Let $(y_n)_{n \ge 1}$ be a sequence of real numbers in $l^1$, that is $$ \sum_{n=1}^\infty |y_n| < +\infty. $$ Let $(x_n)_{n \ge 1}$ be a sequence of real numbers, where $$ x_1 \in \mathbb{R}, \qquad x_2 = \frac{7x_1 - y_1}{6} $$ and $$ (1+3^{n+1})x_n - 3^n x_{n-1} - 2\cdot 3^n x_{n+1} = y_n, \quad n \ge 2. $$
Hence, given $(y_n)_{n\ge 1}$, the sequence $(x_n)_{n\ge 1}$ is determined uniquely by its value $x_1$.
Does there exsits $x_1$ such that $(x_n)_{n \ge 1}$ is in $l^1$?
I tried to do some numerical analysis, and for $$ y_n = \frac{1}{n^2} $$ I suspect that the answer is yes, and $x_1 \sim 0.2496721891826\dots$.