While I was submitting a few sequences to the OEIS, I noticed an asymmetrical pattern involving the rightmost digits of an interesting set of well-known integer sequences.
Let $a \in \mathbb{Z}^+$, $n \in \mathbb{N}_0$, and assume that $[n]$ indicates the $n$-th hyperoperator (i.e., $[n]=[0] \Rightarrow$ zeration, $[n]=[1] \Rightarrow$ addition, $[n]=[2] \Rightarrow$ multiplication, $[n]=[3] \Rightarrow$ exponentiation, $[n]=[4] \Rightarrow$ tetration, $[n]=[5] \Rightarrow$ pentation, and so forth).
Now, since we are also assuming radix-$10$ (the usual decimal numeral system), it follows that
$a[0]a \pmod {10} = 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, \dots$ has a period length $10$.
$a[1]a \pmod {10} = 2, 4, 6, 8, 0, 2, 4, 6, 8, 0, 2, 4, 6,\dots$ has a period length $5$.
$a[2]a \pmod {10} = 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, 4, 9, 6, 5, 6,\dots$ has a period length $10$.
$a[3]a \pmod {10} = 1, 4, 7, 6, 5, 6, 3, 6, 9, 0, 1, 6, 3, 6, 5, 6, 7, 4, 9, 0, 1, 4, 7, 6, 5,\dots$ has a period length $20$.
$a[4]a \pmod {10} = 1, 4, 7, 6, 5, 6, 3, 6, 9, 0, 1, 6, 3, 6, 5, 6, 7, 4, 9, 0, 1, 4, 7, 6, 5,\dots$ has a period length $20$.
$\cdots$
$a[k]a \pmod {10} = 1, 4, 7, 6, 5, 6, 3, 6, 9, 0, 1, 6, 3, 6, 5, 6, 7, 4, 9, 0, 1, 4, 7, 6, 5,\dots$ has a period length $20$, for any $k \in \mathbb{N}-\{0,1,2\}$.
Let $b$ be an integer greater than $a$. Then, in radix-$g$, we can say that $(g=10 \wedge n>2) \Rightarrow a[n]b \neq b[n]a$, whereas $(g=10 \wedge n<3) \Rightarrow a[n]b = b[n]a$.
Consequently, given any $n \in \mathbb{N}_0$, $(g=10$ $\wedge$ (last digit period length $> g))\Rightarrow a[n]b \neq b[n]a$ holds for (almost) all $a$ and $b$.
Question: For which values of $g$ the last statement above is also true?