Generating functions of Collatz iterates?

Question

Let $C(n) = n/2$ if $n$ is even and $3n+1$ otherwise be the Collatz function.

We look at the generating function $f_n(x) = \sum_{k=0}^\infty C^{(k)}(n)x^k$ of the iterates of the Collatz function.

The Collatz conjecture is then equivalent to: For all $n$:

$$f_n(x) = p_n(x) + x^{d+1} \frac{1+4x+2x^2}{1-x^3}$$ where $d$ is the degree of the polynomial $p_n(x)$ with natural numbers as coefficients.

I have computed some of these generating functions.

Let

$$F_n(x) = (f_n(x), f_{C^{(1)}(n)}(x),\cdots,f_{C^{(l)}(n)}(x))$$

where $l$ is the length of the Collatz sequence of $n$ ending at $1$.

The vector $F_n(x)$ when plugging in for $x$ a rational number, seems to parametrize an algebraic variety. Assuming that the Collatz conjecture is true. Can it be explained if or why this vector parametrizes an algebraic variety?

Here is an example for $n=3$:

The variety is given by the equations:

G^3 - H^3 - 4*G^2 + 4*G*H + H^2 + 4*G - 8*H = 0
A*C - B^2 + 10*B - 3*C = 0
F^2 - G*H - 4*F + G = 0
F*G - H^2 - 2*F + H = 0
F*H - G^2 + 2*G - 4*H = 0 
E - H - 7 = 0

and it is parametrized by:

A = (7*x^7 + 14*x^6 + x^5 + 2*x^4 - 13*x^3 - 5*x^2 - 10*x - 3)/(x^3 - 1)
B = (7*x^6 + 14*x^5 + x^4 + 2*x^3 - 16*x^2 - 5*x - 10)/(x^3 - 1)
C = (7*x^5 + 14*x^4 + x^3 - 8*x^2 - 16*x - 5)/(x^3 - 1)
D = (7*x^4 + 14*x^3 - 4*x^2 - 8*x - 16)/(x^3 - 1)
E = (7*x^3 - 2*x^2 - 4*x - 8)/(x^3 - 1)
F = (-x^2 - 2*x - 4)/(x^3 - 1)
G = (-4*x^2 - x - 2)/(x^3 - 1)
H = (-2*x^2 - 4*x - 1)/(x^3 - 1)

where $F_3(x) = (A,B,C,D,E,F,G,H)$

Here is some Sagemath script which does the computations. You can change the number $N=3$ in the script, but for $N=7$ it already takes a long time to compute the Groebner basis.

Edit: Furthermore, the point $(n,C^{(1)}(n),\cdots,C^{(l)}(n))$ seem to always be a rational point of this variety. Example:

G^3 - H^3 - 4*G^2 + 4*G*H + H^2 + 4*G - 8*H = 0
A*C - B^2 + 10*B - 3*C = 0
F^2 - G*H - 4*F + G = 0
F*G - H^2 - 2*F + H = 0
F*H - G^2 + 2*G - 4*H = 0
E - H - 7 = 0
.....
A = 3
B = 10
C = 5
D = 16
E = 8
F = 4
G = 2
H = 1

This last observation can be explained if the previous is true, because we can substitute $x=0$:

$$\forall m \text{ we have : } f_m(0) = m$$

and hence:

$$F_n(0) = (n,C^{(1)}(n),\cdots,C^{(l)}(n))$$

is a rational point on the variety.

Answer 1

Without assuming the Collatz conjecture, it can be shown that the generating functions satsify certain polynomial equations:

Observe that for all $n$:

$$f_{C(n)}(x) = \frac{f_n(x)-n}{x}$$

hence:

$$f_{C^{(2)}(n)}(x) = \frac{f_{C(n)}(x)-C(n)}{x}$$

Solving for $x$ and equating the two identities and letting $x_k:=f_{C^{(k)}(n)}(x)$, we find the polynomial equation:

$$\forall k=0,1,2,\cdots \text{ we have }: x_k x_{k+2}-C^{(k)}(n) x_{k+2}-x_{k+1}^2+C^{(k+1)}(n)x_{k+1} = 0$$

which according to Wolfram Alpha each of these equations represent a "one-sheeted hyperboloid":

Wikipedia Mathworld