Let $f(n)=ab^{-n}:\Bbb N\to\Bbb C^\times$
Find some $a,b\in\Bbb C^\times$ such that $f$ has the following properties:
$f(n)$ as a sequence indexed by $n$ has precisely two positive integer subsequences $f(6n)$ and $f(6n+2)$ and every other sequence $f(6n+x\in\{1,3,4,5\})$ contains no positive integers.
$f(x\in\{0,2\})=f(x+6)$
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Here's as much as I can do. For the ordered pair $(a,b)$, I have that:
Setting $(a,b)$ to be the primitive 2nd root of unity $(-1,-1)$ gives a sequence that alternates between positive integers and not: $f(n+2)=f(n)$.
So I suspect the solution involves generalising this idea by setting $b$ to be a primitive sixth root of unity. I think this because there are exactly two primitive sixth roots of unity and they're separated by exponents $2$ mod $6$, as the solution requires.
Let $\phi$ be the primitive sixth root of unity. Then some function $f$ which was was integer if and only if $\phi^{n-1}$ was a primitive root of unity would give us the required result. One approach I tried is letting $a=\phi^{-1}$ and $b=\phi$, but that doesn't quite work
I'm not quite sure what $\mathbb{C^\times}$ means, but I assume it's something like the set of complex numbers apart from $0$. Please correct me if this is inaccurate, especially if it affects the answer.
Assume that for some $a,b\in\Bbb C^\times$ your property $1$ holds for $n = 1$. Then you have, for some positive integers $m_1$ and $m_2$, that
$$f(6) = m_1 \tag{1}\label{eq1A}$$
$$\begin{equation}\begin{aligned} f(8) & = f(6)(b^{-2}) \\ & = m_1(b^{-2}) \\ & = m_2 \end{aligned}\end{equation}\tag{2}\label{eq2A}$$
This gives that
$$b^{-2} = \frac{m_2}{m_1} \tag{3}\label{eq3A}$$
Next, you have
$$\begin{equation}\begin{aligned} f(10) & = f(8)(b^{-2}) \\ & = m_2\left(\frac{m_2}{m_1}\right) \\ & = \frac{m_2^2}{m_1} \end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Since $f(10)$ is not a positive integer (as $10 = 6(1) + 4$), this means $\frac{m_2}{m_1}$ is a positive rational, non-integral number. Thus, there's at least one prime factor $p$ of $m_1$ where $m_1$ has more factors of $p$ than $m_2$. Let $i$ be the number of these factors of $p$ in $m_1$, and $j$ be the number of factors of $p$ in $m_2$. Then we have
$$j \lt i \tag{5}\label{eq5A}$$
Also, since \eqref{eq4A} is not a positive integer, we also have
$$2j \lt i \tag{6}\label{eq6A}$$
However, note that
$$\begin{equation}\begin{aligned} f(12) & = f(10)(b^{-2}) \\ & = \left(\frac{m_2^2}{m_1}\right)\left(\frac{m_2}{m_1}\right) \\ & = \frac{m_2^3}{m_1^2} \end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Property $1$ for $n = 2$ says this is a positive integer. This means that, regarding the factors of $p$ in the numerator and denominator,
$$3j \ge 2i \tag{8}\label{eq8A}$$
However, adding \eqref{eq5A} and \eqref{eq6A} gives
$$3j \lt 2i \tag{9}\label{eq9A}$$
This contradicts \eqref{eq8A}, showing that property $1$ cannot hold.