We have that $ P(2^n) = n $ for $ 0 \le n \le 2011 $. Let $ Q(x) = P(2x) - P(x) - 1 $. Then \begin{align*} Q(2^n) &= P(2^{n + 1}) - P(2^n) - 1 \\ &= n + 1 - n - 1 \\ &= 0 \end{align*}for $ 0 \le n \le 2010 $. Since $ Q(x) $ has degree 2011, \[Q(x) = c(x - 1)(x - 2)(x - 2^2) \dotsm (x - 2^{2010})\]for some constant $ c $. Also, $ Q(0) = P(0) - P(0) = -1 $. But \[Q(0) = c(-1)(-2)(-2^2) \dotsm (-2^{2010}) = -2^{1 + 2 + \dots + 2010} c = -2^{2010 \cdot 2011/2} c,\]so $ c = \frac{1}{2^{2010 \cdot 2011/2}}, $ and \[Q(x) = \frac{(x - 1)(x - 2)(x - 2^2) \dotsm (x - 2^{2010})}{2^{2010 \cdot 2011/2}}.\]Let \[P(x) = a_{2011} x^{2011} + a_{2010} x^{2010} + \dots + a_1 x + a_0.\]Then \[P(2x) = 2^{2011} a_{2011} x^{2011} + 2^{2010} a_{2010} x^{2010} + \dots + 2a_1 x + a_0,\]so the coefficient of $ x $ in $ Q(x) $ is $ 2a_1 - a_1 = a_1 $. In other words, the coefficients of $ x $ in $ P(x) $ and $ Q(x) $ are the same. We can write $ Q(x) $ as \[Q(x) = (x - 1) \left( \frac{1}{2} x - 1 \right) \left( \frac{1}{2^2} x - 1 \right) \dotsm \left( \frac{1}{2^{2010}} x - 1 \right).\]The coefficient of $ x $ in $ Q(x) $ is then \begin{align*} 1 + \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{2010}} &= \frac{1 + 2 + 2^2 + \dots + 2^{2010}}{2^{2010}} \\ &= \frac{2^{2011} - 1}{2^{2010}} \\ &= 2 - \frac{1}{2^{2010}}.\end{align*}The final answer is then $ 2 + 2 + 2010 = \boxed{2014} $.