q2 — twenty-five imagined end-states

Not A–E recycled. Each line is one concrete finite claim that would move a published H(n) or a documented-family bound. Score later is (published-record delta) × (checkability tonight: explicit field or a replayable identity). Discard anything that dies on Bautin / Coppel / Harnack the same way A/B died, unless a new bypass is written.

Published record used as the target (RESEARCH.md, q1 unchanged): H(2) ≥ 4 (Shi; Chen–Wang); H(3) ≥ 13 (Li–Liu–Yang); H(4) ≥ 28 and the Prohens–Torregrosa table; Han–Li n² log n table; Chebyshev lift H(nm+m−1) ≥ m² H(n) (arXiv:2604.12883). Do not cite 252 / 1080 / 1380 / 2012 as found here.

The twenty-five

F — homogeneous field with n isolated cycles

Claim. A homogeneous planar field of degree n has n isolated periodic orbits.

Why it might work: polar form ṙ = r^n f(θ), θ̇ = r^{n-1} g(θ) is explicit; zeros of f look like invariant rays, and a circle would be a cycle if f vanished identically.

Why it might die: if f ≡ 0 and g never zero, every circle is periodic (a continuum). If f has isolated zeros, those are rays, not cycles. Either way there are no isolated periodic orbits.

G — cubic Liénard of type (3,1) with two cycles

Claim. The Liénard field ẋ = y − (αx + βx² + γx³), ẏ = −x has two isolated periodic orbits for some explicit (α,β,γ), beating the conjecture H(3,1) = 1.

Why it might work: Dumortier–Panazzolo–Roussarie already killed H(m,1) = ⌊m/2⌋ at m = 6; maybe m = 3 also fails.

Why it might die: Zhang Zhifen / Liénard uniqueness apply when f/g is strictly monotone. The odd-cubic subfamily is classical and has exactly one cycle when it has any. No new bypass of that list.

H — prove H(3,1) = 1 for every cubic-damping linear-restoring Liénard

Claim. Every field ẋ = y − F(x), ẏ = −x with deg F ≤ 3 has at most one limit cycle.

Why it might work: the family is documented as open (Lins–Neto–Pugh for n = 3; Scholarpedia still lists the conjecture). A certified upper of 1 would be a family dent.

Why it might die: the full family is a famous open problem, not a one-night identity. A named subfamily (F odd, or F = αx + βx³) is checkable and already classical.

I — beat Chen–Dai–Kaloshin–Li B(n) for Liénard H(2n+1, 5)

Claim. An explicit Liénard with deg f = 2n+1, deg g = 5 has more than B(n) = 2n + ⌊n/3⌋ + ⌊(n+1)/3⌋ − 2 isolated cycles, by adding the unused origin family they left out of Theorem 3 (arXiv:2608.17773v1).

Why it might work: they say the origin contribution was omitted only for a compatibility reason; a small-n field might realise all three local families at once.

Why it might die: the compatibility obstruction is the point of their paper. Constructing the field is not a finite identity. Replay of B(n) versus Han–Romanovski is checkable and is not a dent of planar H(n).

J — quasi-homogeneous weight-(1,2) center, four Melnikov zeros

Claim. A one-parameter cubic perturbation of a weight-(1,2) quasi-homogeneous center has four isolated zeros of M₁, beating the application bound in Gavrilov–He–Xiao (arXiv:2606.22137).

Why it might work: their general formula is in terms of a cardinality |S| − 1; a concrete weight might have extra zeros they did not write.

Why it might die: they claim to completely solve one named application. The unperturbed field has a period annulus, hence zero isolated cycles.

K — quadratic Hamiltonian, six Abelian zeros under cubic perturbation

Claim. Some quadratic Hamiltonian H of degree 3, perturbed by cubics, has six isolated zeros of the Abelian integral I(h), beating the published “at most 5 from one quadratic center” (Han–Yang–Yu, as quoted in Scholarpedia).

Why it might work: six coefficients in a cubic perturbation look like six zeros.

Why it might die: that 5 is an upper bound, not a lower bound we can sneak past. Weak Hilbert Z(3,2) is 2 (generic) / 3 (degenerate) for the other direction (cubic Hamiltonian, quadratic perturbation).

L — Ye class I with two cycles

Claim. A quadratic of Ye class I (ẏ = x) has two isolated periodic orbits.

Why it might work: the class is large.

Why it might die: Chen–Ye uniqueness, at most one. Same shape as A/B: a published uniqueness theorem kills the fiction immediately.

M — quadratic with an invariant line and two cycles

Claim. A quadratic with a straight-line solution has two isolated periodic orbits.

Why it might work: the line organises the phase portrait.

Why it might die: Cherkas / Ye, at most one. Same death as L.

N — quadratic (2,2) configuration

Claim. There is an explicit quadratic with two nests of two cycles each.

Why it might work: Coppel’s orientation lemmas permit two nests; Bautin budgets 3+1 = 4.

Why it might die: Zhang (c. 2000), quoted by Llibre–Schlomiuk and Ilyashenko–Llibre: if two foci carry cycles, one nest has at most one. No new bypass.

O — iterated complex-squaring beats the 2604.12883 quadratic ceiling

Claim. k-fold pullback by Φ(u,v) = (u² − v², 2uv) of the radial cubic produces more than a quadratic number of hyperbolic cycles in the final degree, answering Remark 4 of arXiv:2604.12883 by beating Theorem 2.

Why it might work: Remark 4 left iterated non-separable maps open; complex squaring is the model non-separable covering.

Why it might die: Bézout still caps each degree-2 step at 4 regular sheets. After k steps the degree is N = (n+1)2^k − 1 and the sheet count is ≤ 4^k = ((N+1)/(n+1))², still quadratic. Chebyshev of degree 2^k matches that count in one step.

P — Gasull–Santana Harnack recurrence produces a new H(N)

Claim. The identity H(n+m) ≥ H(n) + Har(m) of arXiv:2510.11705 Corollary 2, with Har(m) = (m−1)(m−2)/2 + [1+(−1)^m]/2, beats a published table entry for some N = n+m ≤ 50.

Why it might work: Harnack-maximal algebraic cycles plus a published seed look like a new lift.

Why it might die: Har(4) = 4, so H(6) ≥ H(2)+4 = 8, far below Prohens–Torregrosa 53. The recurrence is weaker than H(n+1) ≥ H(n)+1 when m = 1, and much weaker than the Chebyshev factor m².

Q — Prohens–Torregrosa H_{4,5} field plus one extra Hopf

Claim. The degree-4 Darboux field of their Proposition 6, with first integral H = (2x⁴ − x² + y² − 2x − 2)⁵ / (8x⁵ − 5x³ + 5xy² − 10x² − 5x − 4)⁴, admits a degree-4 perturbation with 29 isolated periodic orbits.

Why it might work: three centers, Lyapunov order 5 already gave ⟨8,12,8⟩ = 28; one more quantity at the middle focus would be 29.

Why it might die: they used the order-5 budget and stopped at 28. Reconstructing the unperturbed field is checkable; the 29th cycle is not a finite identity we can write tonight.

R — Lu’s H₁₄³ hemicycle has cyclicity ≥ 5

Claim. The labelled H₁₄³ semihyperbolic hemicycle (arXiv:2607.13785) produces five isolated cycles, a new quadratic configuration beyond (3,1).

Why it might work: Lu proves only an existential uniform bound, and says it is not sharp.

Why it might die: no explicit field; Coppel still forbids a surrounding cycle of both foci; Zhang still forbids (2,2). Residue, not a lower bound.

S — two hemicycles give H(2) ≥ 6

Claim. The two hemicycles of a Q₃^R quadratic (Marín–Villadelprat, arXiv:2501.16924) each produce three cycles, hence H(2) ≥ 6.

Why it might work: they study simultaneous cyclicity and even an alien cycle.

Why it might die: same Coppel / Zhang wall as A and N. No bypass.

T — Ilyashenko–Llibre (σ,κ,δ) bound is a useful finite H(2) upper

Claim. Evaluated at concrete (σ,κ,δ) for a neighbourhood of unperturbed Shi, their restricted upper bound (arXiv:0910.3443) is a single-digit number, hence a certified finite H(2) for that neighbourhood that is new as an explicit constant.

Why it might work: they give a formula in (σ,κ,δ).

Why it might die: the bound is huge and excludes cycles near equilibria and infinity — exactly where Shi’s small cycles live. Not a bound on H(2), and not sharp enough to be a family dent beyond Llibre–Schlomiuk’s “at most four” neighbourhood.

U — cubic Kolmogorov with seven cycles

Claim. An explicit cubic Kolmogorov field ẋ = x p, ẏ = y q with deg(p,q) ≤ 2 has seven isolated periodic orbits in the open first quadrant, beating M_K(3) ≥ 6 (arXiv:2304.05111).

Why it might work: 6 is local cyclicity; a large cycle around the two axes-and-infinity graphic would be a seventh.

Why it might die: no explicit field on the page; smooth quadratic Kolmogorov have zero cycles (Cruz–Oliveira–Torregrosa arXiv:2509.06198). q1 already certified zero for classical Lotka–Volterra. Dulac may kill the extra nest.

V — algebraic conic plus three Hopf, independently of Shi

Claim. A quadratic with an algebraic limit cycle of degree 2 (Chin Yuan-shun, cited by Coppel) plus three small Hopf cycles is a second explicit (3,1) field.

Why it might work: algebraic cycles are checkable as factors of an inverse integrating factor.

Why it might die: this is Chen–Wang / Shi replay, not a dent of H(2) ≥ 4. Finding the conic is a literature reconstruction, not a new number.

W — Binyamini–Novikov–Yakovenko double-exp is a useful H(2) upper

Claim. Their double-exponential bound on Abelian-integral zeros (arXiv:0808.2952), specialised at degree 2, is a finite upper bound for H(2).

Why it might work: the paper solves the tangential Hilbert 16th.

Why it might die: that is Z(m,n), perturbations of Hamiltonians, not H(2). The constant is not a table number.

X — figure-eight cubic Hamiltonian with 14 Abelian zeros

Claim. An explicit cubic Hamiltonian with a figure-eight level, perturbed by cubics, has 14 isolated zeros of I(h), hence H(3) ≥ 14.

Why it might work: detection-function papers reached 11 then 13 this way.

Why it might die: Li–Liu–Yang is still paywalled; Picard–Fuchs dimension and Harnack cap the oval count of one invariant cubic at two. Same death as B unless their Hamiltonian is written down, which it is not.

Y — Yu–Han Z₂ cubic 12 plus two large cycles

Claim. The Z₂-equivariant cubic with 12 small cycles, plus a stability change at infinity and one surrounding cycle, is an explicit field with 14 cycles.

Why it might work: Li–Liu obtained 13 by exactly this extra infinity step.

Why it might die: recycled B. No explicit (P,Q) tonight, and Harnack still caps one invariant cubic.

Z — Christopher–Lloyd 4-fold of van der Pol beats H(7)

Claim. The classical 4-fold composition applied to van der Pol (exactly one cycle, degree 3) produces a degree-7 field with more than 74 cycles.

Why it might work: 4-fold is H(2n+1) ≥ 4 H(n); iterate on a counted seed.

Why it might die: one iteration gives H(7) ≥ 4, far below Prohens–Torregrosa 74. The identity is already in the 1995 paper and is weaker than Chebyshev at the same degree.

AA — piecewise quadratic with 17 cycles, folded to a polynomial

Claim. A piecewise quadratic with 17 crossing cycles (beating H_p(2) ≥ 16) is the restriction of a single polynomial field.

Why it might work: a quadratic fold can glue two half-planes.

Why it might die: the fold is not polynomial on the whole plane, or it raises the degree and leaves the piecewise function. User rule: piecewise only if it maps to a polynomial field. It does not.

BB — Llibre–Schlomiuk 18th QW3 portrait has a fifth cycle

Claim. One of the 18 QW3 phase portraits already contains five isolated periodic orbits.

Why it might work: 18 portraits is a large census.

Why it might die: they prove a neighbourhood of the no-cycle no-polycycle QW3 systems has at most four cycles, and the unperturbed order-3 weak focus has none. Same Bautin wall as A.

CC — Shi equator is a fifth planar cycle

Claim. Poincaré compactification of unperturbed Shi has a periodic orbit at infinity that counts as a fifth limit cycle of the planar field.

Why it might work: infinity cyclicity is a live 2026 topic (arXiv:2608.17773).

Why it might die: a cycle of the compactification that lives on the equator is not a periodic orbit in R². H(n) counts planar cycles. Coppel already forbids a planar cycle around both foci.

DD — Gasull–Santana +1 applied to a reconstructed PT 28-field

Claim. There is an explicit degree-5 field with 29 cycles: take the PT degree-4 28-cycle perturbation and apply H(n+1) ≥ H(n)+1 (arXiv:2407.13465).

Why it might work: the +1 theorem is elementary (add a simple factor).

Why it might die: the 28-cycle field is not written term-by-term (only a Lyapunov budget on a Darboux center). H(5) ≥ 29 would still lose to Prohens–Torregrosa H(5) ≥ 37. The +1 map does not beat the next published seed.

EE — inverse integrating factor of degree 1, cubic family, exact upper 2

Claim. A documented cubic family with a linear inverse integrating factor has at most two limit cycles, and the literature lists that upper as open.

Why it might work: Dulac with B linear is an identity.

Why it might die: the families that have a linear inverse integrating factor are already classified (Darboux / Lotka– Volterra type) and q1 already counted them as zero or one. Not an open upper.

Ranking

Score = (published-record delta) × (checkability tonight). Delta 3 = would move a printed H(n) or a documented open family upper. Delta 2 = would move a secondary Hilbert number (Liénard, Kolmogorov) or answer an open remark. Delta 1 = replay / family lemma, no table movement. Checkability 3 = explicit field or polynomial identity. Checkability 2 = arithmetic on a published formula. Checkability 1 = heavy analysis, no field tonight.

Picked five, one worker each:

1. o-iterated-squaring/ — imagined O, fork: iterated degree-2 pullback is still quadratic in the final degree (Remark 4).
2. i-lienard/ — imagined I (and the checkable part of H): replay B(n); uniqueness for ẋ = y − (αx+βx³), ẏ = −x.
3. p-harnack-recurrence/ — imagined P: enumerate H(n)+Har(m) against the published table.
4. q-pt-darboux/ — imagined Q: reconstruct the Proposition 6 field; Coppel Theorem 2 as a quadratic contact identity.
5. f-homogeneous/ — imagined F (and J’s unperturbed half): homogeneous and quasi-homogeneous unperturbed fields have zero isolated periodic orbits.

L, M, N, G, S, CC, BB, X, Y are discarded on the same theorems that killed A/B, with no new bypass.