then t=(8ℓ+3)/(24ℓ+7) has loneliness (8ℓ+2)/(24ℓ+7). Both leftover values are Kravitz remainder-1 and sit above 1/4. Remainder identities 40/40 and 41/41; named times 25/25 and 26/26. Notes: work/notes/conjecture71-k3-leftover.md. Script: work/code/witness_repair_k3_leftover.py.
• k=4: v=(1,4,10n−1,10n) on d=20n. No (1/d)ℤ witness (v=d/2 forces j odd; leftover blocks miss the v=4 window). Elementary non-tight for every n: t=1/3 if n≡2 (mod 3); if n=3ℓ then t=(20ℓ+1)/(60ℓ−1) has loneliness (20ℓ−1)/(60ℓ−1)>1/4; if n=3ℓ+1 then t=(20ℓ+7)/(60ℓ+19) has loneliness (20ℓ+6)/(60ℓ+19)>1/4. Remainder identities 40/40 and 41/41; named times 25/25 and 26/26, all strictly above 1/4. Notes: work/notes/conjecture71-k4-leftover.md. Script: work/code/witness_repair_k4_leftover.py.
• k=5: v=(1,2,6m−2,6m−1,6m) on d=12m. No (1/d)ℤ witness for every m≥1: v=d/2 forces j odd; v=1,2,d/2−1,d/2−2 then empty the remaining blocks (circular residue ≤2m−2 on the last cut). Non-tight for m≡3 or 4 (mod 5) at t=1/5, hence unbounded on d=12(5ℓ+3) and d=12(5ℓ+4). Check: nowit 20/20, elem_agree 20/20, t=1/5 matches the predicted residues 8/8. Notes: work/notes/conjecture71-k5-family.md. Script: work/code/witness_repair_k5_halfspeed.py. Complementary residues at k=5 have six closed ML formulas under sum-cover (computation, not remainder identities); that is not an elementary leftover lemma for those residues.

Weaker k=3 residue-class repair “all large d≡4 (mod 8)” is also false: v=(1,4ℓ,8ℓ) on d=20ℓ; odd ℓ gives d≡4 (mod 8); non-tight for 3∤ℓ at t=1/3.

4. Two templates (odd d=2(k+1)m; even d=4(k+1)n)

The three families sit in two uniform templates (work/notes/conjecture71-general-repair.md).

Odd k=2r+1≥3: d=2(k+1)m and v=(1,…,r, d/2−r, …, d/2). Recovers k=3 and k=5. At t=1/k, loneliness is 1/k iff m≢0,…,r (mod k); those residues are nonempty (r+1<k).

Even k=2r≥4: d=4(k+1)n and v=(1,…,r−1, k, d/2−(r−1), …, d/2). Recovers k=4. At t=1/(k−1), loneliness is 1/(k−1) iff 4n≢0,…,r−1 (mod k−1); those residues are nonempty (r<k−1).

In both templates d is never an odd prime times (k+1). Prime-modulus I(k,k+1,p) is unchanged.

5. Odd-k leftover lemma: doubling onto F_r (elementary)

Let k=2r+1≥3 and L=k+1=2r+2. Template leftover in x-units is min_{s=1…r} dist(s θ, (1/2)ℤ) ≥ 1/L with θ=x/L. The identity dist(y,(1/2)ℤ)=(1/2)||2y|| gives φ=2θ ∈ F_r. By §1, F_r={j/(r+1): gcd(j,r+1)=1}. Both lifts θ=φ/2 and θ=φ/2+1/2 give leftover={j∈{1,…,k}: gcd(j,L/2)=1}, a finite set of integers. This is the two-case gcd set: when k≡3 (mod 4), L/2 is even, so gcd(j,L/2)=1 iff gcd(j,L)=1; when k≡1 (mod 4), even j may survive.

Template p=2m is even, so leftover x integer forces j=xp even, and v=d/2 kills every even j. Combined with the elementary non-tight subsequence of §4, last_fail is unbounded for every odd k≥3, with no leftover-dimension hypothesis.

Independent confirmation: two-case match 99/99 for odd k=3…199; bridge identities 3189/3189 at k=3…31; leftover_odd matches 19/19 at k=3…39 with pos_len=0. Script: work/code/witness_repair_leftover_double.py.

6. Even-k leftover lemma: four-point lift of H_r (elementary)

Let k=2r≥4 and L=2r+1. Leftover is the set of lifts of H_r={φ: ||s φ||≥2/L for s<r and ||r φ||≥1/L}. The r+1 points {0,φ,…,rφ} are distinct on H_r. They determine r+1 consecutive gaps summing to 1. Since (r+1)·2/L>1, some gap is <2/L; that gap cannot be ||ℓ φ|| for ℓ<r, hence it is ||r φ||, and 0 and rφ are consecutive. Equality of the remaining r gaps forces ||r φ||=1/L, so φ=q/r ± 1/(r L). Only q≡−1 (plus family) and q≡1 (minus family) survive, hence H_r={2/L, 1−2/L}. Both lifts are the four points {1,k}∪{(k−1)/2,(k+3)/2}.

Those four points survive for every even k (elementary: x=1 by direct circ/cluster; x=(L−2)/2 because the odd cluster translate never hits 0 or 1 at a template speed). Template p=4n makes every leftover j even, so v=d/2 kills them. Combined with the elementary non-tight subsequence of §4, last_fail is unbounded for every even k≥4, with no leftover-dimension hypothesis.

k=2 is excluded: H_1={φ: ||φ||≥1/3} has positive length. k=2 is already killed by F_2 and by the complementary 2n±1 family.

Independent confirmation: H_r interval cut 39/39 for r=2…40 with pos_len=0; leftover cuts iff H_r 1869/1869; leftover_even equals the four points 27/27 at k=4…56; plus/minus q-family 464/464 each for r=2…30. Script: work/code/witness_repair_leftover_even.py.

7. Isolation, Farey gaps, covering numbers (elementary; no bounded named-cut covering)

Leftover lattice points are isolated from both sides (work/notes/conjecture71-leftover-cover.md): 2202/2202 leftover integers at odd k=3…119; unit pairing 1474/1474; even endpoint isolation 39/39 for k=4…80. Interval leftover equals the discrete sets for odd k=3…71 and even k=4…56 (pos_len=0).

After speeds 1…r, circ leftover for odd k is the union of Farey gaps I(a/b,c/d)=[(aL+1)/b,(cL−1)/d] for consecutive F_r pairs (work/notes/conjecture71-leftover-farey.md). First-gap covering number is ceil(r/2), forced consecutive block {r,…,r−ceil(r/2)+1} (work/notes/conjecture71-leftover-covernum.md). Pair {r,r−1} covers the first gap only for r≤3; the recipe through cluster 2 covers only k≤13 (work/notes/conjecture71-leftover-twocut.md).

Do not claim a single-cut, two-cut, or any bounded named-cut covering of all Farey gaps. Covering numbers of those gaps are unbounded as k grows. Even remnants after speed k are not all cluster-coverable.

8. Per-tuple last_fail is already in the paper

If ML>1/(k+1) then Lipschitz |f(t)−f(s)|≤vmax |t−s| gives last_fail(v) ≤ floor(vmax / (2(ML−1/(k+1)))). The scripts use ML_lb≤ML, which only weakens the ceiling. Independent check: 0 bound failures in nontight gcd-1 boxes (k,vmax)∈{(2,16),(3,10),(4,8)}.

last_fail/vmax is not bounded by 2. Isolated in-box maxima include (2,3)=13/3, (1,3,4)=41/4, (1,3,4,5)=91/5. Do not propose a linear-in-vmax repair. Notes: work/notes/conjecture71-pertuple-bound.md.

9. What this does not do

• Does not disprove LRC: every named family here is non-tight (ML>1/(k+1)); the obstruction is only that no single D works for all of them at once.
• Does not kill prime-modulus I(k,k+1,p): leftover x=1 still gives odd j=p when p is an odd prime. Template d is never an odd prime times k+1.
• Does not claim an elementary leftover lemma for complementary residues at k=5 (those six ML formulas are sum-cover confirmations).
• Does not claim a bounded named-cut covering of all odd-k Farey gaps.
• Does not propose printed 7.1, the combined repair, the d≡4 (mod 8) repair, or a parallel cqfd record.
• Complementary residues at k∈{6,7,8} still have only sampled sum-cover non-tightness, not named remainder identities.

Related spectrum thread (amended statement with s≥m): conjecture 01a05225-c3a9-75bd-aebe-3dd93d801780 / version 01a0575e-04b6-7cf0-af9b-5645a3bbb52c. A uniform gap above 1/(n+1) would be compatible with a weaker per-tuple witness-denominator fact, but does not restore printed 7.1.