A similar Somos sequence problem prove $A_{2n}B_{n+3}$ is integer sequence I have read some interesting with Somos sequence,Recently I met a similar question：

For a fixed positive integer $k$, there are two sequences $A_n$ and $B_n$.
  They are defined inductively, by the following recurrences.
  $$A_1 = k,A_2 = k,A_{n+2} = A_{n}A_{n+1}$$
  $$B_1 = 1,B_2 = k, B_{n+2} = \frac{B^3_{n+1}+1}{B_{n}}$$
  Prove that for all positive integers $n$, $A_{2n}B_{n+3}$ is an integer.

I try to Solve this problem,following is my some works.
since
$$A_{n+2}=A_{n}A_{n+1}\Longrightarrow \ln{A_{n+2}}=\ln{A_{n}}+\ln{A_{n+1}}$$
so we have
$$A_{n}=k^{f_{n}}$$
where $f_{n}$ be the Fibonacci sequece,which satisfies $f_{1}=f_{2}=1,f_{n}=f_{n-1}+f_{n-2}$
for $B_{n}$ It is difficult to handle, even for a general term 
If we choose to use mathematical induction
(1):if  $n=1$ since
$$B_{3}=k^3+1,\Longrightarrow B_{4}=\dfrac{(k^3+1)^3+1}{k}$$
so we have
$$A_{2}B_{4}=(k^3+1)^3+1$$ is integer
if $n=2$,we have
$$B_{5}=\dfrac{B^3_{4}+1}{B_{4}}=\dfrac{\left(\dfrac{(k^3+1)^3+1}{k}\right)^3+1}{k^3+1}=\dfrac{((k^3+1)^3+1)^3+k^3}{(k^3+1)k^3}$$
and $A_{3}=k^2,A_{4}=k^3$,so we have
$$A_{4}B_{5}=\dfrac{((k^3+1)^3+1)^3+k^3}{k^3+1}=\dfrac{(k^3+1)^9+3(k^3+1)^6+3(k^3+1)^3+(k^3+1)}{k^3+1}
=(k^3+1)^8+3(k^3+1)^5+3(k^3+1)^2+1\in N^{+}$$
Now Assmue that $A_{2n}B_{n+3}(n\le k)$ is integer,or $n^{f_{n}}B_{n+3}(n\le k)$ is integrthen consider $$A_{2k+2}B_{k+4}=(k+1)^{f_{k+1}}\cdot\dfrac{B^3_{k+3}+1}{B_{k+2}}$$
Now the problem is equlivant to  prove
$$\dfrac{(B^3_{k+3}+1)(k+1)^{f_{k+1}}}{B_{k+2}}\in N^{+}~~~?$$
 A: Considering the sequence (for $n \ge 2$):
$$
A_{2(n-1)} B_{(n-1)+3}, \space A_{2n} B_{n+3}, \space A_{2(n+1)} B_{(n+1)+3} \space \rightarrow \space A_{2n-2} B_{n+2}, \space A_{2n} B_{n+3}, \space A_{2n+2} B_{n+4}
$$
According to definitions, we have:
$$
{ A_{2n+3} = A_{2n+2} A_{2n+1} = (A_{2n+1} A_{2n}) A_{2n+1} = A^2_{2n+1} A_{2n} = A^3_{2n} A^2_{2n-1} \brace A_{2n+3} = A_{2n+2} A_{2n+1} = A_{2n+2} (A_{2n} A_{2n-1}) = A_{2n+2} A^2_{2n-1} A_{2n-2} }
$$
$$
\Rightarrow A^3_{2n} = A_{2n+2} A_{2n-2}
$$
And,
$$
B_{n+4} = \frac{B^3_{n+3}+1}{B_{n+2}} \Rightarrow B^3_{n+3}+1 = B_{n+4} B_{n+2}
$$
Thus,
$$
A^3_{2n} (B^3_{n+3}+1) = (A_{2n+2} A_{2n-2}) (B_{n+4} B_{n+2})
$$
$$
\Rightarrow (A_{2n} B_{n+3})^3 + A^3_{2n} = (A_{2n+2} B_{n+4}) (A_{2n-2} B_{n+2})
$$
By checking the first two terms: $(A_{2n-2} B_{n+2} = A_2 B_4 \in N^{+}), \space (A_{2n} B_{n+3} = A_4 B_5 \in N^{+})$ and because all $A_k \in N^{+}$
$ \Rightarrow $ Every next term in the sequence is an integer.
$$
(A_{2n} B_{n+3})^3 + A^3_{2n} = (A_{2n+2} B_{n+4}) (A_{2n-2} B_{n+2}) \Rightarrow (Int)^3 + (Int)^3 = (A_{2n+2} B_{n+4}) \cdot (Int)
$$
$$
\Rightarrow A_{2n+2} B_{n+4} \in N^{+}
$$
