Let be $\{X_n\}_{n\in \mathbb{N}}$ random walk on $\mathbb{Z}$. Let be
$$P(X_{n+1} = k + a| X_n = k)= p_a$$ for $a\in \mathcal{A} \subset \mathbb{Z}$. Let say that $X_0 = 0$. I am interested in probabilities $$\tilde{p}_k = P(\exists n\in \mathbb{N}_0. X_n =k\;|\; X_0=0)\text{.}$$
If $\mathcal{A} = \{-1,1\}$, then this problem can be solved if one writes down recurrence relation
$$\tilde{p}_k = p_{-1}\tilde{p}_{k+1} +p_1\tilde{p}_{k-1}$$ with $\tilde{p}_0=1$ and solves it by finding $\lambda$'s satisfying
$$\lambda^k = p_{-1}\lambda^{k+1} +p_1\lambda^{k-1}\text{.}$$
If we devide the upper equation by $\lambda^{k-1}$, we get a quadratic equation, which can be easily solved. We should consider $3$ different cases:
$p_1 = p_{-1} \; (\Leftrightarrow \lambda_{1,2} = 1)$
$p_1 > p_{-1}$
$p_{1} < p_{-1}$
In the case 1. ($\lambda_1 = \lambda_2 = 1$) we should (in general) find such constants $A^+$, $B^+$, $A^-$ and $B^-$ that
$$\tilde{p_k} = A^+ + B^+k,\;k\geq 0 $$ $$\tilde{p_k} = A^- + B^-k,\;k\leq 0$$
In the other two cases $\lambda_1\neq \lambda_2$ and it holds
$$\tilde{p}_k = A^+\lambda_1^k + B^+\lambda_2^k,\; k\geq 0$$ $$\tilde{p}_k = A^-\lambda_1^k + B^-\lambda_2^k,\; k\leq 0$$
In the case one, we can conclude from simetry that $A^+ = A^- = A$ and $B^+ = B^- = B$. Since $1\geq \tilde{p}\geq 0$, it follows $B=0$ and $A =\tilde{p}_0 = 1$. It turns out that it is enough to solve case 2 (in the third, we just change the roles of positive and negative numbers). By help of $\tilde{p}_0=1$ and strong law of large numbers, we can determine $A^{+,-}$ and $B^{+,-}$.
Since $\sum p_a = 1$, one $\lambda_i$ is $1$ for every set $\mathcal{A}$. So the problem occurs when we get an equation of a degree $4$ or more (Cardano's and Ferrari's formulas are quite complicated). This happens when
$$M:= \max \{|a_i -a_j|\; \mid a_i,a_j \in \mathcal{A}\}>3\text{.}$$
My questions were
$1$. How to compute $\tilde{p}_k$ when $M>3$? Are there any other approaches?
$2$. I assume that $|\lambda_i|\leq1$ should hold for every $\lambda_i$. How to prove that?
I can prove $2$) if all $\lambda_i\in \mathbb{R}$ knowing $0\leq \tilde{p}_k\leq 1$ for all $k\in \mathbb{Z}$ (To be more precise: there are nonzero constants only in front of the $|\lambda_i |\leq1$). Thanks in advance.
The second was answered. And the assumpzion is wrong. There is still the first.
The case $\mathcal{A} =\{-3,3,6\}$ with $M=9$ can be reduced to $\tilde{\mathcal{A}} = \{-1,1,2\}$, since $\text{GCD}(\mathcal{A}) = 3$. So from now on, we can assume that $\text{GCD}(\mathcal{A}) = 1$.