Lemma 2. For an odd prime $p$ such that $p|k$, $(c_m, p) = 1$ ($-1 \le m \le n+1$).

*Proof.* Suppose there is an odd prime $p$ such that $p|(c_m, c_{m+1})$. Then Lemma 1 implies $p|k$, which contradicts Lemma 2. $\square$

Lemma 4. For an odd prime $p$ such that $p^m||c_n$ with $m \ge 1$, $p^m|(c_{n+1}^3 + k^{3f_{2n+2}})$.

*Proof.* From Lemma 3, $p \nmid c_{n-1}$. From this computation
$$\begin{align*} c_{n+2} &= \frac{c_{n+1}^3 + k^{3f_{2n+2}}}{c_n} = \frac{c_{n-1}^3 c_{n+1}^3 + c_{n-1}^3 k^{3f_{2n+2}}}{c_{n-1}^3 c_n} = \frac{(c_n^3 + k^{3f_n})^3 + c_{n-1}^3 k^{3f_{2n+2}}}{c_{n-1}^3 c_n} \\ &= \frac{c_n^9 + 3c_n^6 k^{3f_{2n}} + 3c_n^3 k^{6f_{2n}} + k^{9f_{2n}} + c_{n-1}^3 k^{3f_{2n+2}}}{c_{n-1}^3 c_n} \\ &= \frac{c_n^9 + 3c_n^6 k^{3f_{2n}} + 3c_n^3 k^{6f_{2n}} + k^{3f_{2n+2}}(c_{n-1}^3 + k^{3f_{2n-2}})}{c_{n-1}^3 c_n} \tag{9} \\ &= \frac{c_n^9 + 3c_n^6 k^{3f_{2n}} + 3c_n^3 k^{6f_{2n}} + k^{3f_{2n+2}} c_{n-2} c_n}{c_{n-1}^3 c_n} \\ &= \frac{1}{c_{n-1}^3} (c_n^8 + 3c_n^5 k^{3f_{2n}} + 3c_n^2 k^{6f_{2n}} + k^{3f_{2n+2}} c_{n-2}) \end{align*}$$
the lemma follows. $\square$

For a positive integer $n$, if $2^\alpha||n$ with a nonnegative integer $\alpha$, we define $v(n) = \alpha$. For a positive rational number $q = \frac{m}{n}$ with integers $m$ and $n$, define $v(q) = v(m) - v(n)$.

Lemma 5. If $k \equiv 0 \pmod 4$, $v(c_n) = f_{2n}$ for $n \ge 1$.

*Proof.* We get $v(c_1) = 1 = f_2$ and $v(c_2) = v(c_1^3 + k^3) - v(b_3) = 3v(c_1) = 3 = f_4$. Suppose $v(c_n) = f_{2n}$ and $v(c_{n+1}) = f_{2n+2}$. Since $v(c_{n+1}^3) = 3f_{2n+2} < 3f_{2n+2}v(k) = v(k^{3f_{2n+2}})$, Lemma 1 implies $v(c_{n+2}) = v(c_{n+1}^3 + k^{3f_{2n+2}}) - v(c_n) = 3f_{2n+2} - f_{2n} = f_{2n+4}$. $\square$

Lemma 6. Suppose $k \equiv 2 \pmod 4$ and If $n \equiv 2 \pmod 3$, then $v(c_n) > f_{2n}$. If $n \not\equiv 2 \pmod 3$, then $v(c_n) = f_{2n}$.

*Proof.* Since $c_{-1} = 1$, $v(c_{-1}) = 0 > f_{-2} = -1$. Since $c_0 = k^3 + 1$, $v(c_0) = 0 = f_0$. From $c_1 = (k^3 + 1)^3 + 1 \equiv 2 \pmod 8$, we have $v(c_1) = 1 = f_2$.
Since $v(c_1) = v(k) = 1$ and $v(b_3) = 0$, $c_2 = \frac{c_1^3 + k^3}{b_3}$ implies $v(c_2) > 3 = f_4$.

Suppose $v(c_{3m}) = f_{6m}$ and $v(c_{3m+1}) = f_{6m+2}$. Since $v(c_{3m+1}^3) = 3f_{6m+2}$ and $v(k^{3f_{6m+2}}) = 3f_{6m+2}$, we get $v(c_{3m+1}^3 + k^{3f_{6m+2}}) > 3f_{6m+2}$. Thus $v(c_{3m+2}) = v(c_{3m+1}^3 + k^{3f_{6m+2}}) - v(c_{3m}) > 3f_{6m+2} - f_{6m} = f_{6m+4}$.

Suppose $v(c_{3m+1}) = f_{6m+2}$ and $v(c_{3m+2}) > f_{6m+4}$. Since $v(c_{3m+2}^3) > 3f_{6m+4}$ and $v(k^{3f_{6m+4}}) = 3f_{6m+4}$, we have $v(c_{3m+2}^3 + k^{3f_{6m+4}}) = 3f_{6m+4}$. Therefore, $v(c_{3m+3}) = v(c_{3m+2}^3 + k^{3f_{6m+4}}) - v(c_{3m+1}) = 3f_{6m+4} - f_{6m+2} = f_{6m+6}$.

Suppose $v(c_{3m-3}) = f_{6m-6}$, $v(c_{3m-2}) = f_{6m-4}$, $v(c_{3m-1}) \ge f_{6m-2} + 1$ and $v(c_{3m}) = f_{6m}$. From the equation (9), we have
$$\begin{align*} v(c_{3m+1}) &= v(c_{3m}^3 + k^{3f_{6m}}) - v(c_{3m-1}) \\ &= v(c_{3m-1}^8 + 3c_{3m-1}^5 k^{3f_{6m-2}} + 3c_{3m-1}^2 k^{6f_{6m-2}} + k^{3f_{6m}} c_{3m-3}) - 3v(c_{3m-2}) \\ &= v(k^{3f_{6m}} c_{3m-3}) - 3v(c_{3m-1}) = 3f_{6m} + f_{6m-6} - 3f_{6m-4} = f_{6m+2}. \end{align*}$$
Thus we prove the lemma. $\square$

From Lemma 5 and Lemma 6, we prove that if $2^m \mid c_n$, then $2^m \mid (c_{n+1}^3 + k^{3f_{2n+2}})$. With Lemma 4, it follows that $c_n \mid (c_{n+1}^3 + k^{3f_{2n+2}})$ and $c_{n+2}$ is a positive integer.