q3 — ten new imagined end-states

Not A–E and not the q2 five (F homogeneous n-cycles, I Liénard B(n), O iterated squaring, P Harnack recurrence, Q PT+29). Each line is one concrete finite claim.

What q2 left on the table: an explicit degree-4 Darboux seed with three linear centers; a quasi-homogeneous Hamiltonian ẋ=2y, ẏ=−x³ with a period annulus; z↦z² is 2-to-1 (so the next non-separable test is z↦z³); Li–Liu–Yang still paywalled, so a 14th cubic cycle has to be written from an explicit Hamiltonian we own.

The ten

FF — two-well cubic Hamiltonian, 14 Abelian zeros

Claim. The cubic Hamiltonian field of H = y²/2 + (x²−1)²/4, perturbed by an explicit cubic, has 14 isolated zeros of I(h) across the two small wells and the large annulus, hence H(3) ≥ 14.

Why it might work: the Hamiltonian is written down (paywall bypass). Two wells plus a large annulus is the detection-function geometry that produced 11 then 13 in the literature.

Why it might die: Picard–Fuchs dimension is small; typical first-order counts are 2+2+1. Harnack still caps one invariant cubic at two ovals. Same shape as B unless I(h) actually has 14 zeros.

GG — L1 at the three PT centers has rank 29

Claim. A general degree-4 perturbation of the reconstructed H_{4,5} Darboux field has first Lyapunov quantities at (0,0) and (1, ±2) whose linear parts, together with the three traces, give 29 independent conditions, hence H(4) ≥ 29.

Why it might work: the seed is now an explicit polynomial field. First-order Lyapunov quantities are the same L1 as q1 line E after translation to each focus.

Why it might die: Prohens–Torregrosa already ran this through order 5 and stopped at 28. First-order they quote 22, not 29.

HH — perturb ẋ=2y, ẏ=−x³, four zeros of M1

Claim. A one-parameter cubic perturbation of the quasi-homogeneous center ẋ=2y, ẏ=−x³ has four isolated zeros of the first Melnikov function, beating the application bound in Gavrilov–He–Xiao (arXiv:2606.22137).

Why it might work: q2 certified the unperturbed period annulus. M1 is an explicit Abelian integral on H=x⁴+4y².

Why it might die: they claim to completely solve one named application. M1 may have a known small upper bound (2 or 3).

II — complex cube attains 9 regular sheets

Claim. Φ(u,v)=(u³−3uv², 3u²v−v³) (z↦z³) pulls the radial cubic back to a degree-11 field with nine hyperbolic cycles, matching Chebyshev T3 without being separable.

Why it might work: degree 3, Bézout ceiling 9, same budget as T3.

Why it might die: z↦z³ is 3-to-1 on the plane, not 9-to-1 (same CR obstruction as z↦z²).

JJ — Z(2,n) = floor((n−1)/2) as a polynomial identity

Claim. For every quadratic Hamiltonian, the Abelian integral of a degree-n perturbation is a polynomial in the energy of degree at most floor((n−1)/2), and that degree is attained, so the weak Hilbert number Z(2,n) equals that count.

Why it might work: Scholarpedia states Z(2,n)=⌊(n−1)/2⌋ because M(h) is a polynomial. Checkable by expanding ∮ P dy − Q dx on H=y²/2 + x²/2 (or a quadratic H).

Why it might die: classical, not a dent of H(n). Still a replayable weak-Hilbert identity the folder does not have.

KK — constructive +1: degree-4 field with two hyperbolic cycles

Claim. The Gasull–Santana construction H(n+1)≥H(n)+1 applied to the radial cubic produces an explicit degree-4 field with two hyperbolic isolated periodic orbits.

Why it might work: their +1 is elementary (a simple factor / a new nest at a new focus). The radial cubic is counted.

Why it might die: 2 does not beat H(4)≥28. Useful only as an explicit +1 map.

LL — cubic with an invariant line and three cycles

Claim. An explicit cubic with a straight-line solution has three isolated periodic orbits.

Why it might work: Ye / Cherkas uniqueness is for quadratics.

Why it might die: a linear Dulac factor along the line may still cap the count at 1.

MM — Abel form of the odd-cubic Liénard has two zeros

Claim. The Cherkas / Abel transform of ẋ=y−(αx+βx³), ẏ=−x has two positive zeros for some (α,β).

Why it might die: q2 line I already has uniqueness for that family. Recycled.

NN — PT inverse integrating factor gives an extra algebraic cycle

Claim. The inverse integrating factor of the H_{4,5} field has an isolated oval that is a fifth algebraic limit cycle after a small perturbation.

Why it might die: a first integral means continua, not isolated algebraic cycles. Same as Q’s unperturbed half.

OO — quadratic Hamiltonian + cubic perturbation, five simple zeros

Claim. Some explicit cubic perturbation of a quadratic Hamiltonian has five simple zeros of I(h), attaining the published local max (Han–Yang–Yu).

Why it might work: the upper 5 is attained in the literature; an explicit I(h) would be a replayable seed.

Why it might die: five from one center is not H(3)≥14, and constructing the 5-zero integral may be long.

Ranking

Official three: ff-two-well/, gg-pt-lyapunov/, hh-qh-melnikov/. Two more started so the menu is not left idle: ii-complex-cube/, jj-weak-hilbert/.

Worker pool raised. Leftover menu executed: kk-plus-one/, ll-invariant-line/, oo-five-zeros/. Recycled without a second copy: MM (same family as I), NN (same first integral as Q). Four extra live lines: pp-christopher-lloyd/, qq-t2-radial/, rr-kolmogorov/, ss-cubic-l2/, and tt-radial-factor/.

PP — Christopher–Lloyd (u², v²) four-fold

Claim. The map (x,y)↦(u²,v²) applied to a first-quadrant translate of the radial cubic is a degree-7 field with four hyperbolic cycles that beats H(7)≥74.

Why it might work: Gasull–Santana §4 and Christopher–Lloyd 1995; four sign combinations, N=2n+1.

Why it might die: 4 does not beat 74. Same sheet count as T2.

QQ — T2 of the radial cubic, four ovals

Claim. The T2 pullback of the §6 radial cubic is an explicit degree-7 field with four hyperbolic cycles beating H(7)≥74.

Why it might work: q1 already has the identity; this writes the field.

Why it might die: same 4 at N=7; 4 < 74.

RR — cubic Kolmogorov with 7 first-quadrant cycles

Claim. An explicit cubic Kolmogorov field has 7 isolated cycles in x>0, y>0, beating M_K(3)≥6.

Why it might die: a weighted Dulac still kills the named family.

SS — L2 of the two-well cubic jet gives extra cycles

Claim. V2 at the FF wells is a new small-cycle source, hence H(3)≥14.

Why it might die: L1≠0 already, so the order is 1.

TT — (x²+y²) times the radial cubic has two cycles

Claim. Y=(x²+y²)X is degree 5 with two hyperbolic cycles (H(n+2)≥H(n)+1 made explicit).

Why it might die: unperturbed Y still has exactly one circle.

All imagined dents dropped. Forks kept; see README.md and ATTACK.md. No published H(n) moved.