Reposted to MathOverflow because the bounty on this post expired, with no solutions or comments received.
For any discrete subset $S$ of $\mathbb{R}^d$, consider a digraph formed by placing an edge from each point in $S$ to a nearest neighbor. The outdegree of each vertex is $1$; the indegree is at most the kissing number of dimension $d$. Now consider the random graph process that arises when $S$ is sampled from a Homogeneous Poisson Point Process (HPPP). I suspect it's hard to say much about this process for $d>1$. But for $d=1$ the random digraph may be represented as a doubly infinite string of $0$'s and $1$'s representing left- and right-pointing edges, respectively.
I'm interested in $P(0|x)$, the conditional probability of $0$ given the succeeding string $x$. For example, $P(0|x) = 1/(s+2)$ when $x=000...$ is a run of $s$ $0$'s. In particular, let $a(s) = P(0|x)$ when the string $x=101010...$ has length $s$. The first few values are: $$ a(s)=\frac{1}{2}, \frac{2}{3}, \frac{16}{25}, \frac{61}{96}, \frac{272}{427},\frac{1385}{2176} $$ for $s=0$ to $5$. I computed these values up to $a(36)$ and noticed some intriguing patterns. My question is how to prove what the numerics suggest. The first problem is
(a) Prove that $\underset{s\rightarrow\infty}{\lim} a(s) =2/\pi$.
The convergence is fast. The error $|2/\pi - a(s)|$ falls by a factor of $3$ for each $s$, and it is less than $10^{-18}$ for $s=36$. After finding the correct coefficient for the leading error and subtracting it off, the remaining error falls by a factor of $5$ for each $s$. Pulling this thread to its conclusion yields the following formula for $a(s)$: $$ a(s) = \frac{2}{\pi} \left(1 - \frac{4(-1)^s}{3^{s+3}} - \frac{4}{5^{s+3}} - \frac{8(-1)^s}{7^{s+3}} + \frac{4}{9^{s+3}} - \frac{12(-1)^s}{11^{s+3}} + \cdots\right). $$ With more terms, the pattern in the coefficients becomes evident, leading to the following conjecture:
(b) Prove that $$ a(s) = \frac{2}{\pi} \sum_{j=0}^{\infty} \frac{(-1)^{js}c_{2j+1}}{(2j+1)^{s+3}} $$ where $c_n$ is a multiplicative function which is zero for even $n$ and $$ c_{p^{\alpha}} = -4\lfloor p/4\rceil\chi(p)^{\alpha+1}\;\text{ with}\;\;\chi(4k\pm 1)=\pm 1$$ for odd prime $p$ and $\alpha \ge 1$. Here $\lfloor x\rceil$ denotes the nearest integer to $x$.
Alternatively, because $c_n$ is multiplicative, we may write an Euler formula for $a(s)$: $$ a(s) = \frac{2}{\pi} \prod_{\,p>2\\\text{prime}} g\left(s,\chi(p)p\right)\;\;\text{where}\;\; g(s,p) \doteq \frac{p^{s+3}-p}{p^{s+3}-1}. $$