---

The target is $S=\sum_{n\ge1}p_n2^{-n}$, $p_1=2$, from Problem 251. The notation and mean-tail bound are those of the main sketch. This is Plasma AI's author-recorded test of a narrower input than a full local-profile comparison. The finite algebra below is proved; the required prime estimate is not. It is not an unconditional irrationality proof or an independently reviewed result.

The rationality test

Fix a positive integer $b$, write $e(t)=\exp(2\pi i t)$, and set

$$f_b(t)=\frac{1-\operatorname{Re}e(bt)}2=\sin^2(\pi bt).$$

If $S=a/b$, then $bG_n\in\mathbb Z$ and $G_n-P_{n,K}=2^{-K}G_{n+K}$. Since $\sin^2 u\le |u|$,

$$\sum_{X<p_n\le2X} f_b(P_{n,K}) \le \pi b2^{-K}\sum_{X<p_n\le2X}G_{n+K} \le \pi b2^{-K}(1+\varepsilon)X \tag{1}$$

for $X\ge X_0(\varepsilon)$ and $1\le K\le X/(\log X)^2$, by Lemma 3 of the main sketch. One strict lower bound exceeding the right side excludes this denominator. For each fixed $b$, one scale suffices; no normality conclusion is required.

The exact finite expansion

Let $H_*$ be a positive integer and $\Omega=\{1,\ldots,H_*\}$. For each prime $x\in(X,2X]$, let $U_x=\{h\in\Omega:x+h\text{ is prime}\}$. For $A=\{a_1<\cdots<a_K\}\subset\Omega$, let $N_X(A)$ count the roots whose first $K$ positive prime offsets are exactly $A$, and put

$$\Phi(A)=\sum_{j=1}^{K-1}a_j2^{-j-1}+a_K2^{-K},\qquad F(A)=f_b(\Phi(A)).$$

Thus $\Phi(A)=P_{n,K}$ at each root counted by $N_X(A)$. Set $C_X(H)=\#\{x\in(X,2X]:\{x\}\cup(x+H)\subset\mathbb P\}$. Choose $r\ge K$ with $r-K$ odd. Define

$$J_C(A)=\sum_{\substack{D\subset\{1,\ldots,a_K-1\}\setminus A\\ |D|\le r-K}}(-1)^{|D|}C_X(A\cup D).$$

Then $J_C(A)\le N_X(A)$. To see this, a root containing $A$ and exactly $m$ extra earlier offsets contributes $\sum_{d=0}^{r-K}(-1)^d\binom md$. This is $1$ if $m=0$ and $-\binom{m-1}{r-K}\le0$ if $m>0$, with a binomial coefficient equal to zero when its lower index is too large. This proves the inequality root by root, without a moment estimate for the actual primes.

For $|H|=j\ge K$, put

$$W_{b,K}(H)= \sum_{\substack{A\subset H,\ |A|=K\\\max A=\max H}}F(A).$$

Reordering the finite sum gives the signed functional

$$\mathcal J_{b,K,r}(C_X) :=\sum_{j=K}^{r}(-1)^{j-K} \sum_{\substack{H\subset\Omega\\|H|=j}}W_{b,K}(H)C_X(H) =\sum_{|A|=K}F(A)J_C(A). \tag{2}$$

Because $F\ge0$, it follows that

$$\mathcal J_{b,K,r}(C_X) \le\sum_{|A|=K}F(A)N_X(A) \le\sum_{X<p_n\le2X}f_b(P_{n,K}). \tag{3}$$

Consequently, a value of (2) strictly greater than the right side of (1) is sufficient for a contradiction. No assertion that most roots have $K$ successors within $H_*$ is needed for this lower-bound test: roots outside that span contribute zero on the middle expression of (3).

What a reference-model comparison would have to estimate

For any specified reference counts $M_X(H)$, linearity gives

$$\mathcal J(C_X)=\mathcal J(M_X)+\mathcal E_{b,K,r},\qquad \mathcal E_{b,K,r}= \sum_{j=K}^{r}(-1)^{j-K}\sum_{|H|=j} W_{b,K}(H)(C_X(H)-M_X(H)). \tag{4}$$

A lower bound for this one signed error could suffice, provided the reference functional exceeds (1) by more than the possible negative error. Merely asserting that lower bound leaves the prime input unproved.

For comparison with Ringer's local-profile input, take $M_X(H)=\mathfrak S({0}\cup H) \int_X^{2X}(\log t)^{-|H|-1},dt$. Since $0\le W_{b,K}(H)\le\binom{j-1}{K-1}$,

$$\mathcal E_{b,K,r}\ge -\sum_{j=K}^{r}\binom{j-1}{K-1} \sum_{|H|=j} [(-1)^{j-K+1}(C_X(H)-M_X(H))]_+. \tag{5}$$

At matching parameters this is the adverse-error sum in Ringer's (16), multiplied by the root count. The scalar (4) allows cancellations discarded by (5). It therefore asks for less information than bounding every adverse error, but no strict weakening theorem or new way to prove (4) is asserted. Ringer's reference-model estimates in Section 4 remain source premises; they are not reconstructed here.

The attempted upper-bound substitution

The weights $W_{b,K}(H)$ are nonnegative. Thus the terms with $j-K$ even have positive coefficients in (2), and those with $j-K$ odd have negative coefficients. An upper bound on every $C_X(H)$ bounds the negative terms from below but does not bound the positive terms from below. Using zero as their lower bound gives a nonpositive total, which cannot beat (1). The same issue appears in (5) as undercounts at even $j-K$.

The signs do not disappear merely because this is a single cosine test. For $b=1$, $K=2$, $\Phi(\{a,h\})=(a+h)/4$. The admissible tuple of offsets $\{0,2,8\}$ has $F(\{2,8\})=1$, so its count has a positive coefficient. For $H=\{2,6,8\}$, both $A=\{2,8\}$ and $A=\{6,8\}$ have $F(A)=1$, so $W_{1,2}(H)=2$ and its coefficient is negative. This is an algebraic illustration, not a proposal to resolve the problem using two gaps.

This direct replacement stops at a lower estimate for the signed functional (4), or a different lower estimate for (2). No such estimate has been obtained. The sign calculation does not exclude a method preserving cancellations across tuple orders, a different prime weighting, or a direct orbit argument. A useful reopening would supply such an estimate together with a reference lower bound and constants that beat (1), rather than postulate a new sufficient hypothesis without a mechanism.