The trimorphic numbers are integers whose cubes end in the digits of the integers themselves, such as ${\sf{49}}^3=1176\sf{49}$, and I have discovered something interesting about such integers that end in $9$ and $1$.
The OEIS sequence A224473 is a sequence of trimorphic numbers congruent to $9\pmod{10}$ and has formula $a_1(n)=2\cdot5^{2^n}-1\pmod{10^n}$ for a natural number $n$. The sequence is $\{9,49,249,1249,\cdots\}$. If we denote $b_1(n)=a_1(n)^2-1$, we see the following. $$b_1(1)=80\\b_1(2)=2400\\b_1(3)=62000\\b_1(4)=12560000\\\cdots$$ That is, $b_1(n)\equiv0\pmod{10^n}$.
Furthermore, A224474 is a sequence of trimorphic numbers congruent to $1\pmod{10}$ with formula $a_2(n)=2\cdot16^{5^n}-1\pmod{10^n}$, and the first few values are $\{1,51,751,8751,\cdots\}$. If we denote $b_2(n)=a_2(n)^2-1$, we observe a similar thing. $$b_2(1)=0\\b_2(2)=2600\\b_2(3)=564000\\b_2(4)=76580000\\\cdots$$ That is, $b_2(n)\equiv0\pmod{10^n}$.
Questions.
How can it be proved that $b_1(n)$ and $b_2(n)$ are divisible by $10^n$?
Is the fact that they are all trimorphic numbers just a coincidence, or does this behaviour occur due to that property?
Let $c_2(n)=2 \cdot 16^{5^n}$. We define: $$d_2(n)=c_2(n)^2-1=(c_2(n)+1)(c_2(n)-1)=4\cdot16^{5^n}\cdot(16^{5^n}-1)$$ Clearly, $2^n$ divides the product as we have a $16^{5^n}$ term here. Moreover, using lifting, we can see: $$\nu_5(16^{5^n}-1)=\nu_5(5^n)+1=n+1$$ Thus, $2^n \cdot 5^n$ divides $d_2(n)$. Now: $$10^n \mid c_2(n)^2 \implies 10^n \mid (c_2(n)-k\cdot10^n)^2\implies 10^n \mid a_2(n)^2-1 \implies 10^n \mid b_2(n)$$ The same lifting can be done for the other case.
These values are set so that we can use lifting to impose the divisibility by $10^n$ condition. This gives them the trimorphic property.
Note: $\nu_p$ denotes the power of prime $p$.