, 2013

Solution:

We have the following:

- $b(n) = 1$ for $n$ between $1$ and $3$.
- $b(n) = 3$ for $n$ between $4^{2} - 3 \cdot 4 = 4$ and $3 \cdot 4^{2} + 3 = 51$. (Since $a \cdot 4^{2} - b \cdot 4 + c$ takes on $3 \cdot 4 \cdot 4$ distinct values over $1 \leq a \leq 3$, $0 \leq b \leq 3$, $0 \leq c \leq 3$, with minimum $4$ and maximum $51$.)
- $b(n) = 5$ for $n$ between $4^{4} - 3 \cdot 4^{3} - 3 \cdot 4 = 52$ and $3 \cdot 4^{4} + 3 \cdot 4^{2} + 3 = 819$.
- $b(n) = 7$ for $n$ between $4^{6} - 3 \cdot 4^{5} - 3 \cdot 4^{3} - 3 \cdot 4^{1} = 820$ and $3 \cdot 4^{6} + 3 \cdot 4^{4} + 3 \cdot 4^{2} + 3 > 2013$.

Thus

$$\sum_{i=1}^{2013} b(i) = 7 \times 2013 - 2 \times (819 + 51 + 3) = 14091 - 2 \times 873 = 14091 - 1746 = 12345.$$