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Parent: #650. Follow-up to PR #669 / #659. No new census. No L=6.
Why
#669 extracted exact integer identities from the committed 4×4 tables and left most of them empirical. Upgrade the cheap ones to combinatorial proofs, and isolate the two-point diamond conjectures.
The reader scripts/wrapping_five_cell_reader.py on PR #669 re-verifies every identity as integers. Use that as the check, not as the proof.
I2 (high-k binomial).b×n(k) = C(N,k) for k ≥ N−L+1 on both geometries.
Axis I3.d0(L) = L (the L full rows) and b×n(2L−1) = L² (one full row plus one full column).
Axis I4. Pre-both-wrap deficit C(N,k) − n×b = d0 + d1 for L ≤ k < 2L−1.
Diamond I3.d0(2L) = 4·C(2L,4), b×b(2L) = 2L, b×n onset 3L−1 with value 4L². These rest on two sizes. Give a geometric reading or explicitly leave them as two-point conjectures. Do not enumerate diamond L=5 (2^50 is not a ticket).
Also record: M_L irreducible over ℚ at the five committed sizes is a factorization fact, not a wrapping theorem, unless you find a reason it must be. Do not spend the ticket on Galois theory.
Tripwire
A "proof" that uses an unstated fact from one L is not a proof. If you cannot prove an item, write OPEN and point at the #669 falsification kit. Do not enumerate the next size.
No Huawei. No STATUS. Do not close #659 / #640. Draft PR against main.
Parent: #650. Follow-up to PR #669 / #659. No new census. No L=6.
Why
#669 extracted exact integer identities from the committed 4×4 tables and left most of them empirical. Upgrade the cheap ones to combinatorial proofs, and isolate the two-point diamond conjectures.
The reader
scripts/wrapping_five_cell_reader.pyon PR #669 re-verifies every identity as integers. Use that as the check, not as the proof.Prove or kill (in this order)
n×b(k) = C(N,k)fork < L. Diamond: fork < 2L. The blocking-line reading is already in the #659 [P2 exact] Integer structure of the five wrapping-type cells #669 note; write it as a proof that does not mention the tables except as a check.b×n(k) = C(N,k)fork ≥ N−L+1on both geometries.d0(L) = L(the L full rows) andb×n(2L−1) = L²(one full row plus one full column).C(N,k) − n×b = d0 + d1forL ≤ k < 2L−1.d0(2L) = 4·C(2L,4),b×b(2L) = 2L,b×nonset3L−1with value4L². These rest on two sizes. Give a geometric reading or explicitly leave them as two-point conjectures. Do not enumerate diamond L=5 (2^50is not a ticket).Also record:
M_Lirreducible over ℚ at the five committed sizes is a factorization fact, not a wrapping theorem, unless you find a reason it must be. Do not spend the ticket on Galois theory.Tripwire
A "proof" that uses an unstated fact from one L is not a proof. If you cannot prove an item, write
OPENand point at the #669 falsification kit. Do not enumerate the next size.No Huawei. No STATUS. Do not close #659 / #640. Draft PR against main.
Wait for
ASSIGNED_MACHINE.Related: #659, #640, #651, PRs #669, #653, #657.