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.

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: $$S(1) = 1,\;\;S(2) = 6,\;\;S(3) = 21,\;\;S(4) = 107,\;\; \ldots $$

Then $E_1$ corresponds to the following set $E_2$:

$${E_2} = \left\{ {\left\lfloor {\frac{{F(1)}}{{S(G(1))}}} \right\rfloor ,\;\;\left\lfloor {\frac{{F(2)}}{{S(G(2))}}} \right\rfloor ,\;\;\left\lfloor {\frac{{F(3)}}{{S(G(3))}}} \right\rfloor ,\;\; \ldots } \right\},$$

where $\left\lfloor x \right\rfloor $ denotes a mathematical floor function.

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?

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.

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: $$S(1) = 1,\;\;S(2) = 6,\;\;S(3) = 21,\;\;S(4) = 107,\;\; \ldots $$

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.

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?

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.

Let $G(x)$ denote the number of different 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: $$S(1) = 1,\;\;S(2) = 6,\;\;S(3) = 21,\;\;S(4) = 107,\;\; \ldots $$

Then $E_1$ corresponds to the following set $E_2$:

$${E_2} = \left\{ {\left\lfloor {\frac{{F(1)}}{{S(G(1))}}} \right\rfloor ,\;\;\left\lfloor {\frac{{F(2)}}{{S(G(2))}}} \right\rfloor ,\;\;\left\lfloor {\frac{{F(3)}}{{S(G(3))}}} \right\rfloor ,\;\; \ldots } \right\},$$

where $\left\lfloor x \right\rfloor $ denotes a mathematical floor function.

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?

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.

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: $$S(1) = 1,\;\;S(2) = 6,\;\;S(3) = 21,\;\;S(4) = 107,\;\; \ldots $$

Then $E_1$ corresponds to the following set $E_2$:

$${E_2} = \left\{ {\left\lfloor {\frac{{F(1)}}{{S(G(1))}}} \right\rfloor ,\;\;\left\lfloor {\frac{{F(2)}}{{S(G(2))}}} \right\rfloor ,\;\;\left\lfloor {\frac{{F(3)}}{{S(G(3))}}} \right\rfloor ,\;\; \ldots } \right\},$$

where $\left\lfloor x \right\rfloor $ denotes a mathematical floor function.

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?