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Source. The seed family and equation (21), printed pp. 89–90 (PDF pp. 5–6). The source states that the weighted estimate is not hard to prove and omits its details. The following is a full construction of a subfamily satisfying its conditions.

Statement. There is a constant $C_0>0$ such that, for every sufficiently large $x$, a family $\mathcal A_x$ of distinct odd square-free integers exists with:

1. $x^{1/2}<a<x^{3/4}$ for all $a\in\mathcal A_x$;
2. every $a$ is divisible by $3,5,7,11$;
3. any consecutive prime factors $q<q'$ with $q'>11$ satisfy $q'<q^{5/4}$;
4. $\sum_{a\in\mathcal A_x}1/a>(\log x)^{-C_0}$.

Every such integer belongs to the sequence in equation (19).

A fixed initial chain

Put $\lambda=9/8$ and $\rho=11/10$, so $1<\rho<\lambda$ and $\rho\lambda=99/80<5/4$. Choose a fixed prime $Q\ge17$ sufficiently large for

$$\sum_{Y<p\le Y^\rho}\frac1p\ge b:=\frac12\log\rho>0 \qquad(Y\ge Q^\lambda), \tag{1}$$

which follows from the exact prime estimates on the inputs page. In particular $b<1$.

Let $a_0$ be the product of all odd primes at most $Q$. Its prime factors start with $3,5,7,11$. The next two steps are $13<11^{5/4}$ and $17<13^{5/4}$. Thereafter Bertrand's postulate gives the next prime $q'<2q<q^{5/4}$ for $q\ge17$. Thus this fixed prefix satisfies the required consecutive-prime condition.

For $j\ge1$ let

$$u_j=\lambda^j\log Q,\qquad \mathcal P_j=\{p\text{ prime}:e^{u_j}<p\le e^{\rho u_j}\}.$$

These intervals are pairwise disjoint and above $Q$, because $\rho<\lambda$. If $p_j\in\mathcal P_j$, then $p_1<Q^{5/4}$, and

$$\log p_{j+1}\le\rho\lambda u_j<\frac54u_j <\frac54\log p_j.$$

So appending one prime from each successive interval preserves the consecutive-prime condition.

Choosing the number of intervals

Write $X=\log x$ and define

$$A_m=\log a_0+\sum_{j=1}^m u_j.$$

Choose the least integer $m\ge1$ such that $A_m>X/2$; it exists, and for large $x$ the fixed value $A_0$ is below $X/2$. The sum is geometric, so

$$m=\frac{\log X}{\log\lambda}+O(1).$$

Every product $a=a_0p_1\cdots p_m$ has $A_m<\log a\le\rho A_m$. Also

$$A_m=\lambda A_{m-1}+\lambda\log Q-(\lambda-1)\log a_0 \le\frac{\lambda X}{2}+O(1).$$

Consequently $\log a\le(\rho\lambda/2)X+O(1)<3X/4$ eventually. The products lie in $(x^{1/2},x^{3/4})$, are odd and square-free, and their prime factors identify the tuple uniquely.

By (1), their reciprocal sum is

$$\frac1{a_0}\prod_{j=1}^m\left(\sum_{p\in\mathcal P_j}\frac1p\right) \ge\frac{b^m}{a_0} = (\log x)^{\log b/\log\lambda+o(1)}.$$

Any fixed $C_0>-\log b/\log\lambda$ proves the required bound for all sufficiently large $x$.

Membership in the original sequence

The first four primes satisfy equation (19): $7<3\cdot5$ and $11<3\cdot5\cdot7$. Their product $1155$ exceeds $11^2$. Inductively, suppose the product $D$ through the previous largest prime $q$ exceeds $q^2$. The next prime satisfies $q'<q^{5/4}<D$, so its equation (19) inequality holds, and the new product $Dq'$ exceeds $(q')^2$. This proves every later inequality and membership in $\mathcal N$.

Source precision. The source's seed description does not explicitly exclude a factor 2, although its assertion that all seeds satisfy $p_1=3,p_2=5$ requires this. The odd subfamily constructed here has the stated reciprocal mass and supplies the needed correction. The explicit intervals fill the omitted same-paper count; they are not claimed to reproduce a construction written in the source.