Return to Answer

The orbit of $1/3$ is infinite. You can show this via the $3$-adic valuation $\nu_3$.

Let us show by induction that $\nu_3(x_n) = -2^{n}$: We have $\nu_3(x_0) = \nu_3(1/3) = -1 = -2^0$.

Now $\nu_3(x_{n+1}) = \nu_3(4 x_{n+1} (1 - x_{n+1})) = \nu_3(4)+ \nu_3(x_n) + \nu_3(1 - x_n)$.

We have $\nu_3(4) = 0$, and since $\nu_3(x_n)$ is negative by induction, we have $\nu_3(1 - x_n) = \nu_3(x_n)$, so overall $\nu_3(x_{n+1}) = 2\nu_3(x_n) =-2^{n+1}$.

From this is is clear that the orbit must be infinite.

A similar argument shows that $p/q$ has infinite orbit for $p, q$ coprime, $q$ divisible by some odd prime.

The orbit of $1/3$ is infinite. You can show this via the $3$-adic valuation $\nu_3$.

Let us show by induction that $\nu_3(x_n) = -2^{n}$: We have $\nu_3(x_0) = \nu_3(1/3) = -1 = -2^0$.

Now $\nu_3(x_{n+1}) = \nu_3(4 x_{n} (1 - x_{n})) = \nu_3(4)+ \nu_3(x_n) + \nu_3(1 - x_n)$.

We have $\nu_3(4) = 0$, and since $\nu_3(x_n)$ is negative by induction, we have $\nu_3(1 - x_n) = \nu_3(x_n)$, so overall $\nu_3(x_{n+1}) = 2\nu_3(x_n) =-2^{n+1}$.

From this is is clear that the orbit must be infinite.

A similar argument shows that $p/q$ has infinite orbit for $p, q$ coprime, $q$ divisible by some odd prime.

The orbit of $1/3$ is infinite. You can show this via the $3$-adic valuation $\nu_3$.

Let us show by induction that $\nu_3(x_n) = -2^{n+1}$: We have $\nu_3(x_0) = \nu_3(1/3) = -1$.

Now $\nu_3(x_{n+1}) = \nu_3(4 x_{n+1} (1 - x_{n+1})) = \nu_3(4)+ \nu_3(x_n) + \nu_3(1 - x_n)$.

We have $\nu_3(4) = 0$, and since $\nu_3(x_n)$ is negative by induction, we have $\nu_3(1 - x_n) = \nu_3(x_n)$, so overall $\nu_3(x_{n+1}) = 2\nu_3(x_n) =-2^{n+2}$.

From this is is clear that the orbit must be infinite.

A similar argument shows that $p/q$ has infinite orbit for $p, q$ coprime, $q$ divisible by some odd prime.

The orbit of $1/3$ is infinite. You can show this via the $3$-adic valuation $\nu_3$.

Let us show by induction that $\nu_3(x_n) = -2^{n}$: We have $\nu_3(x_0) = \nu_3(1/3) = -1 = -2^0$.

Now $\nu_3(x_{n+1}) = \nu_3(4 x_{n+1} (1 - x_{n+1})) = \nu_3(4)+ \nu_3(x_n) + \nu_3(1 - x_n)$.

We have $\nu_3(4) = 0$, and since $\nu_3(x_n)$ is negative by induction, we have $\nu_3(1 - x_n) = \nu_3(x_n)$, so overall $\nu_3(x_{n+1}) = 2\nu_3(x_n) =-2^{n+1}$.

From this is is clear that the orbit must be infinite.

A similar argument shows that $p/q$ has infinite orbit for $p, q$ coprime, $q$ divisible by some odd prime.

The orbit of $1/3$ is infinite. You can show this via the $3$-adic valuation $\nu_3$.

Let us show by induction that $\nu_3(x_n) = -2^{n+1}$: We have $\nu_3(x_0) = \nu_3(1/3) = -1$.

Now $\nu_3(x_{n+1}) = \nu_3(4 x_{n+1} (1 - x_{n+1})) = \nu_3(4)+ \nu_3(x_n) + \nu_3(1 - x_n)$.

We have $\nu_3(4) = 1$, and since $\nu_3(x_n)$ is negative by induction, we have $\nu_3(1 - x_n) = \nu_3(x_n)$, so overall $\nu_3(x_{n+1}) = 2\nu_3(x_n) =-2^{n+2}$.

From this is is clear that the orbit must be infinite.

A similar argument shows that $p/q$ has infinite orbit for $p, q$ coprime, $q$ divisible by some odd prime.

The orbit of $1/3$ is infinite. You can show this via the $3$-adic valuation $\nu_3$.

Let us show by induction that $\nu_3(x_n) = -2^{n+1}$: We have $\nu_3(x_0) = \nu_3(1/3) = -1$.

Now $\nu_3(x_{n+1}) = \nu_3(4 x_{n+1} (1 - x_{n+1})) = \nu_3(4)+ \nu_3(x_n) + \nu_3(1 - x_n)$.

We have $\nu_3(4) = 0$, and since $\nu_3(x_n)$ is negative by induction, we have $\nu_3(1 - x_n) = \nu_3(x_n)$, so overall $\nu_3(x_{n+1}) = 2\nu_3(x_n) =-2^{n+2}$.

From this is is clear that the orbit must be infinite.

A similar argument shows that $p/q$ has infinite orbit for $p, q$ coprime, $q$ divisible by some odd prime.