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This proves L230 at author-recorded standing. All logarithms are natural.

Write $D=\{n:a_n\ne p_n\}$. Since $D$ has density zero, for every fixed $\eta>0$ and all sufficiently large $n$ there is an unchanged index in each of $[(1-\eta)n,n]$ and $[n,(1+\eta)n]$. Monotonicity and $p_n\sim n\log n$ then imply $a_n\sim p_n$. The series converges, and for some integer $B\ge1$,

$$a_n\le Bn\log(n+1),\qquad p_n\le Bn\log(n+1) \quad(n\ge1). \tag{1}$$

We first fix $B$ and take its countable union at the end.

Define integer gap changes, without halving them, by

$$v_n=(a_{n+1}-a_n)-(p_{n+1}-p_n),\qquad n\ge1.$$

Their support $E$ is contained in $D\cup\{n:n+1\in D\}$, so $|E\cap[1,N]|=o(N/\log\log N)$. Positivity of both sets of gaps and telescoping give, for $N\ge3$,

$$\sum_{n=1}^N|v_n| \le a_{N+1}-a_1+p_{N+1}-p_1 \le6BN\log N. \tag{2}$$

Fix $\epsilon>0$ and an integer $N_0\ge16$, and restrict to the class with (1) and

$$|E\cap[1,N]|\le\frac{\epsilon N}{\log\log N} \quad(N\ge N_0). \tag{3}$$

For each fixed $\epsilon$, every sequence in the class with fixed $B$ satisfies (3) for some $N_0$.

Put $M=\lceil6BN\log N\rceil$ and $\rho_N=\epsilon/\log\log N$. For all sufficiently large $N$, $0<\rho_N<1/4$. An integer vector $(v_1,\ldots,v_N)$ with support of size $t$ has at most $\binom Nt$ choices of support, $2^t$ choices of signs and $\binom Mt$ choices of positive absolute values with sum at most $M$. Thus (2)--(3) give the upper bound

$$\sum_{0\le t\le\lfloor\rho_NN\rfloor} \binom Nt\,2^t\binom Mt. \tag{4}$$

The summand for $t=0$ is $1$. For $1\le t\le\rho_NN$, using $M\le(6B+1)N\log N$ and $\binom ut\le(eu/t)^t$ gives

$$\binom Nt\,2^t\binom Mt \le \left(2e^2(6B+1)\log N\,(N/t)^2\right)^t. \tag{5}$$

The logarithm of the right side is increasing as a function of $t/N$ on $(0,1/4]$. It is therefore at most

$$\begin{aligned} N\rho_N\log\frac{2e^2(6B+1)\log N}{\rho_N^2} &=N\frac{\epsilon}{\log\log N} \left(\log\log N+2\log\log\log N+O_{B,\epsilon}(1)\right)\\ &=(\epsilon+o_{B,\epsilon}(1))N. \end{aligned} \tag{6}$$

There are at most $N+1$ summands in (4) and finitely many possibilities for $a_1$ under (1). The initial value and vector $(v_n)$ determine the position prefix. Thus the number of prefixes through index $N$ in this class is at most

$$\exp\bigl((\epsilon+o_{B,\epsilon}(1))N\bigr). \tag{7}$$

Each prefix determines an interval of possible series values of length at most $C_BN\log N\,2^{-N}$, by (1). For any $s>\epsilon/\log2$, the sum of the $s$th powers of these covering lengths is bounded above by

$$\exp\bigl((\epsilon-s\log2+o_{B,\epsilon,s}(1))N\bigr) \longrightarrow0.$$

The definition of Hausdorff dimension gives a bound of $\epsilon/\log2$ for the values in the class (1), (3). Its countable union over $N_0$ includes the full class with fixed $B$ and the required little-oh support bound, so it has the same dimension bound. Letting $\epsilon\downarrow0$ proves dimension zero for that class. Taking the countable union over $B$ proves the assertion about $V$. A fixed exceptional set with the stated size bound gives a subset of $V$; such a subset cannot contain an interval of positive length. $\square$