Definitions: let $n \in \mathbb{N}_{>0} \cup \{ \infty \}$ and let $E_n$ be the set of sequences $(d_i)_{i=1}^n$ such that $d_1=1$, $d_i$ is an even integer (for $i > 1$) and $0<d_i \le i$. Let $d_i^{r+1} = \vert d_{i+1}^{r}-d_{i}^{r} \vert $, with $d_i^{1} = d_i$, $r<n$ and $1 \le i \le n-r+1$. The sequence $(d_i)_{i=1}^n$ is a Gilbreath sequence if $\forall r \le n $ then $d_1^{r} = 1$.
Examples: $(1,2,2,a,b,c) \in E_6$ is a Gilbreath sequence $\forall a,b,c \in \{ 2,4 \}$.
$(1,2,2,2,2,6) \in E_6$ is not a Gilbreath sequence, because $d_1^6 = 3$ (it's the first counter-example).
Gilbreath's conjecture states that the sequence $(g_i)_{i=1}^{\infty}$ of difference of consecutive prime numbers (i.e. $g_i = p_{i+1} - p_i$) is a Gilbreath sequence (we assume that $g_i \le i$, see here).
Definitions: let $n \in \mathbb{N}_{>0}$, $G_n = \{ s \in E_n \ \vert \ s \text{ is a Gilbreath sequence}\}$ and $\alpha_n = \frac{\#G_n}{\#E_n}$.
The sequence ($\alpha_n$) is decreasing and bounded below, so convergent.
Let $\alpha = lim_{n \to \infty} (\alpha_n)$. Let's call $\alpha$ the Gilbreath ratio.
Remark: $\#E_{2n+1} = (n!)^2 $ and $\#E_{2n+2} = n! (n+1)!$
Questions: $\alpha > 1/2$? $\alpha \ge 4/5$? What are the first (five) digits of $\alpha$? Is there a formula for $\#G_n$?
$\normalsize{ \begin{array}{c|c} n &5&6&7&8&9&10&11&12 \newline \hline \#G_n &4&11&32&124&492&2433&12076&72010 \newline \hline \#E_n &4&12&36&144&576&2880&14400&86400 \newline \hline \alpha_n \simeq &1&0.9167&0.8889&0.8611&0.8542&0.8448&0.8386&0.8334 \end{array} } $
$\normalsize{ \begin{array}{c|c} n &13&14&15&16&17&18 \newline \hline \#G_n &430942&3009200&21032105&167985502&1342885879&12074497923 \newline \hline \#E_n &518400&3628800&25401600& 203212800&1625702400&14631321600 \newline \hline \alpha_n \simeq &0.8313&0.8292&0.8280&0.8266&0.8260&0.8252 \end{array} } $
Some "data" given by the Monte Carlo method:
$\alpha_{100} \simeq 0.82270 \pm 1.2 \cdot 10^{-4}$
$\alpha_{200} \simeq 0.82266 \pm 7 \cdot 10^{-5}$
$\alpha_{300} \simeq 0.82264 \pm 6 \cdot 10^{-5}$
$\alpha_{10^3} \simeq 0.8215 \pm 1.3 \cdot 10^{-3}$
$\alpha_{10^4} \simeq 0.820 \pm 5 \cdot 10^{-3}$
$\alpha_{10^5} \simeq 0.81 \pm 4 \cdot 10^{-2}$
