[pinterbanget]

This answers #4600's open question 1 (@petalas) completely, within #4280's row model: exactly two short rows realize c ≠ 0 if and only if gcd(c, m) ≥ 3. No new κ.

Necessity (written proof)

If gcd(c, m) = 2, #4600's condition forces both short rows to have length m/2. Then:

• (r₁ − r₂)·m/2 ≡ 0, so r₁ + r₂ is even;
• 2B(m/2) ≡ 0.

Together these give c = −(r₁ + r₂)·B(m/2) ≡ 0. Units are already excluded by #4600.

Sufficiency (explicit construction, every m)

Let g = gcd(c, m), d = m/g and c = g·r. Take:

• λ = (m, m−d, d, 0, …);
• ρ = (m−g, g, 0, …);
• β = ((g+1)^r, g^(d−r), M, 1^r, 0^(d−r)), where M is the balanced block of length (g−2)d and sum m − 2r.

Then B(d) = m + r, B(m−d) = 2m − r and B(m) = 2m, all congruences hold, and the coefficient is g·r = c.

Checks

• Construction: verified for all 4,626 pairs (m, c) with m ≤ 200.
• Exact enumeration of the whole profile model for m ≤ 12 agrees.
• The annihilator condition alone is not sufficient. For example, m = 6 with c = 2 or 4, and m = 12 with c = 2 or 10.
• Prove the two-short-row connector obstruction for every modulus #4600's family m = 8t, c = 4t is checked for t ≤ 64.

research/two-short-row-coefficients/verify.py uses the standard library only and takes about 3 s.

Limitation

The row model and its converse are inherited from #4280. Call counts beyond #4600's are not claimed.

Credit: petalas (Nick Petalas) for #4600 and #4280. Ryanson Jonathan, with Claude Code (Claude Opus 5.5) assistance.

Head: 47cc973

🤖 Generated with Claude Code