An interesting series related to primes satisfying $\sum_n x_{nk} = 0$ for all $k$ Consider the series $\sum x_n$ where if $n$ is composite, say $n=p_1^{k_1}....p_m^{k_m}$ then $x_n=x_{p_1}^{k_1}....x_{p_m}^{k_m}$ and if $n$ is prime, $x_p=-x_1-...-x_{p-1}$. I want to determine whether an $x_1\neq 0$ exists such that $\sum x_n=0$. It feels like this will be true for all $|x_1|<1$ or none(more clear after getting a feel for the series). This would have the interesting consequence that $\sum x_{kn}=0$ for all integers $k$.
I have also written/copied from the internet a generator for the sequence in python, found here https://pastebin.com/jcqvSS0W
The sequence feels more related to prime gaps rather than anything to do with primes being prime, which makes me suspect this is a difficult question.
 A: I'll abbreviate $x_1$ to $x$ and I'll write $f(n) = \sum_k f_{k,n} x^k$. I will show the following:
Key Lemma: The sum $\sum_{n=1}^{\infty} |f_{k,n}|$ is finite, and is $O(8^n)$.
Since all the $f_{k,n}$ are integers, this in particular means only finitely many of the summands are nonzero. In other words, for $k$ fixed, $f_{k,n}$ is zero for $n \gg 0$. I can give a direct proof of this fact, and can in fact show that $f_{k,n}=0$ for $n > 2^{2k-1}$, but I will omit this, since I need to prove the Key Lemma anyway and my answer will be long enough.
Using the Key Lemma, if $|x| < 1/8$, then $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} |f_{k,n}| |x|^k$ is convergent. Sums of positive terms may be rearranged freely, so $\sum_{n=1}^{\infty} \sum_{k=1}^{\infty}  |f_{k,n}| |x|^k$ is also convergent. As $|f(n)| \leq \sum_{k=1}^{\infty}  |f_{k,n}| |x|^k$, this shows that $\sum_{n=1}^{\infty} f(n)$ is absolutely convergent.
I do not think that the bound $1/8$ is close to the truth, but it is as hard as I am willing to work tonight. As I noted in the comments above, we do NOT have convergence for $x>2^{-1/6} \approx 0.89$.
Proof of the Key Lemma: Put $A_k = \sum_{n=2}^{\infty} |f_{k,n}|$ (omitting the $n=1$ term doesn't change the value of $A_k$ for $k>1$, and is convenient). Assume inductively we know that $A_1$, $A_2$, ..., $A_{k-1}$ are finite. We will prove the bound $A_k \leq 2 \sum_{i=1}^{k-1} A_i A_{k-i}$.
Let $p_1 < p_2 < p_3 < \cdots$ be the prime numbers. So
$$\sum_{n=2}^{\infty} |f_{k,n}| = \sum_{i=1}^{\infty} |f_{k,p_i}| + \sum_{i=1}^{\infty} \sum_{m=p_i +1}^{p_{i+1}-1} |f_{k,m}| \qquad (\clubsuit).$$
We can also remove the $|f_{k,p_1}|$ term: Since $p_1=2$ and $f(2)=-x$, we have $f_{k,p_1}=0$ for $k>1$.
Now, for $i \geq 2$, we have
$$f_{k,p_i} = - \sum_{m=1}^{p_i-1} f_{k,m} = - \sum_{m=p_{i-1}+1}^{p_i-1} f_{k,m} \ \text{since} \ \sum_{m=1}^{p_{i-1}} f_{k,m}=0.$$
So
$$|f_{k,p_i}| \leq \sum_{m=p_{i-1} +1}^{p_i-1} |f_{k,m}|.$$
We can thus turn the first sum in $(\clubsuit)$ into a copy of the second sum, giving:
$$A_k  \leq 2 \sum_{i=1}^{\infty} \sum_{m=p_i +1}^{p_{i+1}-1} |f_{k,m}| = 2 \sum_{m \ \text{composite}} |f_{k,m}| \qquad (\diamondsuit).$$
Now, for each composite $m$, choose a nontrivial factorization $m = a_m b_m$.
Then
$$f_{k,m} = \sum_{i=1}^{k-1} f_{i,a_m} f_{k-i, b_m}.$$
Each pair $(a,b)$ can be $(a_m, b_m)$ for at most one value of $m$ (namely, $ab$), so we have the very sloppy bound:
$$\sum_{m \ \text{composite}} |f_{k,m}| \leq \sum_{a=2}^{\infty} \sum_{b=2}^{\infty} \sum_{i=1}^{k-1}  |f_{i,a}|\ |f_{k-i,b}|  = \sum_{i=1}^{k-1} A_i A_{k-i} \qquad (\heartsuit).$$
Combining $(\diamondsuit)$ and $(\heartsuit)$, we deduce $A_k \leq 2 \sum_{i=1}^{k-1} A_i A_{k-i}$, as promised.
So $A_k \leq B_k$ where $B_k$ is defined by the recursion
$$B_k = 2 \sum_{i=1}^{k-1} B_i B_{k-i} \qquad (\spadesuit)$$ with initial condition $B_1 = A_1 = 1$.
The recursion $(\spadesuit)$ can be solved exactly, to give a power of $2$ times a Catalan number:
$$B_n = 2^n \frac{1}{2n} \binom{2n-2}{n-1}.$$
Bounding $\binom{2n-2}{n-1} \leq 2^{2n-2} = O(4^n)$ gives $B_n = O(8^n)$, as desired. $\square$
A: Put $r=|x_1|$ for simplicity. Here is a significantly shorter proof than my previous one, with the same conclusion: If $|r|<1/8$, then the sum converges (and does so absolutely).  Moreover, we can bound the sum by $r+\tfrac{1}{4} (1 - \sqrt{1-8r})$.
Put $F(N) = \sum_{n=2}^N |f(n)|$. Notice that, if $p<q$ are two consecutive primes, then
$$\sum_{n=1}^p f(n) = \sum_{n=1}^q f(n) = 0$$
so
$$f(q) = - \sum_{n=p+1}^{q-1} f(n).$$
Summing this equation up over all primes up to $N$, we deduce
$$\sum_{q=2,\\ q\ \text{prime}}^n f(q) = f(2) - \sum_{c=4,\\ c \ \text{composite}}^N f(c).$$
Therefore,
$$F(N) = \sum_{n=2}^N |f(n)| \leq |f(2)| + 2 \sum_{c=4,\\ c \ \text{composite}}^N |f(c)| = r + 2 \sum_{c=4,\\ c \ \text{composite}}^N |f(c)|.$$
Now, for $c$ any composite $\leq N$, let $p$ be the least prime divisor of $c$; then $p \leq \sqrt{N}$ and $c/p \leq N/2$. So
$$\sum_{c=4,\\ c \ \text{composite}}^N |f(c)| \leq F(\sqrt{N}) F(N/2)$$
and we deduce that
$$F(N) \leq r + 2 F(\sqrt{N}) F(N/2). \qquad (\ast)$$
Now note that, if $y \leq \tfrac{1}{4} (1 - \sqrt{1-8r})$, then $r+2y^2$ is also $\leq \tfrac{1}{4} (1 - \sqrt{1-8r})$. So $(\ast)$ allows us to inductively bound $F(N) \leq  \tfrac{1}{4} (1 - \sqrt{1-8r})$ for all $N$. Remembering that $|f(1)|=r$, we get the bound $r+\tfrac{1}{4} (1 - \sqrt{1-8r})$ for the sum starting at $n=1$.
