Consider about the 10-adic integer $x = \dots1787109376$ satisfies $x^2 = x$ and the 10-adic integer $y = \dots8212890625$ satisfying $y^2 = y$.
Define $a(n)$ as the $n$th digit of $x$:
$$a(0) = 6, \quad a(1) = 7, \quad a(2)= 3, \quad \dots $$
Define $b(n)$ as the $n$th digit of $y$:
$$b(0) = 5, \quad b(1) = 2, \quad b(3)= 6, \quad \dots $$
Define $c(n) = a(n) - b(n)$. (See this.)
$$c(0) = 1, \quad c(1) = 5, \quad c(3) = -3, \quad \dots $$
Let $d(n) = \sum_{k=0}^n c(k).$
$$d(0) = 1, \quad d(1) = 6, \quad d(3) = 3, \quad \ldots, \\ \quad d(595) = 0, \quad d(596) = 1, \quad d(597) = 8, \quad \ldots $$
I calculated $d(n)$ until $n \leq 15000$ and made the following predictions, is this correct?
Prediction. When $n > 597$, $d(n) < 0$.
P.S.
\begin{align*} d( 1000) &= -181, \\ d( 2000) &= -105, \\ d( 3000) &= -387, \\ d( 4000) &= -121, \\ d( 5000) &= -453, \\ d( 6000) &= -393, \\ d( 7000) &= -523, \\ d( 8000) &= -539, \\ d( 9000) &= -689, \\ d(10000) &= -791, \\ d(11000) &= -635, \\ d(12000) &= -765, \\ d(13000) &= -779, \\ d(14000) &= -717, \\ d(15000) &= -1363. \end{align*}
Given that the digits should behave like random numbers (and hence $d(n)$ would look like a sample path of a symmetric random walk), I am very skeptical about the idea that $d(n)$ will eventually assume a constant sign.
Indeed, using the first $100,000$ digits of $x$ and $y$, the graph of $d(n)$ turns out to be as follows:
(After $n = 597$, the first time $d(n)$ becomes positive is $n = 37790$.)