TITLE: Two sequences $a_{2n}=a_n+1, a_{2n+1}=a_{n}+2$ and $b_{3n}=b_n+1, b_{3n+1}=b_n+2, b_{3n+2}=b_n+3$
QUESTION [5 upvotes]: Let us consider two  sequences $$a_{2n}=a_n+1, a_{2n+1}=a_{n}+2, a_1=1,a_2=2$$ and  $$b_{3n}=b_n+1, b_{3n+1}=b_n+2, b_{3n+2}=b_n+3, b_1=1,b_2=2,b_3=2.$$
Prove that $a_{2^n} < b_{2^n}$ for $n>3.$
My attempts to prove it by mathematical induction were without success.

REPLY [2 votes]: Write $N$ in binary (base 2).  Chop digits from the end, one at a time.  Every time, you add $1+digit$ to the final value.
So $a_n=$(number of digits) + (sum of digits)-1 in binary.
Similarly, $b_n=$(number of digits)+(sum of digits)-1 in base 3.
On the other hand, I don't know why $a_{2^n}=n+1$ would be less than $b_{2^n}$, when $2^n$ has fewer digits in base 3.