Walkthrough — Hilbert 16(b), H(n)

Discovery notes, not a cleaned proof. Beats: refs/walkthrough-style.md.

• Problem: problems/hilbert16-limit-cycles
• Model: Grok 4.6
• Date: 2026-08-27
• Problem status: open. No published H(n) moved. The second and third campaigns did not move a table entry either.

0. What was actually missing

A finite handle on H(n): either an explicit field that beats a published lower bound, or an exact lemma a stranger can replay. The asymptotic n² log n story and the Ilyashenko/Écalle finiteness theorems do not move a table entry. The degree of freedom is a concrete polynomial field, or a pullback identity, or a restricted upper bound — not a new status noun.

1. Named false starts

Five imagined end-states were written first (ATTACK.md), then attacked backwards.

• Quadratic with five cycles. The extra μ y² term that was supposed to create a second nest at (0,1) is the same term that destroys Shi’s order-3 jet (primitive L1 = 27μ) and moves the second equilibrium. Forcing a weak focus at (0,1) by δ = −5 either centers the origin or leaves it at order 1. Bautin then budgets 1+3 = 4, which is Shi’s number, not 5. A surrounding fifth cycle is excluded for every quadratic by Coppel: a quadratic limit cycle surrounds a unique singularity.
• Cubic with fourteen cycles. Li–Liu–Yang is paywalled. There is no explicit (P,Q) on the page to which one extra nest can be attached. Local M(3) ≥ 12 is not a field. Harnack caps a single invariant cubic at two ovals.

2. The useful failure

The 5-cycle and 14-cycle stories failed for algebraic reasons, not for lack of integration time. That is useful: it says where not to spend a numeric search. The leftover exact objects are the Shi jet (why the small cycles in (3,1) can exist at all), van der Pol’s Liénard data (a cubic that is completely counted), and the Chebyshev pullback (how a counted seed replicates).

3. The click

Two clicks, both after the fictions were dropped.

First: the 2026 Chebyshev paper’s §6 cubic is not a curiosity. In polar coordinates it is ṙ = r(ρ²−r²), θ̇ = −1, so uniqueness is a polynomial identity, and the T3-preimage of the circle is nine ovals for free. That is a checkable field of degree 11 with nine hyperbolic cycles, and it is also the uniqueness lemma of line D.

Second: Remark 4 of that paper left non-separable maps open. Bézout already caps regular real fibers at m², and the adjugate pullback has the same degree budget as Chebyshev. So one-step replication cannot beat m² by changing the map. The open question that remains is iteration, not a single covering.

4. The argument

• Unperturbed Shi: build a Poincaré–Lyapunov F through degree 8 with the even-degree gauge yⁿ = 0. Then dF/dt = V3 r⁸ + O(9), V1 = V2 = 0, V3 = 35625/8. Cross-check Li Chengzhi L3 = 57000 = (64/5) V3. Two real equilibria; (0,1) is a strong focus.
• van der Pol: F(x) = μ(x³/3 − x) meets Liénard’s list; the first-order Melnikov on the linear center is πμ h(2−h).
• Chebyshev: Φ(u,v) = (Tm(u), Tm(v)), Y = (Tm′(v) P∘Φ, Tm′(u) Q∘Φ), deg Y = nm+m−1, m² sheets. Table 1 arithmetic on the published seeds matches; no missed factorization beats 252/1080/1380/2012.
• Restricted upper bounds: polar identities in Z[x,y,ρ]; Hamiltonian dH/dt = 0; Lotka–Volterra weighted Dulac.
• L1 by polar averaging and by Poincaré V1; same primitive polynomial. Lean checks T2, adj, and four points on two quadrics.

5. Computer search

No SAT. No ODE integrator used as a bound. The machines did exact algebra: sympy for the jets, identities, and table; gcc over Q for dF/dt; rustc as a second expansion of the same polynomials; lean 4.32.0 for the tiny adj lemma. Random degree-m maps were sampled only as a Bézout sanity check (never above m²); that sample is not a proof.

Replay: problems/hilbert16-limit-cycles/compute/run_all.sh.

6. What is proved vs still open

Proved here, and replayable:

• Unperturbed Shi is a weak focus of order 3 at the origin, V3 = 35625/8, with a second real focus at (0,1).
• van der Pol (μ>0) satisfies Liénard uniqueness.
• H(nm+m−1) ≥ m² H(n) as an identity, and the 2604.12883 Table 1 arithmetic on the cited seeds.
• The radial cubic has exactly one periodic orbit.
• Quadratic Hamiltonians have no limit cycles; Lotka–Volterra has none in the open first quadrant.
• One-step polynomial pullback of degree m has at most m² regular sheets. The first Lyapunov quantity of a quadratic focus is the displayed L1.

Not proved, and not claimed:

• H(2) ≥ 5, H(3) ≥ 14, or any movement of a published H(n).
• Finiteness of H(n).
• The full Bautin theorem.
• The Prohens–Torregrosa and Han–Li centers (seeds are cited, not reconstructed).
• That the 2026 lifts H(14) ≥ 252 etc. are independent of those unreplayed seeds.

Second campaign (same day)

Twenty-five new fictions (compute/q2/ideas.md). Five scored and attacked. All five imagined dents died; five forks lived.

0. What was actually missing

A handle that is not Chebyshev one-step and not Shi/van der Pol again: either a new seed, a new lift that beats m², or a certified upper the literature treats as open. The degree of freedom is still an explicit field or a polynomial identity.

1. Named false starts

• Homogeneous n-cycles: scaling produces a continuum.
• Liénard beat of B(n): no field; full H(3,1)=1 needs oddness.
• Iterated z↦z² past the quadratic ceiling: two sheets, not four.
• Harnack recurrence: 1225 pairs, no table beat.
• PT+1: their order-5 budget is already 28.

Discarded without a worker, same deaths as A/B: Ye class I two cycles, invariant-line two, (2,2), Liénard two, two hemicycles, Shi equator as a planar cycle, QW3 fifth, figure-eight 14, Yu–Han 12+2.

2. The useful failure

Complex squaring is the model non-separable covering, and it is worse than Chebyshev, not better. Remark 4’s remaining half (iteration) does not get you past m² by changing the map to z↦z². The PT seed, once written as a primitive degree-4 field, is a concrete Darboux center a later line can perturb.

3. The click

Two. First: z↦z² is 2-to-1 on the plane because Cauchy–Riemann ties the components; Bézout’s 4 is a ceiling, not a count this map attains. Second: −B+4xA in the PT first integral cancels from degree 5 to degree 3, so the Darboux field is genuinely degree 4 and has three linear centers.

4. The argument

• Homogeneous: P(λx,λy)=λ^n P; compact orbits scale.
• Odd-cubic Liénard: dE/dt=−αx²−βx⁴; Liénard list when α<0<β.
• Φ=(u²−v²,2uv): adj identity, deg 7 for the radial cubic, two regular preimages, 2^k=(N+1)/(n+1).
• Har(m) table vs the q1 seeds.
• PT primitive field; Coppel: quadratics restrict to degree ≤2 on a line.

5. Computer search

No SAT. No integrator as a bound. sympy / rustc identities and integer tables. Replay: compute/run_all.sh (q1 then q2).

6. What is proved vs still open

Proved here, extra to q1: homogeneous / unperturbed quasi-homogeneous fields have no isolated cycles; B(n) thresholds; the odd-cubic Liénard energy identity; iterated squaring sheet counts; Harnack arithmetic; the explicit PT H_{4,5} Darboux field and the quadratic contact identities.

Not proved: any movement of a published H(n). Finiteness of H(n). The 28-cycle perturbation of that Darboux field. Li–Liu–Yang’s 13 (still paywalled).

Third campaign (same day)

Ten new fictions (compute/q3/ideas.md). Official three, plus two extras so the menu was not idle. All five imagined dents died; five forks lived.

0. What was actually missing

An explicit cubic that is not Li–Liu–Yang’s paywalled field, or a perturbation of the PT seed we now own, or a non-separable covering that actually attains m². The degree of freedom is still an explicit polynomial or a counted Abelian integral, not a new status noun.

1. Named false starts

• Two-well 14 Abelian zeros: the integral is elliptic; one sample and L1 prove I ≢ 0, not 14 roots.
• PT L1 rank 29: unperturbed L1 is 0 (Darboux center); they already stopped at 28.
• Four zeros of M1 on ẋ=2y, ẏ=−x³: scaling caps a cubic at two first-order zeros; the named family has one.
• Holomorphic cube, nine sheets: z↦z³ is 3-to-1 (same CR as O).
• Beat Z(2,n): the formula is a theorem.

2. The useful failure

The two-well Hamiltonian is the geometry people cite for 11 then 13, and first-order on that well does not produce 14. The PT seed’s L1_E is a trap: at (1,±2) it is ±9/2 and the cubic jet cancels it. Evaluating only the q1 quadratic polynomial would have falsely called those points order-1 foci.

3. The click

Two. First: after the q1 focal scaling, both wells of the named van der Pol perturbation have the same L1 = √2 μ, and I(0) = 4μ/15 is elementary on the figure-eight, so the obstruction is not “no formula” but “the formula is too small.” Second: z↦z³ has the same degree budget as T3 and is strictly weaker, so the next non-separable test after squaring is also a miss.

4. The argument

• Two-well: H = y²/2 + x⁴/4 − x²/2; dH/dt identities; figure-eight reduction to 2/15; L1 cubic correction.
• PT: translate three centers; L1 = L1_E + cubic; cancel.
• Quasi-homogeneous: weights (1,2); the moment of x^k scales as λ^{k+3}; I(h)=0 iff α = C √h.
• Cube: CR, fibre 3, N=3n+2, T3 attains 9 at the same N.
• Weak Hilbert: Ĩ(h)=h p(2h) attains floor((n−1)/2).

5. Computer search

No SAT. No integrator as a bound. sympy / rustc identities and integer boxes. Replay: compute/run_all.sh (q1, q2, q3).

6. What is proved vs still open

Proved here, extra to q1 and q2: the two-well classification and I(0); L1 = 0 at the three Darboux centers and two L1(μ) polynomials; first-order cyclicity ≤ 1 for Q=μ(α−x²)y on H=x⁴+4y²; holomorphic-cube sheet counts; the radial family attaining Z(2,n) for n=1…10.

Not proved: any movement of a published H(n). Finiteness of H(n). Fourteen cubic cycles. A 29th quartic cycle. Li–Liu–Yang’s 13 (still paywalled).

Worker pool (same day)

The leftover q3 menu and four extras ran so no line sat idle. All imagined dents died.

0. What was actually missing

An explicit +1 or 4-fold field we own, not only the identity. The degree of freedom is still a written polynomial.

1. Named false starts

• Gasull–Santana +1: second cycle is an unwritten Hopf.
• Cubic + line, three cycles: Dulac kills the named field.
• Five Abelian zeros: Z(2,3)=1, or else H(3)≥5 < 13.
• (u²,v²) or T2 beating H(7)≥74: four ovals, not 74.
• Kolmogorov 7: Dulac, 0 in the first quadrant.
• L2 extra wells: order is already 1.
• (x²+y²)X two cycles: still one circle.

2. The useful failure

(u²,v²) attains 4 sheets at N=2n+1, which holomorphic z↦z² does not. The obstruction in O and II was Cauchy–Riemann, not Bézout. Christopher–Lloyd uses the four sign combinations.

3. The click

Translate the radial circle into the first quadrant, then square each coordinate. Four ovals, degree 7, written term-by-term. Same count as T2, different map, still far from 74.

4. The argument

• KK: line 4x−15y misses the translated circle (15>0).
• LL: Dulac 1/y on ẋ=16y+16x+x³, ẏ=16xy.
• OO: Ĩ(h)=h(α−2h); Area on the cubic nest.
• PP: u̇=v P(u²,v²), four quadrants, f'(ρ)=−1/2.
• QQ: T2 field, 8+8 monomials.
• RR: B=x^{-2}y^{-1}, div(BX)/B=−1.
• SS: V2=−√2 μ(23μ²+18)/96.
• TT: ṙ=r³(ρ²−r²).

5. Computer search

No SAT. No integrator as a bound. sympy / rustc. Replay: compute/run_all.sh.

6. What is proved vs still open

Proved here, extra: the degree-4 line product with one circle; a cubic with invariant line and 0 cycles; 1 and 0 Abelian zeros on the two OO families; an explicit Christopher–Lloyd degree-7 four-oval field; an explicit T2 degree-7 four-oval field; a cubic Kolmogorov Dulac; V2 at the two-well; the degree-5 radial product with one circle.

Not proved: any movement of a published H(n).