jmc

We have that $P(2^n)=n$ for $0\le n\le 2011.$

Let $Q(x)=P(2x)-P(x)-1.$ Then $$\begin{array}{rl}Q(2^n)& =P(2^{n+1})-P(2^n)-1\\ & =n+1-n-1\\ & =0\end{array}$$for $0\le n\le 2010.$ Since $Q(x)$ has degree 2011, $$Q(x)=c(x-1)(x-2)(x-2^2)\cdots (x-2^{2010})$$for some constant $c.$

Also, $Q(0)=P(0)-P(0)=-1.$ But $$Q(0)=c(-1)(-2)(-2^2)\cdots (-2^{2010})=-2^{1+2+\cdots +2010}c=-2^{2010\cdot 2011/2}c,$$so $c=\frac{1}{2^{2010\cdot 2011/2}},$ and $$Q(x)=\frac{(x-1)(x-2)(x-2^2)\cdots (x-2^{2010})}{2^{2010\cdot 2011/2}}.$$Let $$P(x)=a_{2011}{x}^{2011}+a_{2010}{x}^{2010}+\cdots +a_{1}x+a_0.$$Then $$P(2x)=2^{2011}{a}_{2011}{x}^{2011}+2^{2010}{a}_{2010}{x}^{2010}+\cdots +2a_{1}x+a_0,$$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 $$Q(x)=(x-1)\left(\frac{1}{2}x-1\right)\left(\frac{1}{2^2}x-1\right)\cdots \left(\frac{1}{2^{2010}}x-1\right).$$The coefficient of $x$ in $Q(x)$ is then $$\begin{array}{rl}1+\frac{1}{2}+\frac{1}{2^2}+\cdots +\frac{1}{2^{2010}}& =\frac{1+2+2^2+\cdots +2^{2010}}{2^{2010}}\\ & =\frac{2^{2011}-1}{2^{2010}}\\ & =2-\frac{1}{2^{2010}}.\end{array}$$The final answer is then $2+2+2010=\boxed{2014}.$