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Point 1. is easy to satisfy even with nontrivial polynomials, since if you know $t$ modulo $2^n$ then you know for sure what the first n transformations will be.

For example. I could take $f=2^{20}x+174762$ and $g=6x+2$. Indeed I choose f such that $3(2f(n)+1)+1)$ is always divisible by $2^{20}$, meaning that the first 21 transformations applied to 2f(n)+1 are

1. 3x+1 cause 2f(n)+1 is odd
2. x/2 repeated 20 times cause $3(2f(n)+1)+1)=2^{20}(6n+1)$ is divisible by $2^{20}$.

After this we will have reached k = 6n+1 which is smaller than g(n)=6n+2.

Point two is likely very hard and I have no idea how to tackle that. Indeed answering 1 with polynomials f and g with $\deg g < \deg f$ will already be very difficult. Although when taking $f$ to be the similar looking non polynomial $f=2(4^x-1)/3$ one can take $g=2$ for silly reasons.

Point 1. is easy to satisfy even with nontrivial polynomials, since if you know $t$ modulo $2^n$ then you know for sure what the first n transformations will be.

For example. I could take $f=2^{20}x+174762$ and $g=6x+2$. Indeed I choose $f$ such that $3(2f(n)+1)+1)$ is always divisible by $2^{20}$, meaning that the first 21 transformations applied to $2f(n)+1$ are

1. $3x+1$ because $2f(n)+1$ is odd;
2. $x/2$ repeated 20 times because $3(2f(n)+1)+1)=2^{20}(6n+1)$ is divisible by $2^{20}$.

After this we will have reached $k = 6n+1$ which is smaller than $g(n)=6n+2$.

Point 2. is likely very hard and I have no idea how to tackle that. Indeed answering 1. with polynomials $f$ and $g$ with $\deg g < \deg f$ will already be very difficult. Although when taking $f$ to be the similar looking non polynomial $f=2(4^x-1)/3$ one can take $g=2$ for silly reasons.

Point 1. is easy to satisfy even with nontrivial polynomials, since if you know $t$ modulo $2^n$ then you know for sure what the first n transformations will be.

For example. I could take $f=2^{20}x+174762$ and $g=6x+2$. Indeed I choose f such that $3(2f(n)+1)+1)$ is always divisible by $2^{20}$, meaning that the first 21 transformations applied to 2f(n)+1 are

1. 3x+1 cause 2f(n)+1 is odd
2. x/2 repeated 20 times cause $3(2f(n)+1)+1)=2^{20}(6n+1)$ is divisible by $2^{20}$.

After this we will have reached k = 6n+1 which is smaller than g(n)=6n+2.

Point two is likely very hard and I have no idea how to tackle that. Indeed answering 1 with polynomials f and g with $\deg g < \deg f$ will already be very difficult.

Point 1. is easy to satisfy even with nontrivial polynomials, since if you know $t$ modulo $2^n$ then you know for sure what the first n transformations will be.

For example. I could take $f=2^{20}x+174762$ and $g=6x+2$. Indeed I choose f such that $3(2f(n)+1)+1)$ is always divisible by $2^{20}$, meaning that the first 21 transformations applied to 2f(n)+1 are

1. 3x+1 cause 2f(n)+1 is odd
2. x/2 repeated 20 times cause $3(2f(n)+1)+1)=2^{20}(6n+1)$ is divisible by $2^{20}$.

After this we will have reached k = 6n+1 which is smaller than g(n)=6n+2.

Point two is likely very hard and I have no idea how to tackle that. Indeed answering 1 with polynomials f and g with $\deg g < \deg f$ will already be very difficult. Although when taking $f$ to be the similar looking non polynomial $f=2(4^x-1)/3$ one can take $g=2$ for silly reasons.

Point 1. is easy to satisfy even with nontrivial polynomials, since if you know $t$ modulo $2^n$ then you know for sure what the first n transformations will be.

For example. I could take $f=2^{20}x+174762$ and $g=6x+2$. Indeed I choose f such that $3(2f(n)+1)+1)$ is always divisible by $2^{20}$, meaning that the first 21 transformations applied to 2f(n)+1 are

1. 3x+1 cause 2f(n)+1 is odd
2. x/2 repeated 20 times cause $3(2f(n)+1)+1)=2^{20}(6n+1)$ is divisible by $2^{20}$.

After this we will have reached k = 6n+1 which is smaller than g(n)=6n+2.

Point two is likely very hard and I have no idea how to tackle that.

Point 1. is easy to satisfy even with nontrivial polynomials, since if you know $t$ modulo $2^n$ then you know for sure what the first n transformations will be.

For example. I could take $f=2^{20}x+174762$ and $g=6x+2$. Indeed I choose f such that $3(2f(n)+1)+1)$ is always divisible by $2^{20}$, meaning that the first 21 transformations applied to 2f(n)+1 are

1. 3x+1 cause 2f(n)+1 is odd
2. x/2 repeated 20 times cause $3(2f(n)+1)+1)=2^{20}(6n+1)$ is divisible by $2^{20}$.

After this we will have reached k = 6n+1 which is smaller than g(n)=6n+2.

Point two is likely very hard and I have no idea how to tackle that. Indeed answering 1 with polynomials f and g with $\deg g < \deg f$ will already be very difficult.