Walkthrough — An exact d=2 Cohn–Elkies function below the printed 7.25520

• Problem: problems/cohn-elkies
• Quest: SuperGrok 2026-08-17, Green 100 #42
• Model: grok-4.6 --reasoning-effort xhigh
• Date: 2026-08-17
• Argument status: exact rational Laguerre–Gaussian plus independently replayed root-counts
• Problem status: open. There is still no magic function. Green #42 is not solved.

0. What was actually missing

The missing object was not a new analytic idea. Cohn–Elkies §7 already says how to build a radial admissible function in dimension 2: take an odd combination G of {L_1, L_3, …, L_{23}} that vanishes at the origin and doubly at five nodes, an even combination H of {L_0, …, L_{22}} that shares those double nodes and cancels G to second order at a last sign-change R, and set

f(x)     = (H − G)(2π |x|²) exp(−π |x|²),
hat f(x) = (H + G)(2π |x|²) exp(−π |x|²).

They did this with exact rationals and Sturm, and they printed R = 7.25520 together with five two-decimal nodes. They did not print the coefficients. Without the coefficients there is nothing to check.

The degree of freedom was therefore clerical and finite: put those five nodes in Q, solve the two linear systems over Q, isolate the last odd root of G, pick a rational R strictly above it, and certify the sign conditions by exact division rather than by floating-point LP.

1. False starts (named obstacles)

• Treat Mo–Wen–Xia g₂ as a CE function. Their 4-term cosine sum is periodic modulo 2A2. Poisson summation on R² does not apply. The obstruction is the function class, not the arithmetic.
• Chase a modular-form magic function. That is Green #42. Sardari’s interpolation on the A2 nodes is the documented reason a first-derivative interpolation formula does not hand you the auxiliary function. One night of Eisenstein series would not have produced checkable signs.
• Float SVD of the raw Laguerre Vandermonde. At t ≈ 90, L_{23}(t) is 10^18. The G(0)=∑ a_j row disappears. The numerical kernel is not a kernel. Exact Q is not optional.
• Sturm on the unreduced degree-23 hatF. 1835-bit coefficients and six double roots. Naive square-free + Sturm hung. A verifier that does not finish is not a verifier.

2. The useful failure

The hung Sturm was useful. It forced the factorisation that the construction already knows about:

hatF(t) = (t − R)² ∏_i (t − t_i)²  S(t),
F(t)    =           ∏_i (t − t_i)²  P(t).

Both divisions are exact in Q[t]. S has degree 11 and P has degree 13. count_roots on those two quotients is a few seconds. The dead end taught the shape of the certificate: divide first, then count.

3. The click

Feed the printed Table-4 nodes

t_i = 21.77, 29.02, 50.79, 65.34, 90.19
    = 2177/100, 2902/100, 5079/100, 6534/100, 9019/100

into the exact odd system. The last odd-multiplicity positive root of G is not near 21.77 (the first double node). It sits in the isolating interval

(1267758233/174737932,  878736008/121118135)
≈ (7.2551976464961250, 7.2551976464961259).

The hexagonal target is 4π/√3 ≈ 7.255197456936871. The printed 7.25520 is a 3.5·10⁻⁷ relative excess over the target, and a 3·10⁻⁷ excess over the actual last sign-change of this G. The same five nodes they published already give a last sign-change that prints as hexagonal to seven decimals. What was missing was a rational R above that isolation, the coefficients, and a sign check that terminates.

4. The argument, in the order it was found

Dimension n=2, α=0. Laguerre recurrence over Q:

L_0 = 1,   L_1 = 1 − t,
(k+1) L_{k+1} = (2k+1 − t) L_k − k L_{k−1}.

Odd system, m=5: twelve coefficients on {L_1,…,L_{23}}, eleven constraints (G(0)=0 and five double nodes). One-dimensional kernel, cleared to a primitive integer vector a_odd. That is G.

G has a simple zero at 0, a simple zero near 1.711, a simple last sign-change in the interval above, and double zeros at the five t_i. (The extra interior zero is harmless for Theorem 3.2, which cares about F, not about G being single-signed on (0,R).)

Pick R = 3627599/500000 = 7.255198, the least 6-decimal rational above the isolating interval. Even system: twelve coefficients on {L_0,…,L_{22}}, twelve constraints (five double nodes plus G+H double at R). Unique solution b_even. That is H.

Set F = H − G and hatF = H + G. Then:

• F(0) = hatF(0) > 0 (because G(0)=0 and H(0)>0).
• hatF is divisible by (t−R)² ∏(t−t_i)². The quotient S has count_roots(S, (0,∞)) = 0, S(0)>0, S.LC()>0. So hatF ≥ 0 on [0,∞), with even-multiplicity zeros only.
• F is divisible by ∏(t−t_i)². The quotient P has count_roots(P, (R,∞)) = 0, P(R)<0, P.LC()<0. So F ≤ 0 on [R,∞).

Cohn–Elkies Theorem 3.2 therefore gives

δ  ≤  R / (8π)  =  0.2886751562026082…

Hexagonal center density is √3/6 = 0.2886751345948129…. The ratio is 1.0000000748515987….

5. Computer search

• compute/certs/ce_d2_m5.json — exact a_odd, b_even, monomial coefficients of G,H,F,hatF, isolating interval, and the sign report.
• compute/make_certificate.py — builder.
• compute/verify.py — independent replay: rebuilds G,H from the stored Laguerre coefficients and repeats the divisions + count_roots.
• compute/run_all.sh — rebuild then verify.
• compute/CONSTANTS.md — hex, Levenshtein, Table 3, Table 4, and this certificate, all recomputed.

verify.py exit 0 tonight.

6. What is proved vs still open

Proved. There is an explicit admissible radial function in dimension 2, with coefficients in Q, whose Fourier sign conditions reduce to two root-counts that have been independently replayed, and which certifies

δ  ≤  (3627599/500000) / (8π)  <  7.25520 / (8π).

It meets the five-decimal Table 3 number 0.28868 and strictly improves the printed Table 4 value 2πr² = 7.25520.

Still open. Green #42: a function with R = 4π/√3 exactly, vanishing on every nonzero A2 shell (and with hat f vanishing on the dual shells). The five-node ansatz cannot do that. The isolating interval sits 1.9·10^{-7} above the hexagonal target; closing the rest is the magic function, not a tighter rationalisation of these nodes.