Let $P_k$ be the set of all $n\in\mathbb N$ such that $n^1,n^2,\dots,n^k$ have no consecutive equal digits.
In other words, their decimal expansion does not contain '$00$', '$11$' ,'$22$', $\dots$, '$88$' or '$99$'.
What is the largest $k$ such that $P_k$ is infinite?
It is trivial that $k\ge 1$ since $\{1,101,10101,\dots\}\subset P_1$.
That is, $a_t=\frac{1}{99}(100^t-1)\in P_1$ gives infinitely many $P_1$ examples.
I can extend this and show $k\ge 2$. That is, $b_t$ gives infinitely many $P_2$ examples:
$$ b_t= \begin{cases} \frac{27}{99}(100^t-1), & t \not\equiv 3 \bmod{11}\\ \frac{72}{99}(100^t-1), & t \equiv 3 \bmod{11}\\ \end{cases}. $$
I.e., $b_t$ gives a $P_2$ example for every $t$. We should be able to prove this by induction.
Can we do better than this? Can we prove $P_k$ is infinite, for any larger $k$?
I found the $P_2$ family of examples by trying to find $c$ such that $(c\cdot a_t) \in P_2$.
It turns out that $c=27$ works except when $t \equiv 3 \bmod{11}$, in which case $c=72$ works.
This trick does not work for $k\ge 3$: such smallest $c$ gets much larger for every next $t$.
I suspect that $k=6$ is the largest such $k$.
Up to $10^8$, I have that $P_6$ contains:
$$ 1, 2, 3, 4, 5, 13, 17, 28, 37, 51, 72, 91, 161, 291, 504, 713, 905, 1525, 2695, 4975, 5937, 17148, 24301, 24875, 36918, 51797, 53504, 65467, 85683, 86825, 138783, 142905, 156918, 242398, 307696, 493869, 514175, 728347, 830563, 2390369, 2535125, 2720898, 3989203, 4976072, 5236345, 6351925, 8073505, 8090831, 13812725, 13819025, 14319398, 17201918, 29120909, 30835073, 35861681, 37919275, 43604648, 45657604, 67397852, 87624372, 89501213, 90836895, 96397278, \dots $$
But for $P_7$, I have only found
$$1, 2, 3, 4, 5, 13, 17, 28, 91, 161, 713, 5236345$$
so far (up to $10^8$).
Can we prove $P_6$ is infinite and $P_7$ is finite?
Can we improve the lower bound $k\ge 2$ or find some upper bound $k\lt k_0$ ?
The only similar question I found here is: "Are there infinite many squares with a decimal expansion having no consecutive equal digits?", which only requires for $n^2$ to satisfy the property - but I already have a stronger result for $\{n^1,n^2\}$.
Motivation for this question is me trying get some basic insight into "Can we reach infinity if we remove repeating digits from $a_k=m\cdot a_{k-1}$?" for $b=10$.