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Exact theorem · latent quotient failure

QA-RA-S6-T15

Exact two-reservoir latency theorem

Exact statement

For every q≥4q\ge4, the equal-length Hamming-adjacent integers aq=29q−1+154a_q=2^{9q-1}+154 and bq=29q−1+1178b_q=2^{9q-1}+1178 have identical binary digit-growth observations for exactly 12q−12=4L/3−1212q-12=4L/3-12 steps, where L=9qL=9q, and differ at the next observation.

StatusProved by an all-parameter symbolic run induction and hostilely audited
External reviewNo documented external or specialist review of this FPRD result is recorded.

Context

The pair differs in one low binary digit. That distinction is dormant through two carry-separation regions even when exact source length is retained.

Hypotheses and scope

  • Base two, canonical finite binary words, and the digit-length-growth observation g are used.
  • q is an integer at least 4.

Proof or evidence

The proof establishes γ(aq)=011101111(10)∞\gamma(a_q)=011101111(10)^\infty and γ(bq)[0,12q−11)=011101111(10)6q−1111\gamma(b_q)_{[0,12q-11)}=011101111(10)^{6q-11}11. Affine run templates close both parity inductions as exact Laurent-polynomial identities; finite replay supplies only base and transcription checks.

Theorem, mechanism, and proof →

Verification notes

The recurrence is stopped before flank overlap, even and odd terminal seams are checked separately, the complete a-branch tail is proved by T^4(U_r)=U_{r+1}, and the first differing index was checked for off-by-one errors.

Limitations

  • The theorem concerns one coarse binary observation, not raw-input computational complexity.
  • It says nothing about eventual palindromes or decimal reverse-and-add.

Open work

Compare the serial-reservoir mechanism with the Stage 2 obstruction-and-receipt calculus without promoting the unproved complete-future shift.