Inspired by Patrick Corn

Since $\sum_{n=1}^\infty H_n x^n \; =\; - \frac{\ln(1-x)}{1-x} \qquad \qquad |x| < 1$ we deduce that $\begin{array}{rcl} \displaystyle \sum_{n=1}^\infty \frac{H_n}{n2^n} & = & \displaystyle -\int_0^{\frac12} \frac{\ln(1-x)}{x(1-x)}\,dx \; = \; -\int_{\frac12}^1 \frac{\ln x}{x(1-x)}\,dx \\ & = & \displaystyle -\sum_{n=0}^\infty \int_{\frac12}^1 x^{n-1} \ln x\,dx \end{array}$ and since $\int_{\frac12}^1 x^{n-1} \ln x\,dx \; = \; \left\{ \begin{array}{lll} \displaystyle -\tfrac12(\ln 2)^2 & \qquad & n = 0 \\ \displaystyle-\frac{1}{n^2} + \frac{\ln 2}{n\,2^n} + \frac{1}{n^2\,2^n} & & n \ge 1 \end{array} \right.$ we deduce that $\begin{array}{rcl} \displaystyle \sum_{n=1}^\infty \frac{H_n}{n2^n} & = & \displaystyle \tfrac12(\ln 2)^2 + \sum_{n=1}^\infty\left\{ \frac{1}{n^2} - \frac{\ln 2}{n\,2^n} - \frac{1}{n^2\,2^n}\right\} \\ & = & \displaystyle \tfrac12(\ln 2)^2 + \zeta(2) - (\ln 2)^2 - \mathrm{Li}_2(\tfrac12) \\ & = & \zeta(2) - \tfrac12(\ln2)^2 - \tfrac{1}{12}\pi^2 + \tfrac12(\ln2)^2 \; =\; \tfrac{1}{12}\pi^2 \end{array}$ making the answer $2 + 12 = \boxed{14}$ .