This is more a puzzle than a research question, a puzzle to me. Perhaps it is straightforward for others.
Imagine Repeatedly interpreting a number expressed with the usual base-$10$ digits as "digits" multiplying powers of $2$ rather than powers of $10$. For example, interpret $n=27$ as $2\cdot 2^1 + 7\cdot 2^0 = 11$. Call this mapping $f(n)$. $f(n)=n$ when $n$ is a single (decimal) digit.
Let $g(n)$ be repeated application of $f(n)$, $g(n) = f^k(n)$ until a single digit is reached. For example, for $g(17355)=3$: $$ 1\cdot 2^4 + 7\cdot 2^3 + 3\cdot 2^2 + 5\cdot 2^1 + 5\cdot 2^0 = (1,7,3,5,5)\cdot(16,8,4,2,1) = 99 $$ $$ (9,9)\cdot(2,1) = 27 $$ $$ (2,7)\cdot(2,1) = 11 $$ $$ (1,1)\cdot(2,1) = 3 $$
$g(17356)=4$: $$ (1,7,3,5,6)\cdot(16,8,4,2,1) = 100 $$ $$ (1,0,0)\cdot(4,2,1) = 4 $$
It is easy to see that $g(n)$ is well-defined, in that it does eventually reach a single digit: $f(n) < n$ for all $n \ge 10$ because $2^k < 10^k$ for $k \ge 1$.
It appears that $g(n)$ is essentially $({n}\mod 8)$. More precisely, let $m= ({n}\mod 8)$. Then I think that: $$g(n) = m \;\; \mathrm{if} \;\; m > 1$$ $$g(n) = m+8 \;\; \mathrm{if} \;\; m \le 1$$ But I do not see a proof.
Q1. Is the above mod-$8$ formula for $g(n)$ correct? Is there a simple proof?
Q2. What is the generalization with base $10$ and base $2$ replaced with base $b_1$ and $b_2$, $b_1 > b_2$?
