= Solution
Use the recurrence exactly as printed. Set $B=A^*A$ and $R=I-B/\alpha$. It gives
$$
x_{n+1}=Rx_n+\frac1\alpha A^*y,\qquad
x_0=(B+\alpha I)^{-1}A^*y.
$$
Repeated substitution yields a function of $A^*y$ alone:
$$
\boxed{x_n=R^n(B+\alpha I)^{-1}A^*y+
\frac1\alpha\sum_{j=0}^{n-1}R^jA^*y.}
$$
The empty sum is zero for $n=0$. This is <Landweber iteration with a Tikhonov initial value>. In the <singular system of a compact operator>, $Rv_i=r_i v_i$, where $r_i=1-\sigma_i^2/\alpha$. Thus
$$
x_n=\sum_i\left[
r_i^n\frac{\sigma_i}{\sigma_i^2+\alpha}
+\frac{\sigma_i}{\alpha}\sum_{j=0}^{n-1}r_i^j\right](y,u_i)v_i.
$$
For every positive <singular value>, the finite geometric sum gives
$$
\frac{\sigma_i}{\alpha}\sum_{j=0}^{n-1}r_i^j
=\frac{1-r_i^n}{\sigma_i}.
$$
Therefore \b[the <closed form of a Tikhonov-initialized Landweber iterate> is]
$$
\boxed{x_n=\sum_i g_{\alpha,n}(\sigma_i)(y,u_i)v_i,\qquad
g_{\alpha,n}(\sigma)=\frac1{\sigma}\left[
1-\frac{\alpha}{\alpha+\sigma^2}\left(1-\frac{\sigma^2}{\alpha}\right)^n
\right].}
$$
In the printed notation, $g_\alpha(y,u_i)$ means the coefficient $g_{\alpha,n}(\sigma_i)(y,u_i)$; its dependence on $n$ and $\sigma_i$ is implicit. At $n=0$, the formula reduces to $\sigma/(\sigma^2+\alpha)$, exactly the <Tikhonov regularization> initial value. At fixed $n$, its apparent singularity at $\sigma=0$ is removable:
$$
g_{\alpha,n}(\sigma)=\frac{(n+1)\sigma}{\alpha}+O(\sigma^3).
$$
Thus each fixed iterate is a bounded reconstruction operator and the series converges in norm.
The terminology in the question requires care. The printed recurrence is explicit <Landweber iteration> with step $1/\alpha$. Convergence of its nonzero modes requires $|1-\sigma_i^2/\alpha|<1$; a sufficient uniform step condition is
$$
\boxed{\alpha>\frac{\|A\|^2}{2}.}
$$
With exact compatible data and this condition, $x_n\to x^\dagger$. At equality the largest singular-value mode can alternate, and smaller $\alpha$ can produce growing modes. With noisy data, <early stopping of Landweber iteration> is needed to prevent recovery of arbitrarily unstable small-singular-value components.
Conventional <Iterated Tikhonov regularization> instead uses the implicit update
$$
(B+\alpha I)x_{n+1}=\alpha x_n+A^*y.
$$
With the same <Tikhonov regularization> initial value its filter would be $[1-(\alpha/(\alpha+\sigma^2))^{n+1}]/\sigma$, not the filter derived above. The solution here retains the authoritative printed update rather than silently changing it.
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