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}