Warming up summation

Let $\displaystyle \text{S} = \sum_{n=1}^{\infty} \sum_{k=n+1}^{\infty} \dfrac{H_{n}}{k^6} = \sum_{n=1}^{\infty} \sum_{k=n}^{\infty} \dfrac{H_{n}}{k^6} - \sum_{n=1}^{\infty}\dfrac{H_{n}}{n^6}$

Since $\displaystyle \sum_{i=k}^{n} \sum_{j=k}^{i} a_{i,j} = \sum_{j=k}^{n} \sum_{i=j}^{n} a_{i,j}$ , we have,

$\displaystyle \text{S} = \sum_{k=1}^{\infty} \sum_{n=1}^{k} \dfrac{H_{n}}{k^6} - \sum_{n=1}^{\infty} \dfrac{H_{n}}{n^6}$

Note that $\displaystyle \sum_{n=1}^{k} H_{n} = (k+1)H_{k} - k$

$\displaystyle \implies \text{S} = \sum_{k=1}^{\infty} \dfrac{(k+1)H_{k} -k}{k^6} - \sum_{n=1}^{\infty} \dfrac{H_{n}}{n^6}$

$\displaystyle = \sum_{k=1}^{\infty} \dfrac{H_{k}}{k^5} - \sum_{k=1}^{\infty} \dfrac{1}{k^5}$

$\displaystyle = \dfrac{\pi^6}{540} - \dfrac{1}{2} \zeta(3)^2 - \zeta(5)$

where the last equality follows from Euler's Sum.

$\implies \text{Ans.} = \boxed{556}$

The sum is $\begin{array}{rcl} \displaystyle S \; = \; \sum_{n=1}^\infty \sum_{k=n+1}^\infty \frac{H_n}{k^6} & = & \displaystyle \sum_{n=1}^\infty \sum_{k=n+1}^\infty \sum_{m=1}^n \frac{1}{mk^6} \; = \; \sum_{m=1}^\infty \sum_{k=m+1}^\infty \sum_{n=m}^{k-1}\frac{1}{mk^6} \\ & = & \displaystyle \sum_{m=1}^\infty \sum_{k=m+1}^\infty \frac{k-m}{mk^6} \; = \; \sum_{k=2}^\infty \sum_{m=1}^{k-1}\left(\frac{1}{mk^5} - \frac{1}{k^6}\right) \\ & = & \displaystyle \sum_{k=2}^\infty \frac{H_{k-1}}{k^5} - \sum_{k=2}^\infty \frac{k-1}{k^6} \; = \; \sum_{k=1}^\infty \frac{H_k}{(k+1)^5} - \zeta(5) + \zeta(6) \\ & = & \displaystyle \sum_{k=1}^\infty \left(\frac{H_{k+1}}{(k+1)^5} - \frac{1}{(k+1)^6} \right) - \zeta(5) + \zeta(6) \; = \; \sum_{k=1}^\infty \frac{H_k}{k^5} - \zeta(5) \; = \; S_{1,5} - \zeta(5) \end{array}$ Using Euler's sum $S_{1,q} \; = \; \sum_{k=1}^\infty \frac{H_k}{k^q} \; = \; (1 + \tfrac12q)\zeta(q+1) - \tfrac12\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$ for any $q \ge 2$ , we deduce that $S \; = \; \tfrac72\zeta(6) - \tfrac12\big[\zeta(2)\zeta(4) + \zeta(3)^2 + \zeta(4)\zeta(2)\big] - \zeta(5) \; = \; \frac{\pi^6}{540} - \tfrac12\zeta(3)^2 - \zeta(5)$ making the answer $6 + 540 + 3 + 2 + 5 = \boxed{556}$ .