Look at the number of divisors for $n=p \pm 2$, a prime $p$ plus or minus $2$: $\tau(p+2)$ and $\tau(p-2)$. Define a function $\Delta(k)$ to be $\{+1,0,-1\}$ if, for $p$ the $k^{\mathrm{th}}$ prime, $\tau(p+2)$ is $\{>,=,<\}$ $\tau(p-2)$ respectively.
For example, the $k=10^{\mathrm{th}}$ prime is $p=29$, and $p+2=31$ has $2$ divisors while $p-2=27$ has $4$ divisors $\{1,3,9,27\}$. So $\Delta(10) = -1$.
Accumulate $\Delta(k)$, executing what appears to be one-dimensional random walk:
$\max = 113$, $\min = -282$. The $500000^{\mathrm{th}}$ prime is $7368787$.
My question is:
Does this $\Delta(k)$ quantity return to zero infinitely often, like a random walk revisiting the origin?