Question 69
For $n \geq 1$, let $a_{n}$ be the largest odd divisor of $n$, and let $b_{n}=a_{1}+a_{2}+\cdots+a_{n}$. Prove that $b_{n} \geq \frac{n^{2}+2}{3}$, and determine for which $n$ equality holds. For example,
$$
a_{1}=1,\ a_{2}=1,\ a_{3}=3,\ a_{4}=1,\ a_{5}=5,\ a_{6}=3
$$
thus
$$
b_{6}=1+1+3+1+5+3=14 \geq \frac{6^{2}+2}{3}=12 \frac{2}{3} .
$$
$$
a_{1}=1,\ a_{2}=1,\ a_{3}=3,\ a_{4}=1,\ a_{5}=5,\ a_{6}=3
$$
thus
$$
b_{6}=1+1+3+1+5+3=14 \geq \frac{6^{2}+2}{3}=12 \frac{2}{3} .
$$