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This proves L228 at author-recorded standing. All logarithms without a subscript are natural. Write $h_n=(p_{n+1}-p_n)/2$ for $n\ge2$ and

$$H_m=\sum_{r\ge0}\frac{h_{m+r}}{2^{r+1}}\qquad(m\ge2).$$

We first derive the only distributional estimate used in the construction.

Very small gaps have small upper density

There are absolute constants $A$ and $M_0$ such that, for every $M\ge M_0$ and $0<\epsilon\le1$,

$$\#\{2\le n\le M:h_n\le\epsilon\log n\}\le A\epsilon M. \tag{1}$$

The source premise is the prime-pair bound at the beginning of the proof of Theorem 4.2, p. 10 of Nathan McNew, The convex hull of the prime number graph, author's preprint of the 2018 paper. Its library page identifies the edition and proof location. For $D(t,x)=\#\{p\le x:p,p+t\text{ are prime}\}$, the displayed Brun-sieve bound is, with an absolute constant,

$$D(t,x)\ll \frac{x}{\log^2x} \prod_{\substack{\ell\mid t\\\ell\le x}} \left(1-\frac1\ell\right)^{-1}. \tag{2}$$

All products indexed by $\ell$ in this proof run over primes.

We use it only for positive even $t\le2\log x$, so that the product is $t/\phi(t)$. The statement, including the absolute constant, was inspected on the printed page. The sieve argument behind (2) is not reproved here.

For $B\ge1$, the identity $t/\phi(t)=\sum_{e\mid t}\mu(e)^2/\phi(e)$ gives

$$\sum_{t\le B}\frac{t}{\phi(t)} \le B\sum_{e\ge1}\frac{\mu(e)^2}{e\phi(e)} =B\prod_{\ell\text{ prime}} \left(1+\frac1{\ell(\ell-1)}\right)\ll B. \tag{3}$$

If $\epsilon\log M<1$, the set in (1) is empty. Otherwise its members have $p_{n+1}-p_n\le2\epsilon\log M$. Put $x=p_M\sim M\log M$ and sum (2) over the positive even differences up to $B=2\epsilon\log M\le2\log x$. By (3), the count is at most

$$O\left(\frac{x}{\log^2x}\,\epsilon\log M\right)=O(\epsilon M),$$

with constants independent of $\epsilon$ and $M$ once $M$ is large. This proves (1) uniformly in $\epsilon$. In particular, if $b_n=o(\log n)$ is nonnegative, the indices with $h_n\le b_n$ have density zero: bound $b_n$ eventually by $\epsilon\log n$ and then let $\epsilon\downarrow0$ in (1).

Choosing the positions and their capacity

Fix the moduli $(q_n)$ and a prefix length $N$. Put

$$\Omega_n=\inf_{m\ge n}\log_2\frac{\log m}{q_m}.$$

Since $q_n=o(\log n)$, the numbers $\Omega_n$ are finite, nondecreasing and tend to infinity. Choose $n_0\ge\max(N,3)$ such that $\Omega_{n_0}\ge8$, and set $L_n=\lfloor\Omega_n/4\rfloor$ for $n\ge n_0$. Thus $L_n\ge2$ is nondecreasing and tends to infinity. Define

$$i_0=n_0,\qquad i_t=i_{t-1}+L_{i_{t-1}}\quad(t\ge1),\qquad U_t=2^{L_{i_t}+3}. \tag{4}$$

The set of starts $\{i_t:t\ge1\}$ has density zero, since its successive spacings tend to infinity. We choose distinct, increasing endpoints. Put $j_0=n_0$ and let $j_t$ be the least integer satisfying

$$j_t\ge\max(i_t+1,j_{t-1}+1),\qquad h_{j_t}\ge q_{j_t}U_t+1. \tag{5}$$

Each choice exists. For fixed $t$, $q_jU_t+1=o(\log j)$, and the last consequence of (1) says that almost all sufficiently large $j$ satisfy the second condition in (5). No quantitative bound for the first such endpoint is needed.

For $j\ge i_t$, the definition of $\Omega$ gives

$$q_jU_t\le 2^{3-3\Omega_{i_t}/4}\log j =:\epsilon_t\log j, \qquad \epsilon_t\downarrow0. \tag{6}$$

Define the set of position indices

$$D=\bigcup_{t\ge1}\{i_t+1,i_t+2,\ldots,j_t\}. \tag{7}$$

It has density zero. To prove this, fix an integer $T\ge1$ and discard the finite union of intervals in (7) with $t<T$. If a remaining $n\in D$ is not an endpoint, choose the smallest $t$ with $i_t<n\le j_t$. Every earlier endpoint is below $n$: otherwise its start is smaller than $i_t<n$, so its interval would contain $n$. Therefore $n\ge\max(i_t+1,j_{t-1}+1)$ and $n<j_t$. By the minimality of $j_t$, $h_n<q_nU_t+1$, hence, since these are integers,

$$h_n\le q_nU_t\le\epsilon_t\log n\le\epsilon_T\log n.$$

The endpoint set has density zero because $j_t>i_t$, so its count up to $M$ is at most the count of starts up to $M$. By (1), the upper density of $D$ is at most $A\epsilon_T$. Letting $T\to\infty$ proves the assertion. It also proves

$$\frac{j_t}{i_t}\longrightarrow1: \quad\frac{j_t-i_t}{j_t}\le \frac{|D\cap[1,j_t]|}{j_t}\longrightarrow0. \tag{8}$$

The set $D$, the starts, endpoints and caps depend on the given primes and moduli, not on the value of the series to be attained.

If, additionally, $q_n\le C(\log n)^\alpha$ eventually, where $0\le\alpha<1$, then

$$\Omega_n\ge(1-\alpha)\log_2\log n-\log_2 C$$

for all large $n$. The start spacings are consequently bounded below by a positive constant times $\log\log i_t$. There are $O(M/\log\log M)$ starts up to $M$, by splitting at $\sqrt M$, and at most that many endpoints. By (8), intervals with $i_t<\sqrt M$ end below $2\sqrt M$ for all sufficiently large $M$. An index of $D$ above $2\sqrt M$ which is not an endpoint therefore has, by the same minimal-endpoint argument as above,

$$h_n\le\eta_M\log n,\qquad \eta_M=8\,2^{-3\Omega_{\lfloor\sqrt M\rfloor}/4} =O\left((\log M)^{-3(1-\alpha)/4}\right).$$

Apply the uniform bound (1), with this $\eta_M$, to obtain

$$|D\cap[1,M]| \le2\sqrt M+O(M/\log\log M)+A\eta_M M =O(M/\log\log M). \tag{8a}$$

This proves the additional quantitative clause about $D$.

Filling an interval of values

Define positive real weights

$$w_t=q_{j_t}\left(2^{-(i_t-n_0+1)}-2^{-(j_t-n_0+1)}\right). \tag{9}$$

By (8), $q_{j_t}=o(\log j_t)=o(\log i_t)$, so $w_t\to0$. Since $j_t>i_t$, the factor $1-2^{-(j_t-i_t)}$ lies in $[1/2,1)$. Since the moduli form a divisibility chain and the endpoints increase, $q_{j_{t-1}}\le q_{j_t}$. Consequently, for $t\ge2$,

$$\frac{w_{t-1}}{w_t}\le2^{i_t-i_{t-1}+1} =2^{L_{i_{t-1}}+1}\le2^{L_{i_t}+1}. \tag{10}$$

Also $w_1\ge2^{-L_{n_0}-2}$. For any $b\in[0,1]$ define

$$r_0=b,\qquad u_t=\left\lfloor\frac{r_{t-1}}{w_t}\right\rfloor,\qquad r_t=r_{t-1}-u_tw_t\quad(t\ge1). \tag{11}$$

Then $0\le r_t<w_t$, so $r_t\to0$ and $\sum_{t\ge1}u_tw_t=b$. Moreover $0\le u_1\le2^{L_{n_0}+2}<U_1$, and for $t\ge2$, (10) gives $0\le u_t<w_{t-1}/w_t\le2^{L_{i_t}+1}<U_t$. Thus every digit fits its prescribed capacity.

Set $v_t=q_{j_t}u_t$ and, for all $n\ge1$, put

$$a_n=p_n+2\sum_{i_t<n\le j_t}v_t. \tag{12}$$

This finite sum is nonnegative. It vanishes for $n\le n_0$ and for $n\notin D$. For $n\ge2$, the new halved gap is

$$h'_n:=\frac{a_{n+1}-a_n}{2} =h_n+\sum_{i_t=n}v_t-\sum_{j_t=n}v_t. \tag{13}$$

There is at most one term in each of the two sums. If an endpoint equals $n$, then $v_t<q_nU_t<h_n$ by (5); therefore $h'_n\ge1$, including when a start and an endpoint coincide. Also $a_1=2$, $a_2=3$. This proves strict increase and positivity.

For every active interval in (12), $n\le j_t$, so $q_n\mid q_{j_t}$. It follows that $a_n\equiv p_n\pmod{2q_n}$ for $n\ge2$. Equation (13) also shows $h'_n\equiv h_n\pmod{q_n}$, including at starts, because $i_t<j_t$ and $q_{i_t}\mid q_{j_t}$.

The density-zero conclusion for $D$ implies $a_n/p_n\to1$. Indeed, $a_n\ge p_n$, and for every fixed $\eta>0$ and all large $n$ there is an $m\in[n,(1+\eta)n]$ outside $D$. Monotonicity and the prime number theorem give $a_n\le a_m=p_m\le(1+\eta+o(1))p_n$; let $\eta\downarrow0$.

The changes to gaps are small even on their support. By (6), $v_t\le\epsilon_t\log j_t$, and by (8), $\log j_t/\log i_t\to1$. Thus a positive term in (13) is $o(\log i_t)$ and a negative term is $o(\log j_t)$. There are at most two incident terms at any index, so $h'_n-h_n=o(\log n)$. These changes also have density-zero support, since only starts and endpoints can occur.

Finally, all the terms $v_t$ in (12) are nonnegative. Tonelli's theorem and a finite geometric sum give

$$\begin{aligned} \sum_{n\ge1}\frac{a_n-p_n}{2^n} &=2\sum_{t\ge1}v_t\sum_{n=i_t+1}^{j_t}2^{-n}\\ &=2\sum_{t\ge1}v_t(2^{-i_t}-2^{-j_t})\\ &=2^{2-n_0}\sum_{t\ge1}u_tw_t =2^{2-n_0}b. \end{aligned} \tag{14}$$

Take $\delta=2^{2-n_0}$ and $b=(T-S)/\delta$ for any $T\in[S,S+\delta]$. Equations (12)--(14) prove every clause of L228. Dyadic rationals are dense, so one can choose such a $T>S$ that is dyadic rational. The construction makes no assertion that the altered terms are prime. $\square$