Let $E\_1$ denote the infinite enumerated collection of two-symbol (`0` as blank symbol and `1` as non-blank symbol) one-tape (assuming that the tape is infinite in both directions) Turing machines: $${E\_1} = \{ 1,\;\;2,\;\;3,\;\; \ldots \},$$ 


where each element is simply the index of the corresponding Turing machine. So, we can denote the $x$-th Turing machine by $\text{M}\_x$. 


The exact technical description of the design of these machines can be found in [“The game” section in Wikipedia article about Busy beaver](https://en.wikipedia.org/wiki/Busy_beaver#The_game). 


Let $G(x)$ denote the number of different states (not including the halting states!) of the corresponding $\text{M}\_x$. It means that if there are $${(4n + 4)^{2n}}$$ $n$-state Turing machines, then $G(x) = 1$ if $1 \le x \le 64$, $G(x) = 2$ if $65 \le x \le 20736$ etc. 


Let `(0)` denote an infinite sequence of consecutive blank symbols on the tape. 


Let $T\_1$ denote the configuration when a Turing machine starts in the state $A$ on an infinitely blank tape. Then $T\_2$ denotes the configuration when a Turing machine starts in the state $A$ with its head positioned over the single `1` on the `(0)1(0)` tape; $T\_3$ denotes the configuration when a Turing machine starts in the state $A$ with its head positioned over the leftmost `1` on the `(0)11(0)` tape; $T\_4$ denotes the configuration when a Turing machine starts in the state $A$ with its head positioned over the leftmost `1` on the `(0)111(0)` tape, etc. That is, machines always start in the state $A$ so that the head at the start is positioned over the leftmost non-blank symbol of the tape, and the number of consecutive non-blank symbols on the tape increases by 1. 


We define the following function: $F(x) = 0$ if $\text{M}\_x$ halts on all configurations denoted by $T\_y$ for any $y \ge 1$; otherwise, $F(x) = z$, where $z$ is the **minimal** positive number such that $\text{M}\_x$ does not halt on the configuration denoted by $T\_z$. 


Let $S(x)$ denote the largest number of shifts made by any halting $x$-state 2-symbol Turing machine that starts from an infinitely blank tape. This definition is equal to the definition of “Maximum shifts function” given in [Wikipedia article about Busy beaver](https://en.wikipedia.org/wiki/Busy_beaver#Maximum_shifts_function_S): $$S(1) = 1,\;\;S(2) = 6,\;\;S(3) = 21,\;\;S(4) = 107,\;\; \ldots $$ 


Then $E\_1$ corresponds to the following infinite sequence of integers: 


$${E\_2} = \left\{ {\left\lfloor {\frac{{F(1)}}{{S(G(1))}}} \right\rfloor ,\;\;\left\lfloor {\frac{{F(2)}}{{S(G(2))}}} \right\rfloor ,\;\;\left\lfloor {\frac{{F(3)}}{{S(G(3))}}} \right\rfloor ,\;\; \ldots } \right\},$$ 


where $\left\lfloor x \right\rfloor $ denotes a mathematical [floor function](https://en.wikipedia.org/wiki/Floor_and_ceiling_functions). 


Consider four possibilities: 


**Possibility 1**: $E\_2$ contains finitely many different integers and this amount is denoted by $Y$; then there exists the largest integer in $E\_2$ and this integer is denoted by $Z$. 


*Question if Possibility 1 is true*: is it possible to estimate a lower and upper bounds for $Y$ and $Z$? 


**Possibility 2**: $E\_2$ contains infinitely many different integers. 


*Question if Possibility 2 is true*: is it possible to prove (or back up) this? 


**Possibility 3**: $E\_2$ contains all integers. 


*Question if Possibility 3 is true*: is it possible to prove (or back up) this? 


**Possibility 4**: $E\_2$ contains infinitely many occurrences of any integer. 


*Question if Possibility 4 is true*: is it possible to prove (or back up) this? 



Can all lengths of shortest non-halting inputs of all Turing machines be limited by the Busy Beaver applied to the corresponding numbers of states?
There exists a family of Turing machines $\{ \mathcal{T}\_n : n \in \mathbb{N} \}$ such that:


* $\mathcal{T}\_n$ has $k n$ states where $k$ is some fixed universal constant;
* $F(\mathcal{T}\_n) \geq BB(BB(n))$.


Specifically, $\mathcal{T}\_n$ simulates the $n$-state Busy Beaver in the left half of the tape, keeping track of the number of simulated iterations. Then, upon halting, it has the number $BB(n)$ in a register. 


The machine proceeds to simultaneously simulate (via dovetailing) all Turing machines with at most $BB(n)$ states. If and when the $x$th simulated machine halts at time $y$, the Turing machine checks whether the $y$th input bit is a $0$ or a $1$. If it is a $0$, $\mathcal{T}\_n$ halts. If it is a $1$, it continues running.


To prevent $T\_n$ from halting, it is thus necessary to have $1$ bits in all positions corresponding to halting times of all machines of $\leq BB(n)$ states; consequently, there must be a $1$ in position $BB(BB(n))$ on the input tape.


