Gilbreath's conjecture states that when we take the absolute values of the consecutive differences of prime numbers, as so :
$$ 2, 3, 5, 7,11,13,17,19,23,... \\ 1, 2, 2, 4, 2, 4, 2, 4,... \\ 1, 0, 2, 2, 2, 2, 2,... \\ 1, 2, 0, 0, 0, 0,... \\ 1, 2, 0, 0, 0,... \\ 1, 2, 0, 0,... \\ (...), $$
we will get only 1s in the first column (if we ignore the first term that is a $2$).
My interest lies in the second column, where the $3$ at the beginning is ignored. The conjecture implies that should always be either a $0$ or a $2$. The sequence created by those numbers (A089582 in the OEIS) is the following :
$$ 2, 0, 2, 2, 2, 2, 2, 2, 0, 0, 0, 0, 0, 0, 2, 2, 0, 2, 2, 0, 0, 2, 2, 2, 0, 0, 0, 2, 2, 0, 2, 0, 0, 0, 2, 2, 0, 0, 0, ... $$
We see that if we subtract $1$ from each term, we get a supposedly infinite sequence of $1$s and $-1$s at approximately equal proportions : $$ 1, -1, 1, 1, 1, 1, 1, 1, -1, -1, -1, -1, -1, -1, 1, 1, ... $$ Let $\Delta_\pi(n)$ denote the $n$th term of that sequence (because we can), do the following sums converge? If so, for what values of $s$ and what is their respective values?
$$\sum_{n\in\mathbb{N}} \frac{\Delta_\pi(n)}{n^s}$$ $$\sum_{p\mathrm{\ is\ prime}} \frac{\Delta_\pi(p)}{p^s}$$ $$\sum_{n\in\mathbb{N}} \frac{\Delta_\pi(n)}{(p(n))^s}$$ NB: $p(n)$ denotes the $n$ prime number