@Rama1729 Assuming RH $$\lambda_n=\frac{n\log n}{2}-\frac{\log(2\pi)+1-\gamma}{2}n+n y_n$$ where $(y_n)$ is in $\ell_2$. RH is equivalent to say that $(y_n)$ is in $\ell_2$. This is in my paper Asymptotic of Keiper-Li coefficients, Functiones et approximatio 45 (2011)7--21.

@Rama1729 yes the RH is not required, but the argument is slightly more complex. But apparently you knew the answer.

In arxiv.org/abs/1810.05198 there is a translation of the classical paper of Siegel about Riemann's nachlass, where he consider the Riemann-Siegel expansion. As authors appear as the translators, which I did not consider good, because it is difficult to find by search on the web.

Today appeared in arXiv a paper arXiv:2110.07407v1 "A conjecture of Zagier and the value distribution of quantum modular forms" by Aistleitner and Borda, that have related material and references in particular the paper cited by @Kulikov and others by the same authors.

@jacobul I have ready a modification of my answer to make it more explicit when I see your suggestions. Now I am retired but I used this construction for years in my course of measure theory, suggested by other professor of my University.

@Seva I tried the email address in your web page. But my message is not send. I am not sure what is the problem. Perhaps if you send a mail to me I can answer to it.

@Seva I will send you my computation if I find your email.

@Seva If my computation are correct, $v_p^6$ is a positive real number for $p$ odd. So I take its positive $6$-root and then only remain a sign.

The derivatives of $\zeta(s)$ at a point such as $1-1/B$ behaves in a complicated way. The first few may be dominated by the pole at $0$. So they are near integers if $B$ is an integer. The Taylor series of $f(s)$ is not a good recipe to compute $f(-n)$. Mathematica gives $\zeta^{(4)}(1-1/3)=-5831.99742689039619$ near an integer. mpmath gives -5831.9974268904026. I do not think there is a bug here.

For $n\ge2$ there are $n+2$ sets in this family, $n$ of them contains $1$, and $n\ge (n+2)/2$. So this case is trivial.