Let $\mathbb{N}_{\text{odd}}$ be the set of odd positive integers. For $x_0 \in \mathbb{N}_{\text{odd}}$ consider the set-valued sequence $\{A_n\}_{n=0}^{\infty}$ defined by the formula

$$ A_0 = \{x_0\}, \qquad A_n = \{\delta(3x + 1), \; \delta(3x + 3), \, : \, x \in A_{n-1}\}, \quad n \geq 1, $$

where $\delta(k)$ denotes the largest odd factor of the integer $k$ (e.g., $\delta(1) = \delta(2) = \delta(4) = 1$, $\delta(3) = \delta(6) = \delta(12) = 3$, $\delta(15) = \delta(60) = 15$, etc). Is it true that

$$ A := \bigcup_{n=0}^{\infty} A_n = \mathbb{N}_{\text{odd}}? $$

A variant: Suppose now that $\mathbb{N}_{\pm 1} := \{x \in \mathbb{N} \,:\, x \equiv \pm1 \ (\text{mod}\; 6)\}$. For $x_0 \in \mathbb{N}_{\pm 1}$, $x_0 \ne 1$ consider the set-valued sequence $\{B_n\}_{n=0}^{\infty}$ defined by

$$ B_0 = \{x_0\}, \qquad B_n = \{\delta(3x - 1), \; \delta(3x + 1) \, : \, x \in B_{n-1}\}, \quad n \geq 1 $$ Is it true that

$$ B := \bigcup_{n=0}^{\infty} B_n = \mathbb{N}_{\pm 1}? $$