Who's up to the challenge? 37

Using the generating function of the Harmonic numbers $\sum_{n=1}^\infty H_n x^n \; = \; - \frac{\ln (1-x)}{1-x} \qquad \qquad |x| < 1\;,$ we can write $\sum_{k=1}^\infty \sum_{n=1}^\infty (-1)^{k+n} \frac{H_k H_n}{k+n}a^{k+n} \; = \; \int_0^a \frac{\ln^2(1+x)}{(1+x)^2x}\,dx$ for any $0 < a < 1$ , and hence that $\begin{array}{rcl} \displaystyle \lim_{a \to 1-}\sum_{k=1}^\infty \sum_{n=1}^\infty (-1)^{k+n} \frac{H_k H_n}{k+n}a^{k+n} & = & \displaystyle \int_0^1 \frac{\ln^2(1+x)}{(1+x)^2x}\,dx \\ & = & \displaystyle \int_0^1 \ln^2(1+x)\left( \frac{1}{x} - \frac{1}{x+1} - \frac{1}{(x+1)^2}\right)\,dx \\ & = & \displaystyle \int_1^2 \ln^2x\left(\frac{1}{x-1} - \frac{1}{x} - \frac{1}{x^2}\right)\,dx \\ & = & \displaystyle \sum_{n=3}^\infty \int_1^2 x^{-n} \ln^2 x\,dx \end{array}$ A couple of rounds of integration by parts gives $\int_1^2 x^{-n} \ln^2x\,dx \; = \; \frac{2 - 2^{2-n} - (n-1)(\ln 2)2^{2-n} - (n-1)^2(\ln2)^22^{1-n}}{(n-1)^3}$ for all $n \ge 3$ , and hence $\lim_{a \to 1-}\sum_{k=1}^\infty \sum_{n=1}^\infty (-1)^{k+n} \frac{H_k H_n}{k+n}a^{k+n} \; = \; \tfrac14\zeta(3) - \tfrac13(\ln2)^3 + \tfrac12(\ln2)^2 + \ln2 - 1$ leading to the answer $3 + 4 + 3 + 2 + 3 + 2 + 2 + 2 + 1 + 2 + 1 - 1 \,=\,\boxed{24}$ .