How to derive a recursive formula from the following formula,
$$ u_{n}=a_{n-1}u_{0}+\sum_{k=1}^{n-1}(a_{n-1-k}-a_{n-k})u_{k}+\Gamma(2-\alpha)h^{\alpha}f(t_{n},u_{n})? $$
P.S.: Consider the following initial value problem (IVP) for $\ 0<\alpha<1$, $$^{c}D_{0}^{\alpha}u(t)=f(t,u(t)),\\\ u(0)=u_{0},$$
where $^{c}D_{0}^{\alpha}$, denotes Caputo derivative and $f:\left[0,T\right]\times D\rightarrow R,\,D\subset R$. Divide the interval $\left[0,T\right]$ into $N$ subintervals. Consider an equispaced grid with step length $h=T/N$ with $t_{k}=kh,$ $k=0,1,...N$.
L1 method (The reference [ 1 ]) is used for the numerical evaluation of the fractional derivatives of order $\alpha \ $, $0<\alpha<1$. The approximate $^{c}D_{0}^{\alpha}u(t)$ by the following formula:
$$ ^{c}D_{0}^{\alpha}u(t)\mid_{t=t_{n}}=\sum_{k=0}^{n-1}b_{n-k-1}\left(u_{k+1}-u_{k}\right)=f(t_{n},u_{n}), $$ where $u_{k}$ denotes the approximate value of the solution of IVP at $t=t_{k}$ and $$ b_{n-k-1}=\frac{h^{-\alpha}}{\Gamma(2-\alpha)}\left((n-k)^{1-\alpha}-(n-k-1)^{1-\alpha}\right). $$
The previous equation can be rewritten as: $$ u_{n}=a_{n-1}u_{0}+\sum_{k=1}^{n-1}(a_{n-1-k}-a_{n-k})u_{k}+\Gamma(2-\alpha)h^{\alpha}f(t_{n},u_{n}), $$ where $a_{k}=(k+1)^{1-\alpha}-k^{1-\alpha}$. Note that: $f,u_{0},\alpha$ is known, $u(t_{k})\approx u_{k}$ and $h=T/N$ with $t_{k}=kh,k=0,1,...N$.
My question: How to derive a recursive formula from the last formula? This is easier and useful for programming.