For $f(x)=\sum_{n=0}^\infty a_n x^n$, a real number $R\neq 1$, $g(x)=f(x)-Rf(x^2)$ Abel summable at $x=1$, $g(1)=\lim_{x\to 1^-} g(x)$, the elementary Ramanujan sum of $f(x)$ at $x=1$ is defined by $f(1)=g(1)/(1-R)$ (definition, also here).
For $$ f^{(N)}(x)=\left(x\frac{d}{dx}\right)^N f(x)~, $$ $f^{(N)}(1)=g^{(N)}(1)/(1-2^N R)$.
For $g(x)=x-x^2+x^3-\cdots=x/(1+x)$ ($|x|<1$) and $R=2$, we have $f(x)=x+x^2+x^3+\cdots=x/(1-x)$ ($|x|<1$), $f(1)=-g(1)=-1/2$ and $f^{(N)}(1)=\zeta(-N)$.
What we obtain if we replace $x^2$ with $x^3$ in the definition of the elementary Ramanujan summation? (here) For $g(x)=x-x^2+x^3-\cdots$ and $R=2$, what are $f_2(x)$ and $F_2(s)$ such that $f_2(x)-2f_2(x^3)=g(x)$ and $f_2^{(N)}(1)=F_2(-N)$? what are $f_3(x)$ and $F_3(s)$ such that $f_3(x)-3f_3(x^3)=g(x)$ and $f_3^{(N)}(1)=F_3(-N)$? (here and here)
How are $F_2(s)$ and $F_3(s)$ related to $\zeta(s)$?
Edit 1. In general, for an odd prime number $p$ and $f(x)-Rf(x^p)=g(x)=x-x^2+x^3-\cdots$, the coefficients are given by $a_0=0$ and $$ a_n=(-1)^{n+1}\frac{R^{m_p(n)+1}-1}{R-1}~,~~ (n>0) $$ where $m_p(n)$ is the multiplicity of the prime factor $p$ of $n$.
The alternating series $$ F_p(s)=\sum_{n=1}^\infty a_n n^{-s} $$ converges absolutely at least for $s>1$ and $|R|<p^s$.
The analytic continuation of the series $G_p(s)=\sum_{n=1}^\infty |a_n|n^{-s}$is given by $$ G_p(s)=\frac{1}{1-Rp^{-s}}\zeta(s)~. $$
Edit 2. $G_p(s)$ corresponds to $f(x)-Rf(x^p)=x+x^2+x^3+\cdots$. It generalizes the result about iteration and analytic continuation obtained here.
For $f(x)=\sum_{n=0}^\infty a_n x^n$ and $g(x)=\sum_{n=0}^\infty b_n x^n$, $R\neq 1$ and $k\in\mathbb{N}$, $k>1$, such that $f(x)-Rf(x^k)=g(x)$, the coefficients of $f(x)$ and $g(x)$ are recursively related by $$ a_0=\frac{b_0}{1-R}~,~~ a_{kn+i}=b_{kn+i}\quad (0<i<k),~~ a_{kn+k}=b_{kn+k}+Ra_{n+1}~. $$ The corresponding Dirichlet series $\sum_{n=1}^\infty a_n n^{-s}$ and $\sum_{n=1}^\infty b_n n^{-s}$ satisfy $$ \sum_{n=1}^\infty a_n n^{-s}-R\sum_{n=1}^\infty a_n (kn)^{-s}=\sum_{n=1}^\infty b_n n^{-s}~. $$ Thus $$ \sum_{n=1}^\infty a_n n^{-s}=\frac{1}{1-Rk^{-s}}\sum_{n=1}^\infty b_n n^{-s} $$ and the corresponding analytic continuations $F(s)$ and $G(s)$ satisfy $$ F(s)=\frac{1}{1-Rk^{-s}}G(s)~. $$ For $g(x)=x-x^2+x^3-\cdots$ and $k=p\neq2$, $$ F(s)=F_p(s)=\frac{1}{1-Rp^{-s}}\eta(s)=\frac{1-2^{1-s}}{1-Rp^{-s}}\zeta(s)~. $$