Pascal-ish Triangle

We are considering the standard Euler sum $\sigma_h(1,n) \; = \; \sum_{k=1}^\infty H_k (k+1)^{-n}$ for $n \ge 2$ , so that $S_n \; = \; \sum_{k=1}^\infty H_k k^{-n} \; = \; \sigma_h(1,n) + \zeta(n+1)$ Now the standard Euler sum is $\sigma_h(1,n) \; = \; \tfrac12n\zeta(n+1) - \tfrac12\sum_{k=1}^{n-2} \zeta(n-k)\zeta(k+1)$ and so $S_n \; = \; \tfrac12(n+2)\zeta(n+1) - \tfrac12\sum_{k=1}^{n-2} \zeta(n-k)\zeta(k+1)$ In particular, $S_3 \; = \; \tfrac52\zeta(4) - \tfrac12\zeta(2)^2 \; = \; \tfrac{1}{36}\pi^4 - \tfrac{1}{72}\pi^4 \; = \; \tfrac{1}{72}\pi^4$ making the answer $\boxed{72}$ .