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This elementary conditional, recorded by Plasma AI, gives an alternative precise missing input for Problem 251. It uses the lattice also present in the parity-prediction sketch, without supplying an estimate for primes.

The independent review of 2026-10-16, accepted by the separate grade, found the conditional identities, lattice, matching-tail conclusion, modular-order consequence and qualified fixed-denominator criterion refutation-failed. This records the grader's accepted judgment at its exact conditional scope, not independent acceptance of the required prime-pair existence or of a solution to Problem 251. For the fixed-denominator criterion the smaller starting index must be unbounded; the separate dyadic-denominator case remains open. The word “Equivalently” below denotes contraposition of the order implication under all its inherited hypotheses, not equivalence to the individual-tail sufficient bounds.

The final warning's assertion of a necessarily nonzero finite-window contribution was refuted-as-stated. In the examined argument, the frozen subject has only the grade's authorized body correction from “nonzero” to “potentially nonzero” in that warning; its mathematical formulas, hypotheses and proof are unchanged. The review and grade retain the original assertion and its counterexamples. The paragraph comparing the criterion with the step lemma and the sections "Graded congruences", "Occurrence of equal windows" and "The preceding gap and the reversed constraint" are unreviewed additions. The review does not extend to those additions, other arguments in the parity-prediction sketch or any result behind the contextual obstruction link.

Use the same indexing:

$$S=\sum_{n\ge1}\frac{p_n}{2^n},\qquad g_n=p_{n+1}-p_n,\qquad G_n=\sum_{j\ge0}\frac{g_{n+j}}{2^{j+1}}.$$

The convergent sums satisfy $S=2+G_1$ and $G_{n+1}=2G_n-g_n$. Suppose $S=a/(2^sd)$ in lowest terms, where $s\ge0$ and $d\ge1$ is odd. Then $dG_n$ is an integer for $n\ge s+1$. Since $g_n$ is even for $n\ge2$, the recurrence also gives

$$G_n\in\frac2d\mathbb Z \qquad\bigl(n\ge\max(s+2,3)\bigr). \tag{1}$$

Indeed $dG_{n+1}=2dG_n-dg_n$ is even once $dG_n$ is integral and $n\ge2$.

A repeated-window constraint

Let $m>n\ge\max(s+2,3)$ and $K\ge1$. Suppose

$$g_{n+j}=g_{m+j}\qquad(0\le j<K). \tag{2}$$

If

$$|G_{n+K}-G_{m+K}|<\frac{2^{K+1}}d, \tag{3}$$

then $G_n=G_m$ and $G_{n+K}=G_{m+K}$. When $d>1$, this also implies

$$\operatorname{ord}_d(2)\mid(m-n). \tag{4}$$

To prove this, iterate the tail identity and cancel the equal windows:

$$G_n-G_m=2^{-K}(G_{n+K}-G_{m+K}).$$

By (1) the left side lies in $(2/d)\mathbb Z$, while (3) makes its absolute value less than $2/d$. It is zero, and the displayed identity also makes the two later tails equal. Iterating $G_{n+1}=2G_n-g_n$ from $G_1=S-2$ gives

$$\{G_n\}=\{2^{n-1}S\} =\left\{\frac{2^{n-1-s}a}{d}\right\}.$$

Equality of $G_n,G_m$, together with $\gcd(a,d)=1$ and the oddness of $d$, therefore gives $2^{m-n}\equiv1\pmod d$, proving (4).

All tails are positive, so a sufficient condition for (3) is that both $G_{n+K}$ and $G_{m+K}$ are less than $2^{K+1}/d$. Equivalently, if (2) holds but (4) fails, rationality forces a separation

$$|G_{n+K}-G_{m+K}|\ge\frac{2^{K+1}}d.$$

What would finish this route

For a fixed odd $d>1$, it would suffice to produce such pairs at arbitrarily large indices with equal gap windows, small later tails, and $\operatorname{ord}_d(2)\nmid(m-n)$. Arbitrarily large indices are needed because the binary preperiod $s$ is unknown. This would exclude every rational value whose reduced denominator has odd part $d$. A separate argument is needed for $d=1$, where the congruence gives no obstruction. The step lemma of the sketch makes every odd part $d$ a dyadic case for the gaps $p_{n+t}-p_n$ with $t=\operatorname{ord}_d(2)$; the congruence (4) is information about the original gap sequence that this dyadic form does not use.

No theorem supplying these pairs is proved here. Repetition of short gap patterns alone does not bound the later tails at the needed scale, and a count of repeated windows alone does not put them in different index classes modulo $\operatorname{ord}_d(2)$. The critical-length obstructions still apply: exact cancellation of the window leaves a difference of two infinite tails, and an upper bound for it must be justified. Congruences of the gaps are weaker than (2); replacing equality by congruence introduces a potentially nonzero finite-window contribution that this argument does not control.

Graded congruences

The final warning concerns an unrestricted substitution of congruences for equality. A specified sequence of moduli does control the finite-window term. The following extension is author-recorded, outside the independent review above; it supplies no occurrence theorem for prime windows.

Keep the reduced rational representation and $m>n\ge\max(s+2,3)$, and put $h_i=g_i/2$, $H_i=G_i/2$ and $Q_{i,K}=\sum_{j<K}h_{i+j}2^{-j-1}$. Suppose

$$h_{n+j}\equiv h_{m+j}\pmod{2^{j+1}} \qquad(0\le j<K). \tag{5}$$

Then $I=Q_{n,K}-Q_{m,K}$ is an integer. If $|H_{n+K}-H_{m+K}|<2^K/d$, then

$$H_{n+K}=H_{m+K},\qquad H_n-H_m=I. \tag{6}$$

For $d>1$ this again implies $\operatorname{ord}_d(2)\mid(m-n)$.

Indeed truncation gives

$$H_n-H_m-I=2^{-K}(H_{n+K}-H_{m+K}).$$

The left side lies in $(1/d)\mathbb Z$ by the odd-lattice lemma and the integrality of $I$. The strict bound makes its absolute value less than $1/d$, so it vanishes. Thus (6) holds, and $G_n-G_m=2I\in\mathbb Z$. The fractional-part calculation proving (4) gives the order consequence. Unlike exact matching, (5) need not make the initial tails equal: the integer $I$ need not be zero.

In the original $G$ units the sufficient tail-difference bound is still $2^{K+1}/d$. This is twice the threshold obtained in O5 from only $d(P_{n,K}-P_{m,K})\in\mathbb Z$, where $P=2Q$. The graded condition makes $P_{n,K}-P_{m,K}$ an even integer, which permits use of the finer lattice $(2/d)\mathbb Z$. Equal gap windows are a special case. A single common modulus $2^{K+1}$ for every original gap also implies (5), but asks for stronger congruences than its separate moduli $2^{j+2}$ for $g_{n+j}$. For this index-class route, the missing input is arbitrarily late pairs with these congruences, small following tails and incompatible index classes. The preceding-gap constraint below gives a separate sufficient test that also covers $d=1$.

Occurrence of equal windows

This section is an author-recorded addition, outside the independent review above. It shows that the occurrence part of the sufficient condition is unconditional, so that the index classes are the whole missing input for a fixed odd $d>1$.

Fix an odd $d$ and $A>1$, put $K=\lceil A\log_2\log X\rceil$, and use the notation $I_X=\{n:X<p_n\le2X\}$, $N_X=|I_X|$ of the parity-prediction sketch. Call $n\in I_X$ good when $G_{n+K}<2^K/d$ and $p_{n+K}-p_n\le4KX/N_X$.

Lemma (repeated windows). For large $X$, at least $(3/4-o(1))N_X$ starts in $I_X$ are good, and all but at most $(6\log X)^K=X^{o(1)}$ good starts share their window $(g_n,\ldots,g_{n+K-1})$ with another good start.

Proof. The sketch's mean-tail bound (T), written for $G=2H$, gives $\sum_{n\in I_X}G_{n+K}\le(1+\varepsilon)X$, so at most $(1+\varepsilon)dX/2^K=o(N_X)$ starts have $G_{n+K}\ge2^K/d$, because $2^K\ge(\log X)^A$. Next, $\sum_{n\in I_X}(p_{n+K}-p_n)=\sum_{j<K}\sum_{n\in I_X}g_{n+j}$ is at most $K$ times the length of an interval from $X$ to $2X+o(X)$, so at most $(1/4+o(1))N_X$ starts have span above $4KX/N_X$. A good window consists of $K$ even gaps whose halves are positive integers with sum at most $L=2KX/N_X=(2+o(1))K\log X$. There are at most $\binom{\lfloor L\rfloor}{K}\le(eL/K)^K\le(6\log X)^K$ such windows, and $K\log(6\log X)=O((\log\log X)^2)$. A window occurring at only one good start accounts for one start, which gives the bound. $\square$

For a pair of good starts $m>n$ with equal windows, both later tails lie in $(0,2^K/d)$, so (3) holds. Under $S=a/(2^sd)$ with $n\ge\max(s+2,3)$, the constraint above gives $G_n=G_m$, and for $d>1$ also $\operatorname{ord}_d(2)\mid(m-n)$. Thus a rational value with odd part $d>1$ makes the index class of $n$ modulo $\operatorname{ord}_d(2)$ constant on each window class of good starts, which is all but $X^{o(1)}$ of at least $(3/4-o(1))N_X$ starts. Its refutation needs only one late pair of good starts with equal windows in different index classes. No theorem inspected for this route relates a window of consecutive prime gaps to the index of its first prime modulo a fixed integer, so this remains unknown. For $d=1$ the index congruence gives no constraint; the preceding-gap congruence of the next section gives one for every odd $d$.

The preceding gap and the reversed constraint

This section is an author-recorded addition, outside the independent review above. It replaces the index-class condition, and the separate dyadic case, by a residue of the gap just before the windows. It supplies no occurrence theorem for that residue.

Keep the reduced representation $S=a/(2^sd)$, put $n_0=\max(s+2,3)$, and let $c_i=dH_i$, a positive integer for $i\ge n_0$ by the odd-lattice lemma. For $r\ge1$ write

$$\Psi_r(i)=\sum_{k=1}^{r}2^{k-1}h_{i-k}.$$

Iterating $c_{i+1}=2c_i-dh_i$ gives $c_i=2^rc_{i-r}-d\Psi_r(i)$, hence

$$c_i\equiv-d\,\Psi_r(i)\pmod{2^r}\qquad(i-r\ge n_0). \tag{7}$$

For $r=1$ this says $c_i\equiv h_{i-1}\pmod2$: the parity of the integer carry at $i$ records the class of the preceding gap $g_{i-1}$ modulo $4$.

Proposition (forward windows fix the past). Let $m>n>n_0$, assume the graded congruences (5) and $|H_{n+K}-H_{m+K}|<2^K/d$, so that (6) gives $H_n-H_m=I$. Then for every $r$ with $1\le r\le n-n_0$,

$$\Psi_r(m)-\Psi_r(n)\equiv I\pmod{2^r}. \tag{8}$$

In particular $g_{n-1}-g_{m-1}\equiv2I\pmod4$. For equal windows (2) one has $I=0$, so $g_{n-1}\equiv g_{m-1}\pmod4$. This holds for every odd $d$, including $d=1$, and uses no index class.

Proof. By (6), $c_n-c_m=dI$. Subtracting the two instances of (7) gives $d(\Psi_r(m)-\Psi_r(n))\equiv dI\pmod{2^r}$, and $d$ is invertible modulo $2^r$. At $r=1$ this is $h_{m-1}-h_{n-1}\equiv I\pmod2$. $\square$

With the occurrence lemma, a rational value with any odd part $d$ makes the class of $g_{n-1}$ modulo $4$ constant on each window class of good starts $n>n_0$. One pair of good starts with equal windows and preceding gaps in different classes modulo $4$ refutes every rational value with odd part $d$ whose preperiod satisfies $\max(s+2,3)<n$. More generally a pair satisfying (5) and the later-tail bound refutes it when (8) fails for some $r\le n-n_0$. A refutation of every rational value needs such pairs at unbounded indices for each odd $d$; the constants may depend on $d$.

The same congruence (7) also runs in the other direction.

Proposition (past residues fix the tail). Let $r\ge1$ and $n\ne m$ with $\min(n,m)-r\ge n_0$. Suppose

$$h_{n-k}\equiv h_{m-k}\pmod{2^{r-k+1}}\qquad(1\le k\le r) \tag{9}$$

and $\max(H_n,H_m)<2^r/d$. Then $H_n=H_m$, and consequently $|h_n-h_m|<\max(H_{n+1},H_{m+1})-1$.

Proof. Each term $2^{k-1}(h_{n-k}-h_{m-k})$ is divisible by $2^r$, so $\Psi_r(n)\equiv\Psi_r(m)\pmod{2^r}$, and (7) gives $c_n\equiv c_m\pmod{2^r}$. Both carries lie in $(0,2^r)$, so they are equal. From $H_i=h_i/2+H_{i+1}/2$ one gets $h_n-h_m=H_{m+1}-H_{n+1}$. Every tail exceeds $1$, because every half-gap is at least $1$ and not all of them equal $1$. $\square$

So a rational value makes $H$ constant on each class of (9) among the starts whose tails lie below $2^r/d$. There are at most $2^{r(r+1)/2}$ classes. For $r=\lceil A\log_2\log X\rceil$ with $A>1$ this is $X^{o(1)}$, and the mean-tail bound (T) leaves only $o(N_X)$ starts in $I_X$ with $H_n\ge2^r/d$. Thus all but $X^{o(1)}+o(N_X)$ starts of $I_X$ share their class with another start below the tail bound. A refutation needs one late pair in a common class whose next half-gaps differ by at least the larger following tail minus one. A large difference of next gaps alone is insufficient: both following tails must obey that same inequality, as well as the initial-tail bounds used in (9).

Both forms ask that about $A\log_2\log X$ consecutive gaps on one side of a start leave something free on the other side: the class modulo $4$ of the adjacent gap, or the size of the next gap. These are statements about the joint distribution of consecutive primes across one index at the critical length. The repeated windows supply the pairs, but no information about their neighbours. The inspected cluster and residue-string theorems (Maynard--Tao, Shiu, Banks--Freiberg--Turnage-Butterbaugh) leave the relevant prime positions unprescribed and supply no such joint pattern at growing length. In the prescribed-residue extension, Theorem 1.5 of arXiv:2409.12819v2 supplies at least one word in its specified family for fixed squarefree modulus and fixed word length. Its statement and the fixed-pattern lower count preceding it were inspected on printed pp. 3--4; the proof of Theorem 1.5 was not checked here. It supplies neither an exact gap window nor a quantitative joint count with its adjacent gap at this growing length. Occurrence of fixed windows alone does not suffice: with $K$ fixed, the bound $|H_{n+K}-H_{m+K}|<2^K/d$ asks two following tails to agree within a fixed constant. No inspected occurrence theorem pays that tail condition; the related nested-cluster attempt is recorded in the critical-length obstructions. No inspected theorem supplies either form. This describes the inspected inputs; it is not a proof that the forms are out of reach.