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This is Plasma AI's author-recorded algebraic comparison for the phase-transport route to Problem 251. It has not been independently reviewed. It identifies the best universal bound obtainable by averaging doubled endpoint phases, and the arithmetic information that would make that bound strict. It proves no estimate for the actual prime sequence and does not resolve the irrationality question.

The exact optimization

Let $\lambda=(\lambda_j)_{j\ge0}$ be a finitely supported real sequence with $\sum_j\lambda_j=1$. Put

$$A_\lambda(t)=\sum_{j\ge0}\lambda_j \bigl(1-\cos(2\pi2^jt)\bigr),\qquad t\in\mathbb R.$$

Then

$$\inf_\lambda\sup_{t\in\mathbb R}A_\lambda(t)=\frac32. \tag{H1}$$

The same equality holds if all coefficients are required to be nonnegative. In both cases an optimizer is $\lambda_0=2/3$, $\lambda_1=1/3$, with all other coefficients zero.

Proof. Since $2^j\equiv\pm1\pmod3$, every summand at $t=1/3$ is $3/2$. Normalization therefore gives $A_\lambda(1/3)=3/2$, including for signed coefficients. This proves the lower bound. For the displayed optimizer, put $x=\cos(2\pi t)$ and use $\cos(4\pi t)=2x^2-1$. Directly,

$$\frac23(1-x)+\frac13\bigl(1-(2x^2-1)\bigr) =\frac32-\frac23\left(x+\frac12\right)^2 \le\frac32. \tag{H2}$$

Equality holds exactly when $t\equiv1/3$ or $2/3\pmod1$. $\square$

For any finite list of phases, summing (H2) gives the corresponding ceiling $3/2$ times the length of the list. A list consisting entirely of nontrivial cube-root phases attains it for every normalized mixture. Thus neither more doubled frequencies nor signed coefficients can give a smaller universal ceiling in this class. This is an obstruction to a bound using arbitrary phases; it is not an obstruction to exploiting additional properties of prime tails.

The prime-tail interface

Use the phase-transport notation

$$H_n=\sum_{i\ge0}\frac{(p_{n+i+1}-p_{n+i})/2}{2^{i+1}},\qquad I_X=\{n:X<p_n\le2X\},\qquad N_X=|I_X|.$$

Fix an odd integer $q\ge3$, let $W$ be the odd positive integers not divisible by $q$, and let $w(m)$ be its half-gap tail from $m\in W$. Fix a positive integer multiplier $b$ and suppose the reference is resonant in the precise sense

$$bw(m)\equiv\frac13\ \text{or}\ \frac23\pmod1 \qquad(m\in W). \tag{H3}$$

This hypothesis holds, for example, for $q=3,b=1$, whose two reference tails are $4/3$ and $5/3$. The statements below retain (H3) as a hypothesis rather than infer it for an arbitrary reference. Proposition P9 of the transport note supplies such a reference with $b$ an odd multiple of each fixed odd denominator $d$: writing $d=3^\nu d_0$ with $3\nmid d_0$, it takes $q=2\cdot3^\nu+1$ and $b=d_0(2^{q-1}-1)/3$. This author-recorded construction verifies (H3), without estimating actual prime tails. Put $C_n=bH_n$ and $t_n=bw(p_n)-C_n$, for $X>q$.

For a fixed nonnegative integer $j$, write $Q_j(X)$ for the dephased transport with multiplier $2^jb$, so its carries and gap arguments are also multiplied by $2^j$. The endpoint proof of Proposition P4 in the transport note gives

$$Q_j(X)=N_X-\sum_{n\in I_X}\cos(2\pi2^jt_n)+r_j(X), \qquad r_j(X)=o(N_X). \tag{H4}$$

The telescoping proof applies to each fixed positive multiplier: it uses the recurrence for the carries and the reference tails, containment of every sufficiently large prime in $W$, and its stated boundary estimates. It uses neither oddness of the multiplier nor primality of $q$ for this unconditional endpoint. The comparison uses that endpoint only; it does not require progression equidistribution or a sharper boundary error. All multipliers here are fixed before $X$ tends to infinity. If the frequencies grow with $X$, their endpoint errors need a separate uniform bound; the fixed-parameter identity supplies none.

Define

$$\overline Q(X)=\frac{2Q_0(X)+Q_1(X)}3,\qquad D(X)=\sum_{n\in I_X} \left(\cos(2\pi t_n)+\frac12\right)^2.$$

By (H2) and (H4),

$$\overline Q(X)=\frac32N_X-\frac23D(X)+o(N_X). \tag{H5}$$

In particular, the unconditional ceiling is $3N_X/2+o(N_X)$. A fixed strict saving below $3N_X/2$ on unbounded scales is equivalent, after adjusting a positive margin for the error, to

$$\limsup_{X\to\infty}\frac{D(X)}{N_X}>0. \tag{H6}$$

This demands a positive proportion of energy, not merely one nonzero term. Equivalently, there must be fixed $\eta,\kappa>0$ and unbounded scales on which at least $\kappa N_X$ indices satisfy $|\cos(2\pi t_n)+1/2|\ge\eta$. Indeed the square is at most $9/4$. If $D(X)\ge cN_X$, the indices with square at least $c/2$ number at least $2cN_X/9$, since the other indices contribute at most $cN_X/2$. The converse follows by summing the squares on the specified indices.

The individual zeros need careful interpretation. If $bw(p_n)\equiv\sigma_n/3\pmod1$, where $\sigma_n\in\{1,-1\}$, then

$$\left(\cos(2\pi t_n)+\frac12\right)^2=0 \quad\Longleftrightarrow\quad C_n\equiv0\ \text{or}\ \frac{2\sigma_n}{3}\pmod1. \tag{H7}$$

Thus a zero summand does not by itself imply an integer carry. Conversely, eventually integer carries make every sufficiently late summand zero. No lower bound in (H6) is established here.

Why doubling does not change the rationality sum

The site weight, with its multiplier displayed, is

$$T_{b,q}(B,w)=\sum_{r\ge1} \left(\cos\frac{2\pi(B-2bw)}{2^r} -\cos\frac{2\pi(B-bw)}{2^r}\right).$$

It is absolutely convergent. If $B\in\mathbb Z$ and $bw\equiv\pm1/3\pmod1$, then for every integer $j\ge0$,

$$T_{2^jb,q}(2^jB,w)=T_{b,q}(B,w). \tag{H8}$$

For $j=1$, the first term on the left is $\cos(2\pi(B-2bw))-\cos(2\pi(B-bw))=0$; both cosines are $-1/2$. Reindexing the remaining absolutely convergent terms gives the right side. Iterating is legitimate because $2^jB$ is integral and multiplication by two preserves the two nonzero residues modulo three. This proves (H8).

At a composite site $m$ in the prime gap $(p_r,p_{r+1})$, the transport uses $B=2C_r-b(m-p_r)/2$. If $C_r$ is integral, then so is $B$. Consequently eventually integer carries give $Q_j(X)=Q_0(X)$ exactly for all sufficiently large $X$, for every fixed $j$. Every normalized finite mixture therefore has the same actual site sum under this hypothesis. Its unconditional description may differ, as (H5) shows, but frequency averaging has supplied no arithmetic saving.

The remaining whole-conclusion step

Suppose $S=\sum_{n\ge1}p_n2^{-n}=a/(2^sd)$ with $d$ odd. The canonical sketch's Lemma A gives eventually integer $dH_n$. If $b$ is a fixed integer multiple of $d$ and the chosen reference satisfies (H3), then $C_n$ is eventually integral. Equations (H3)--(H5) force $Q_0(X)=\overline Q(X)=3N_X/2+o(N_X)$. A strict upper bound for this actual-prime sum on unbounded scales would give a contradiction.

For a complete proof this step must work for each odd $d$, with a suitable fixed resonant reference and multiplier divisible by $d$. Constants and the unbounded scales may depend on $d$, and the upper argument may use the eventual integrality forced by the assumed rational value. This note supplies neither that upper argument nor (H6).

The sharp bound (H1) closes only the proposed purely algebraic improvement by normalized doubled phases. Reopening that approach needs an additional restriction on the actual phases strong enough to force (H6), or a signed arithmetic estimate for the transport retaining its prime-gap-selected coefficient and every cofactor. A bound merely attaining the universal ceiling does not fill the strict-saving gap. These conclusions change no other route in the canonical sketch.