When is $(\frac{v_1}{1}+\frac{v_2}{2}+ \dots \frac{v_p}{p})+(\frac{v_1}{p+1}+\frac{v_2}{p+2}+\dots \frac{v_p}{2p})+\dots$ equal to zero? Let $V=\{\vec{v} =(v_1,\dots v_p): v_i\in \mathbb{Z}\}$. 
And define $$\phi(\vec{v},s)= \sum_{n=1}^\infty\frac{v_n}{n^s}$$
Where for all $k=1,2, \dots$ we have $v_{k+p}=v_k$. 
My question 
(1) Is there a nice characterization of all the vectors such that $\phi (\vec{v},1)=0$?
(2) Is there a nice characterization of all the vectors such that $\phi(\vec{v},s)=0$?
Exposition
This is not a topic that has sprung from nowhere; I have asked quite a few questions (here and here) on this topic. In this notation we can rewrite some often discussed functions: 
$\begin{align}
&
\phi((1),s)=\zeta(s), \\
& \phi((1,-1),s)=\eta(s), \\
& \phi((1,0,-1,0),s)=\beta(s)
\end{align}$
Where these functions are the zeta, Dirichlet eta and the Dirichlet  beta function respectively. 
For each $s\in \mathbb{R^{+}}$ we have a mapping  $\phi$ from $V$ into the real numbers(when this sum converges). For $s>1$ we needn't worry about convergence but for $s=1$ it looks like (I am fairly confident) $\sum_{n=1}^p{v_n}=0$ is a necessary and sufficient condition for $\phi(\vec{v},1)\in \mathbb{R}$. 
Then one can ask what is the $\ker(\phi(\vec{v},1))$? That is what vectors are mapped to zero? 
$\begin{align}
& (0) \\
&(1,-3,1,1), \\
&(1,−1,−2,−1,1,2), \\
&(1,1,−5,1,1,−5,1,1,4)  
\end{align}$
Are all vectors in the kernel of $(\phi(\vec{v},1))$. 
The first line should be unsurprising but these others are kind of interesting: 
The second line is 
$$0=\frac{1}{1} -\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\dots $$
We can build these using the techniques laid out by Mats Granvik here. 
Basically we can argue like this: 
We know that 
$$\lim_{s\to 1} \zeta(s)(1-\frac{1}{a^{s-1}})(1-\frac{1}{b^{s-1}}) =ln(a)\times 0 = 0  $$
But note that 
$$ \begin{align}
& \zeta(s)(1-\frac{1}{a^{s-1}})(1-\frac{1}{b^{s-1}}) \\
& = (\sum_{n=1}^{\infty} {\frac{1}{n^s}) (1-\frac{a}{a^s}-\frac{b}{b^s}+\frac{ab}{{(ab)}^s}) } 
\end{align}
$$
Taking $a=2,b=3$ and multiply through 
$$= (\sum_{n=1}^{\infty} {\frac{1}{n^s}})
 -(\sum_{n=1}^{\infty} {\frac{2}{(2n)^s}})  -(\sum_{n=1}^{\infty} {\frac{3}{(3n)^s}})  +(\sum_{n=1}^{\infty} {\frac{6}{{(6n)}^s}}) $$
$\phi([1],s)-\phi([0,2],s)-\phi([0,0,3],s)+\phi([0,0,0,0,0,6],s)$
I will demonstrate how we can add within the vectors (Note that we are breaking some pretty serious rules if we take $s=1$ as all of these sums above diverge.)
$$
\begin{align}
&+\phi([1,1,1,1,1,1],s)\\  
&-\phi([0,2,0,2,0,2],s)\\
&-\phi([0,0,3,0,0,3],s)\\
&+\phi([0,0,0,0,0,6],s)\\
&=\phi([1,-1,-2,-1,1,2],s)\\
\end{align}
$$
I will offer up one other notation which may prove useful. Let 
$(d|n) = \begin{cases} 
1 \text{ when d|n}\\
0 \text{ else }
\end{cases}$
Then this $(1,-1,-2,-1,1,2)$ can be rewritten $1(1|n)-2(2|n)-3(3|n)+6(6|n)$
So we can see this vector $(1,-1,-2,-1,1,2)$ didn't appear of thin air: we can build these. Now if we generalize this technique from two factors to more factors we can build all sorts of vectors which map to zero. In fact, this technique can let find elements in the kernel of $\phi(\vec{v},17)$ if we wanted. We could just multiply $\zeta(s)(1-\frac{1}{a^{s-1}})(1-\frac{1}{b^{s-17}})$
Questions
1) How can I characterize these vectors? 
2) Have we found all the zeros of $\phi(\vec{v},1)$ using this technique? That is, will this technique allow me to find a basis for the $\ker \phi(\vec{v},1)$? and what about $\ker \phi(\vec{v},s)?$ 
Motivations
I think it simply must make sense to explore the underlying algebraic structure of the module $A_s=\{\phi(\vec{v},s): \vec{v}\in V \}$ where we can draw scalars from $\{p/q^s: p,q \in \mathbb{Z} \}$. Imagine for a second that $\phi([1,0,-1,0],2) \in \mathbb{Q}$ which is unknown. Then wouldn't that have some pretty interesting implications about the underlying algebraic structure of $\phi^{-1}(A_2,2)$? I am not really an algebraist and don't know whether it's really possible that Algebra could determine whether Catalan's Constant is irrational or not... but it doesn't seem impossible to me. 
 A: I think this may  be a better way to frame the puzzle and I attempt to answer question $1$. 
Let $V$ be the set of all infinite periodic integer sequences and for each $s\in \mathbb{R^+}$ define a homomorphism $\phi_s:V\to\mathbb{R}$ as follows 
$$ \phi_s\bigg((v_1,v_2, \dots ) \bigg)=\sum_{n=1}^\infty\frac{v_n}{n^s}$$
Note (as we did above) that the domain of $\phi$ is dependent on our selection of $s$
We will define a $W\subset V$ such that $\forall \vec{w}\in W$ 
$\phi_1(\vec{w})=0$ and conjecture that $W$ forms a basis for the $\ker(\phi_1)$
Let $m$ be a positive integer and write out it's prime factorization $m=p_1^{e_1}p_2^{e_2} \dots p_n^{e_n}$, and define a function 
$$g_m(x)=\prod_{j=1}^n{\bigg(1-j(j|x) \bigg)^{e_j}}$$
where $(d|n)= 1_{d\mathbb{N}}(n)$ is $1$ when $d|n$ and otherwise $0$.
Then $ \vec{g_m}:=\bigg((g_m(1),g_m(2), \dots \bigg)$ is a periodic sequence and for non-prime $m$ we will define our set $W$. 
That is, we let $W= \{ \vec{g_m}: 1<m \not \in \mathbb{P} \}$ and then note that by the informal argument above (in the question) that $ [W] \subset \ker(\phi_1)$ where $[ \dot ]$ refers to the span of the set with scalars from the rationals. The selection of $m$ not prime guarantees that the product has at least $2$ non unit factors. Note that when $m$ is the $k$th prime we can say that $\phi_1(g_{p_k})=\ln(k+1)$. 
This then serves to answer my question (1). $[W]$ is the generalization I am after (I think). But I am not sure how I might attempt to show the other direction (which is question (2)): Is it the case that $[W] \supset \ker(\phi_1)$? 
I should really define operations in $V$: vector addition is what you'd expect and vectors interact with rationals as demonstrate in this post. 
