Predictive quotients and delayed information

Latent receipts in binary reverse-and-add

Two binary integers can have the same length, differ in one bit, and produce exactly the same length-growth observations for much longer than their input words—before that hidden bit finally changes one observation. The proof reveals why: one carry seam converts the exhausted hiding region into a second hiding region.

QA-RA-S6-D15 · definition

Observe only the growth heartbeat

For binary reverse-and-add, put

$T(n)=n+\operatorname{rev}_2(n),\qquad g(n)=\ell_2(T(n))-\ell_2(n)\in\{0,1\}.$

The infinite word $\gamma(n)=g(n)g(Tn)g(T^2n)\cdots$ records only whether each update adds a binary digit. Define the first-divergence latency by

$\tau_g(x,y)=\min\{t\ge0:g(T^t x)\ne g(T^t y)\}.$

A two-reservoir receipt amplifier is a family in which a hidden distinction crosses one observationally null transport region, a seam creates a new presentation with a fresh null region, and only the second seam exposes the distinction. The definition tracks a change of presentation—not merely the addition of two waiting times.

QA-RA-S6-T15 · exact theorem

One hidden bit remains dormant for exactly $4L/3-12$ steps

For every $q\ge4$ , define

$a_q=2^{9q-1}+154,\qquad b_q=2^{9q-1}+1178.$

Both words have length $L=9q$ . Since $1178-154=2^{10}$ and bit 10 is absent from 154, their Hamming distance is one. Put $p=011101111$ .

Their exact responses satisfy

$\gamma(a_q)=p(10)^\infty, \qquad \gamma(b_q)_{[0,\,12q-11)}=p(10)^{6q-11}11.$

Consequently $\tau_g(a_q,b_q)=12q-12=\frac43L-12.$

The prefix notation is important: the theorem characterizes the displayed initial segment of $\gamma(b_q)$ , not its unprinted tail. The two responses agree at indices $0,\ldots,12q-13$ . At index $12q-12$ , the $a_q$ branch outputs 0 and the $b_q$ branch outputs 1.

Proof mechanism

The first seam manufactures the second reservoir

The common nine-step setup outputs $p$ . After it, each orbit has two directed flanks of width 20 separated by a zero gap

$G_1=9q-33.$

The exact flank recurrence outputs alternating $10$ pairs. A growth step consumes one gap position and a nongrowth step consumes two, so $6q-22$ alternating steps consume $(3q-11)+2(3q-11)=G_1$ exactly. The recurrence is stopped at the seam; it is never used through overlapping flanks.

The two branches then behave differently. The $a_q$ branch enters the word

$U_r=(10\,1^r01\,0^{r+1})_2, \qquad T^4(U_r)=U_{r+1}, \qquad \operatorname{out}_4(U_r)=0101.$

This proves its complete alternating tail. The $b_q$ seam instead creates a new separated-flank state with gap $G_2=9q-17$ for even $q$ and $G_2=9q-16$ for odd $q$ . Thus the first reservoir has disappeared, but the latent bit has acquired another linear region in which it remains observationally null.

The second flank family obeys the exact renormalization

$s_r=(180\,2^r-3,\;192\,2^r+42,\;r+8), \qquad \Phi^6(s_r)=s_{r+6}, \qquad \operatorname{out}_6(s_r)=101010.$

After $6q-11$ second-reservoir steps, the remaining gap is zero or one according to parity. Seven raw additions at the terminal seam output $0101011$ . Its first six bits continue the common alternation; the seventh activates the hidden receipt. Counting the common observations gives

$9+(6q-22)+6+(6q-11)+6=12q-12.$

QA-RA-S6-C15.1 · receipt-growth corollary

Nine spatial bits buy twelve temporal steps

The theorem gives the exact latency relay

$\tau_g(a_{q+1},b_{q+1})=\tau_g(a_q,b_q)+12.$

If $D_L$ is a response horizon sufficient to determine the complete growth future inside every exact $L$ -bit chart, then

$D_{9q}\ge12q-11=\frac43(9q)-11.$

This is a lower bound for the declared observation interface. It is not an algorithmic running-time lower bound for procedures allowed to inspect the complete input word.

QA-RA-S6-N15 · negative boundary

Latency relay is not complete-future bisimulation

Finite replay suggested that a shifted tail of $\gamma(b_q)$ might equal a shifted tail of $\gamma(b_{q-1})$ . The symbolic proof does not establish that coinductive statement, and the latency theorem does not require it. The exact conclusion is only

$\tau_g(a_{q+1},b_{q+1})-\tau_g(a_q,b_q)=12.$

This distinction prevents a finite-prefix theorem from being silently promoted into equality of two complete response states.

Evidence, provenance, and scope

The all-parameter proof represents parameterized binary run words as Laurent polynomials in $X=2^m$ and checks every bridge and renormalization as an exact identity. Both parity inductions close, and the six values $4\le q<10$ supply the finite base cases. A separate little-endian digit-list implementation replays eleven targeted parameters as a transcription check; it is not substituted for the induction.