Let $s(n)=2n+1$ and $\sigma(n)=\{n,s(n),s^2(n),s^3(n),\ldots\}$,
where $s^3$ denotes functions composition, $s^3(n)=s(s(s(n)))$.
For example $\sigma(11)=\{11,23,47,95,\ldots\}$.
As another example $\sigma(4)=\{4,9,19,39,\ldots\}$.
Question 1. Is there a positive integer $k$ such that every
element of $\sigma(2^k)$ is composite?
In other words, is it possible that, starting with $2^k$, all of its iterates under $s$ are composite?
Questions 2. Is there a positive integer $n$ such that every element of $\sigma(n)$ is composite?
The above may be lacking motivation, I came up with it while thinking of a comment (between Q2 and Q3) in the following MO question, but it is unclear if what I ask would be of any help answering that question ... but I thought I might ask anyway.
For example $s(16)=33$ composite, $s^2(16)=s(33)=67$ prime. So $16$ did not work.
$s(32)=65$ composite, $s(65)=131$ prime, did not work.
If $n=2^{20}=1048576$ then $s^j(n)$ is composite for $j=0,..,7$ but $s^8(n)=268435711$ is prime.
$s^j(58)$ is composite for $j=0,\ldots,11$ but $s^{12}(58)=241663$ is prime, did not work.
Would it ever work for either some $n$ of the form $2^k$, or for any
positive integer $n$?
Does the answer follow from known results, any reference?
We answer the second question.
Riesel showed that there are infinitely many natural numbers $k$ such that $k2^n-1$ is composite for all natural numbers $n$. Such numbers $k$ are called Riesel numbers. The smallest number known to be Riesel is $509203$ but there may be smaller ones.
In your notation, if $k$ is a Riesel number, then all numbers in $\sigma(k-1)$ are composite. For in general $s(k2^m-1)=k2^{m+1}-2+1=k2^{m+1}-1$.