Behavior of $n\sqrt{3}-a_n\sqrt{n}$ as $n\to\infty$, where $a_{n+1}=a_n+n/(a_1+\dots+a_n)$

Question

For a fixed real $a>0$, define the sequence $(a_n)_{n>0}$ by $a_1=a$ and $$a_{n+1}=a_n+\frac{n}{a_1+\dots+a_n}.\qquad(n>0)$$

It is known (see this answer) that $$\lim_{n\to\infty}\frac{a_n}{\sqrt{n}}=\sqrt{3},$$ hence the question: what about $\color{blue}{b_n=n\sqrt{3}-a_n\sqrt{n}}$ as $n\to\infty$?

There is a related question on MSE, asking originally for a "proof" of $b_n\to\sqrt{3}/4$, with some stronger (wrong) claims on the asymptotics of $b_n$ around. But the end of the answer by @Somos captures the main observation that $b_n$ doesn't (seem to) have a limit; instead, it is close to a periodic function of $\log n$, with a pretty big period:

The above depicts $b_n$ for $a=1$, as a function of $\log_2n$, obtained using PARI/GP:

experiment(N, M, a) = {
    my (x = Set(vector(M + 1, k, floor(N ^ ((k - 1) / M)))));
    my (y = vector(#x), u = a * 1.0, v = u, n = 1);
    for (k = 1, #x,
        while (n < x[k], u += n / v; v += u; n += 1);
        y[k] = x[k] * sqrt(3) - u * sqrt(x[k]);
        x[k] = log(x[k]) / log(2));
    write("graph.svg", plothrawexport("svg", x, y, 1))
};

by running experiment(2^27, 2^12, 1). Other values of $a$ behave similarly.