Evaluate $\frac{\partial^2}{\partial t^2} \left[ \prod_{j=1}^k (1+t+\dots+t^{d_j -1}) \right]$ at $t=1$ I need to find a "nice" formula for the evaluation of 
$\frac{\partial^2}{\partial t^2} \left[ \prod_{j=1}^k (1+t+\dots+t^{d_j -1}) \right]$ at t=1, where $d_j \in \mathbb{N}$.

I have already proved that 
\begin{equation}
\frac{\partial}{\partial t} \left[ \prod_{j=1}^k (1+t+\dots+t^{d_j -1}) \right]_{| t=1} =\frac{1}{2} \cdot \prod_{i=1}^k d_i \cdot \left[ \sum_{i=1}^k d_i -k \right]
\end{equation}

and now I'm looking for a similar concise formula for the second derivative at t=1.
Let $$f_i(x)=\sum_{n=0}^{d_i-1}x^n=1+x+\cdots+x^{d_i-1}$$ and let $$\Psi=\prod_{j=1}^kf_j(t)$$
We have that $$\Phi=\frac\partial{\partial t}\Psi=\Psi\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)$$ Therefore $$\frac{\partial^2}{\partial t^2}\Psi=\frac\partial{\partial t}\Phi=\Psi'\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)+\Psi\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)=\Psi\left(\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)^2+\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)'\right)$$
Finally after simplification
$$\Psi\left(\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)^2+\sum^k_{j=1}\frac{f_j(t)f_j''(t)-f_j'(t)^2}{f_j(t)^2}\right)$$
The rest is just computations of the derivative at of $f_j$ at $t=1$ which should simplify nicely in terms of $d_j$.



Further Calculations
$$f_j(1)=d_j$$
$$f_j'(1)=\frac{d_j(d_j-1)}{2}$$
$$f_j''(1)=\frac{d_j(d_j-1)(d_j-2)}{3}$$
Thus you may finally plug and chug.



After the plug and chug I get $$\frac{\partial^2}{\partial t^2}\Psi=\left(\prod^k_{j=1}d_j\right)\left(\left(\sum^k_{j=1}\frac{d_j-1}2\right)^2-\sum^k_{j=1}\frac{(d_j-1)(d_j-5)}{2}\right)$$
I've cheated, I admit:

$$\frac{\partial ^2\left(\prod_j^k \sum _{d=0}^j t^d\right)}{\partial t^2}$$
at $t=1$ seems to equal  

$$\frac{1}{144} (k-1) k (k+1) (9 k+22) (k+1)!$$

I can show you the cheat, but cannot provide a proof.