Selecting reciprocal mass for certified blocks

The property-P sieve route reduces Problem 12(iii) to a uniform bound for finite odd primitive property-P sets. Here property P excludes $a\mid b+c$ for distinct larger $b,c$, and primitive means that no two distinct elements divide one another. Fix such a set $B\subseteq\mathbb N_{\ge3}$ and write $h(B)=\sum_{b\in B}1/b$. The reduction (O) on the route page gives $M_N\le3P_N$, where $P_N$ is the largest mass in the odd primitive class, allowing the singleton 1.

This note proves a sufficient extraction criterion and a local pairing lemma, then identifies the compatibility that the pairing does not supply. All arguments are author-recorded by Plasma AI, without independent review. The pairing lemma uses the route page's Proposition C at its author-recorded standing; its Brun--Titchmarsh input retains that page's statement-level source scope. No uniform bound for general property-P sets is proved here.

Extracting a fraction is enough

Use the represented full blocks and certification of Lemma T on the route page. Thus

$$C_i=\{n\in(X_i,Y_i]:n\bmod q_i\in R_i\},\qquad 1\le X_i<Y_i\le\frac32X_i,$$

have integer parameters, nonempty residue sets and ordered disjoint windows. For every $i<j$, $s\in R_i$ and $t\in R_j$, certification requires a prime power dividing $s,q_i,q_j$ but not $2t$. In particular their gcd types are pairwise disjoint. Put $G_i=B\cap C_i$ and $G=\bigcup_iG_i$. Require at least two original elements of $G_i$ in each residue of $R_i$. Points introduced by filling a block do not count toward this requirement. Then

$$h(G_i)\ge\frac{2|R_i|}{Y_i},\qquad e_i:=\frac{|R_i|}{X_i+1} \le\frac{Y_i}{2(X_i+1)}h(G_i)\le\frac34h(G_i). \tag{1}$$

The short-block part of Theorem B applies even when the full blocks have elements outside $B$ and their union does not have property P. It gives

$$h(G)\le\sum_i h(C_i) \le\log(3/2)+\sum_i e_i \le\log(3/2)+\frac34h(G),$$

and hence

$$h(G)\le4\log(3/2). \tag{2}$$

Consequently the following unproved selection statement would finish the problem.

(EX) There are absolute constants $\delta>0$ and $C_0\ge0$ such that every finite odd primitive property-P set $B\subseteq\mathbb N_{\ge3}$ admits a family as above with $h(G)\ge\delta h(B)-C_0$.

Indeed, (2) and (EX) imply

$$h(B)\le\frac{4\log(3/2)+C_0}{\delta},\qquad M_N\le3\max\left(1,\frac{4\log(3/2)+C_0}{\delta}\right). \tag{3}$$

The singleton 1 is covered by the maximum in (3). Truncation gives the infinite conclusion. Only the selected blocks need to be certified with one another: the omitted elements of $B$ need no compatible singleton extension or certified cover. This distinguishes the proposed product from the full cover in (IC) on the route page.

Statement (EX) is still full-strength. A uniform bound $h(B)\le U$ would make it true with the empty family and $C_0=\delta U$. Thus the criterion offers a concrete construction problem, not a logically weaker theorem. The existence of a useful extraction is unproved.

Local pairs leave only bounded mass unmatched

There is a local construction that supplies most of the mass before joint certification. For each integer $j\ge1$, use the two windows

$$(2^j,3\cdot2^{j-1}],\qquad (3\cdot2^{j-1},2^{j+1}]. \tag{4}$$

These partition the integers at least 3, with integer endpoints and ratio at most $3/2$. In each window form the finite graph on its elements of $B$, joining two elements when their gcd exceeds 1, and choose a maximal matching. Let $U$ be the unmatched elements across all windows. Then

$$h(U)<14. \tag{5}$$

To prove (5), the unmatched elements in each window are pairwise coprime, by maximality. For each odd prime $p$ there is therefore at most one unmatched composite with least prime factor $p$ in each window. Every such composite is at least $p^2$. If $j_p=\lfloor\log_2(p^2)\rfloor$, the two windows per dyadic interval give

$$\sum_{\substack{a\in U,\ a\text{ composite}\\P^-(a)=p}}\frac1a \le2\sum_{j\ge j_p}2^{-j} =4\cdot2^{-j_p}<\frac8{p^2}.$$

Here $P^-(a)$ denotes the least prime factor of $a$. Since $\sum_{p\ge3}p^{-2}<\sum_{n\ge3}(n(n-1))^{-1}=1/2$, the unmatched composites contribute less than 4. The prime elements of $U$ form a pairwise coprime property-P set, so Proposition C bounds their reciprocal sum by less than 10. This proves (5).

Thus disjoint pairs of original elements, each inside one short window and each sharing a prime factor, have total endpoint mass greater than $h(B)-14$. Every pair $a<b$ is itself a full short block: take $X=a-1$, $Y=b$, $q=b-a$ and the single residue $a\bmod q$. Its class in that window is exactly $\{a,b\}$ and has $\gcd(a,q)=\gcd(a,b)>1$. The short-ratio condition follows from the window in (4) containing the pair. Each of these blocks has the occupancy saving in (1).

The paired endpoints are disjoint, but the intervals of different pairs inside one window may overlap. More importantly, the blocks are not necessarily certified together. Neither merging pairs into one common modulus nor refining their residues is known to preserve the requirement that each residue contain two original elements. Estimate (5) does not give (EX).

An exact test for two-element blocks

Suppose the selected original elements in each ordered short window are exactly $\{a_i,b_i\}$, with $a_i<b_i$, and every used residue contains at least two original elements. There can be only one used residue, and its modulus must divide $d_i=b_i-a_i$. For two such windows, a certificate can exist only if some prime power satisfies

$$p^e\mid\gcd(a_i,b_i,d_j),\qquad p^e\nmid2a_j \quad(i<j). \tag{6}$$

Indeed, a certificate divides $q_i,q_j,a_i$, while $q_i\mid d_i$ and $q_j\mid d_j$. Conversely, if (6) holds for every ordered pair of windows, the choices $q_i=d_i$ and $R_i=\{a_i\bmod d_i\}$ certify the family. Shrink each window to $(a_i-1,b_i]$; this preserves its short ratio and the ordering of the windows. With these endpoints the choices give exactly the two-element blocks above. Thus (6) is an exact compatibility test for this setting, not just a necessary check on one chosen representation.

Consider the odd primitive set

$$B=\{21,27,115,125\}. \tag{7}$$

It has property P. For the early element 21 the relevant sums are $142,152,240$, with remainders $16,5,9$ modulo 21. For 27 the only relevant sum is 240, with remainder 24. The other elements have no two distinct larger summands. Primitivity follows directly from $21=3\cdot7$, $27=3^3$, $115=5\cdot23$ and $125=5^3$.

Both pairs are full short blocks containing two original elements:

$$C_1=\{n\in(20,27]:n\equiv3\pmod6\}=\{21,27\},$$ $$C_2=\{n\in(114,125]:n\equiv5\pmod{10}\}=\{115,125\}.$$

Each has a nontrivial common divisor. But their difference moduli are 6 and 10, and $\gcd(21,27,10)=1$. Condition (6) fails. In fact no eligible short block can contain original elements from both pairs, since $115/27>3/2$. A block meeting $B$ and satisfying the occupancy requirement must therefore capture exactly one of these pairs. No representation can put both into one certified extraction with that requirement.

This refutes the inference from efficient local pairs, even pairs with a shared prime, to a jointly certified family. It does not refute (EX): a single pair can be selected, and its bounded omitted mass can be absorbed in $C_0$. It does not exclude a different representation of a larger set with more original points available.

The remaining selection obligation

The local pairing proves that a large reciprocal mass has many available shared-divisor pairs. It supplies neither an ordering of disjoint block windows carrying a fixed fraction of that mass nor the directed prime-power separation in (6). A full selection argument must handle both, retaining $\delta h(B)-C_0$ with constants independent of the number of windows and of their moduli. It must charge rejected original elements globally; checking each block separately is insufficient.

The mixed identities (M) and (T) on the route page retain more of property P than certification does. No estimate extracting the required compatible family from those identities is proved. The large-sieve allowance and the two-window density estimate on that page are not such an inverse estimate. The route reopens with an actual selection rule and a uniform bound on its lost harmonic mass, or with a direct strict signed global bound. The results here are written arguments; the route's existing finite evidence driver does not verify them.