This question is an extension of this question, but more specific.
This paper E. Bradlow et al is a Weibull counting model which I am using to estimate how many failures will happen between preventive maintenance schedules (my data is the system's mean time to failure that is Weibull-distributed with shape parameter less than one).
In page 12, it shows the derivation of the final results:
$$\begin{align} \Pr(N(t)=n) &= \sum_{j=n}^\infty{\frac{(-1)^{j+n}(\lambda t^c)^j \alpha_j^n}{\Gamma(cj+1)}} \\ E(N(t)) &= \sum_{n=1}^\infty{\sum_{j=n}^\infty{\frac{n(-1)^{j+n}(\lambda t^c)^j \alpha_j^n}{\Gamma(cj+1)}}} \end{align}$$
Where $Pr(N(t) = n)$ is the probability that the count will be $n$, and $E(N(t))$ is the expected value. What I'm having trouble understanding is the vector $\alpha^n_j$, which has the following definitions:
$$\begin{align} \alpha_m^l &= \sum_{m=0}^{l-1}{\frac{\Gamma(cm+1)\Gamma(cl-cm+1)}{\Gamma(m+1)\Gamma(l-m+1)}} \\ \alpha_j^0 &= \frac{\Gamma(cj+1)}{\Gamma(j+1)}, \quad j=0,1,2,\ldots \\ \alpha_j^{n+1} &= \sum_{m=n}^{j-1}{\alpha_m^n \frac{\Gamma(cj-cm+1)}{\Gamma(j-m+1)}} \end{align}$$
I understand $\alpha_j^0$ and $\alpha_j^{n+1}$, but $\alpha_j^{n+1}$ uses $\alpha_m^n$ which is my problem. I don't understand that at all. The external subscript is $m$, but internally, $m$ is the one that increments. So what does that mean? Is the $m$ subscript just a marker if which formula to use, without need for a value?
But if so, then in the formula $\alpha_j^{n+1} = \sum_{m=n}^{j-1}{\alpha_m^n \frac{\Gamma(cj-cm+1)}{\Gamma(j-m+1)}}$, since the value of $m$ doesn't matter for $\alpha_m^n$ and since $n$ is already fixed (it doesn't increment), then it's a constant. Does this mean I can take $\alpha_m^n$ out the summation to make it
$$\alpha_j^{n+1} = \alpha_m^n\sum_{m=n}^{j-1}{ \frac{\Gamma(cj-cm+1)}{\Gamma(j-m+1)}}$$
This looks a little wrong to me due to the subscripts and I must be misunderstanding something. Any hints will be greatly appreciated!
