Reciprocal Mersenne Subseries

Possible estimates to prove

A conditional bound along a right-branch run.

Let $r\ge 6$, $L\ge 1$, and suppose that the transitions at rows $r,\dots ,r+L-1$ all take the right branch. Put $w_s=\mathrm{rem}(s)-2^s$ and let $p_s$ be the lower-prefix pulse, so $0\le p_s\le 2(s-2)$. Iterating the exact recurrence $w_{s+1}=4w_s-4-p_s$ gives

The reverse triangle inequality and the finite geometric sum give

This estimate alone does not force a right-branch run to end. For that conclusion one also needs an upper bound on its terminal deviation. More precisely, if $C>0$ and

then division by $4^L$ and taking logarithms give

Every term and hypothesis is explicit here. A separate window estimate must supply the displayed terminal upper bound; it does not follow from the right-branch recurrence alone. This explains the possible connection between the initial deviation and the run length without asserting a sharp threshold, a run-length equivalence, or an unproved bound for all later rows. This conditional calculation is an ordinary argument.

An exact finite phase experiment.

For the $\mathrm{200,000}$ integers ${10}^6\le M<1.2\times {10}^6$, an integer divisor sieve through $\mathrm{1,200,039}$ computes $2^{40}{\mathrm{\Theta }}_{40}(M)$ exactly. Reducing that integer modulo $2^{40}$ gives the following counts and empirical proportions for $\mathrm{dist}({\mathrm{\Theta }}_{40}(M),\mathbb{Z})<t$.

The experiment samples consecutive integers, not reset rows. Its median distance is $0.24470815549\dots$, compared with $1/4$ for a uniform fractional part. A uniform fractional part has probability $2t$ of distance less than $t$, for $0<t\le 1/2$, which explains the comparison in the last row. Agreement of a finite histogram with that model is not a distribution theorem.

Truncation matters at the smallest tested thresholds. With the same starting indices, the counts below $2^{-14}$ are $22,36,37,37$ for $L=16,24,32,40$, respectively. Thus similar central quantiles do not justify ignoring the truncation error in a rare-event test. More generally, for $L^{\prime }\ge L\ge 1$ and $B\ge 0$, if $\tau (M+i)-1\le B$ over $L<i\le L^{\prime }$, then

Distance to the nearest integer is $1$-Lipschitz. A threshold comparison is therefore guaranteed to remain unchanged when the distance at the shorter cutoff is farther than this error from the threshold. This is a sufficient stability test, not a necessary one. The estimate uses a bound over the specified finite range, not an unproved uniform bound on the divisor function.

A probabilistic application requires further assumptions. For example, if events $E_r$ in a specified probability space satisfied $\mathbb{P}(E_r)\le C{2}^{(3-r)/2}$ for every $r\ge R$, the union bound would give

Summability would also imply only finitely many failures almost surely, not zero failures at every index. Neither the finite histogram nor its fit to a uniform-phase model proves these probability bounds, identifies their events with reset failures, or defines a probability for the actual deterministic orbit. In particular, no membership conclusion is drawn from small expected counts computed in such a model. The counts are exact finite computations; the error and probability inequalities are ordinary arguments; an application to resets is unproved.

The reset signs and the exact branch identities. Write $p_s$ for the nonnegative lower-prefix pulse and $w_s=\mathrm{rem}(s)-2^s$. On a middle transition, where the upper carry does not occur, the update is

The natural-number subtraction can be read as signed subtraction because $p_s\le 2(s-2)<2^{s+1}$ for $s\ge 5$. The middle-branch condition is $4\mathrm{rem}(s)<2^{s+1}+4+p_s$, together with failure of the upper carry. Thus a positive middle-reset deviation is exactly the additional inequality $4\mathrm{rem}(s)>p_s$. The pulse bound and branch inequality alone do not prove it: $s=5$, $\mathrm{rem}=1$, $p=5$ satisfy both numerical bounds but give $4\mathrm{rem}-p=-1$. This is a counterexample to those numerical premises, not a claim that the concrete orbit reaches that state.

On an upper transition, let $o_s>0$ be the overshoot and $p_s^{+}\ge 0$ the upper-prefix pulse. The carry condition pays the subtraction, so

This sign already follows from the exact upper-branch identity; only the middle sign needs further orbit information. On a right transition the corresponding formula is $w_{s+1}=4w_s-4-p_s$. These identities separate the sign calculation from the unproved magnitude estimate needed for square-root escape. At the reported middle step $s=14$, for example, $\mathrm{rem}(14)=392$, $p_{14}=3$ and $\mathrm{rem}(15)=34333$, giving $w_{15}=1565$.