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Here is a new difference sequence system, which I came up with today through a greedy trial to imitate Y sequence system originally created by a Japanese googologist Yukito. Although it expands in a different way from Y-sequence system, it looks not so bad.

WARNING: I might update the expansion rule so that it will finally coincide wth that of Y-sequence system. However, since I am satisfied with this method itself, I might not complete the updating.

# Historical Background

Y-sequence system is a difference sequence system which is believed to surpass BM2.3 = BM4 by several googologists, but has not been formalised yet. Therefore several googologists are trying to create a well-defined notation which works in the same way as Y-sequence system. Of course, since Y-sequence system itself is not completely defined, "works in the same way" just means "is consistent with known finitely many expansions in Y-sequence system".

Since Yukito dislikes notations created by others to be called "Y-sequence system", I do not call my system something like "Y-sequence system version blah-blah". Instead, I offer homage to Yukito here, and clarify the credit.

# Convention

For a set $$X$$, I denote by $$X^{< \omega}$$ the set of arrays of elements of $$X$$. I denote by $$\mathbb{N}_+$$ the set of positive integers, by $$Y \subset \mathbb{N}_+^{< \omega}$$ the subset of non-empty sequences whose leftmost entry is $$1$$, by $$\leq_{\textrm{lex}}$$ the lexicographic order on $$\mathbb{N}_+^{\omega}$$, and by $$<_{\textrm{lex}}$$ the strict total order associated to $$\leq_{\textrm{lex}}$$.

For an array $$a$$, I denote by $$\textrm{Lng}(a)$$ the length of $$a$$. For an array $$a$$ and an $$i \in \mathbb{N}$$ smaller than $$\textrm{Lng}(a)$$, I denote by $$a_i$$ the $$(1+i)$$-th entry of $$a$$. An initial segment of an array $$a$$ is an array of the form $$(a_i)_{i=0}^{j-1}$$ for some $$j \in \mathbb{N}$$ smaller than $$\textrm{Lng}(a)-1$$. In particular, the empty array is an initial segment of any array.

For arrays $$a$$ and $$b$$, I denote by $$a \frown b$$ the concatenation of $$a$$ and $$b$$.

# Mountain

An $$a \in \mathbb{N}^{< \omega}$$ is said to be a mountain if $$a_i \leq i$$ for any $$i \in \mathbb{N}$$ smaller than $$\textrm{Lng}(a)$$. I denote by $$M \subset \mathbb{N}^{< \omega}$$ the subset of mountains. For an $$i \in \mathbb{N} \setminus \{0\}$$, I denote by $$\textrm{Trivial}_i \in Y_0$$ the mountain $$(0) \frown (j)_{j=0}^{i-2}$$.

I define a total computable map \begin{eqnarray*} \textrm{Ancestor} \colon \mathbb{N}^{< \omega} & \to & \mathbb{N}^{< \omega} \\ a & \mapsto & \textrm{Ancestor}(a) \end{eqnarray*} in the following recursive way:

1. Put $$L := \textrm{Lng}(a)$$.
2. If $$L = 0$$, then $$\textrm{Ancestor}(a) := ()$$.
3. If $$L \neq 0$$ and $$a_{L-1} > L-1$$, then $$\textrm{Ancestor}(a) := ()$$.
4. If $$L \neq 0$$ and $$a_{L-1} = 0$$, then $$\textrm{Ancestor}(a) = (L-1)$$.
5. Suppose $$L \neq 0$$ and $$0 < a_{L-1} \leq L-1$$.
1. Denote by $$a'$$ the initial segment of $$a$$ of length $$a_{L-1}$$.
2. Set $$\textrm{Ancestor}(a) := \textrm{Ancestor}(a') \frown (L-1)$$.

The map $$\textrm{Ancestor}$$ gives a sequence of "ancestors" of the rightmost node when the input is a mountain. I note that this map is quite different from the ancestor map in the article on the difference sequence system.

I define total computable maps \begin{eqnarray*} \begin{array}{lcrcl} \textrm{PreDifference} & \colon & \mathbb{N}^{< \omega} \times \mathbb{N}^{< \omega} & \to & \mathbb{N}^{< \omega} \\ & & (a,s) & \mapsto & \textrm{PreDifference}(a,s) \\ \textrm{PreMountain} & \colon & \mathbb{N}^{< \omega} \times \mathbb{N}^{< \omega} & \to & M \\ & & (a,s) & \mapsto & \textrm{PreMountain}(a,s) \end{array} \end{eqnarray*} simultaneously in the following recursive way:

1. Put $$L := \textrm{Lng}(a)$$.
2. Suppose $$\textrm{Lng}(s) \geq L$$
1. For each $$i \in \mathbb{N}$$ smaller than $$L$$, define a $$(b_i,c_i) \in \mathbb{N} \times \mathbb{N}$$ in the following recursive way:
1. Denote by $$a' \in \mathbb{N}_+^{< \omega}$$ the initial segment of $$a$$ of length $$i+1$$.
2. Denote by $$s' \in \mathbb{N}_+^{< \omega}$$ the initial segment of $$s$$ of length $$i+1$$.
3. Suppose that there is an entry $$j$$ of $$\textrm{Ancestor}(s')$$ satisfying $$0 \neq a_j < a_i$$.
1. Denote by $$p \in \mathbb{N}$$ the maximum of such a $$j$$.
2. Set $$b_i := a_i - a_p$$.
3. Set $$c_i := i - p$$.
4. Suppose that there is no entry $$j$$ of $$\textrm{Ancestor}(s')$$ satisfying $$0 \neq a_j < a_i$$.
1. Set $$b_i := 0$$.
2. Set $$c_i := 0$$.
2. Set $$\textrm{PreDifference}(a,s) := (b_j)_{j=0}^{L-1}$$.
3. Set $$\textrm{PreMountain}(a,s) := (c_j)_{j=0}^{L-1}$$.
3. Suppose $$\textrm{Lng}(s) < L$$.
1. Set $$\textrm{PreDifference}(a,s) := ()$$.
2. Set $$\textrm{PreMountain}(a,s) := ()$$.

Roughly speaking, $$\textrm{PreDifference}$$ is a function which outputs the difference sequence of the input except that it ignores $$0$$ and reflects the ancestor relation of $$s$$, and $$\textrm{PreMountain}$$ is a function which outputs a hydra whose ancestor relation coincides with the intersection of the ancestor relations of the inputs.

I define total computable maps \begin{eqnarray*} \begin{array}{lcrcl} \textrm{Difference} & \colon & \mathbb{N}^{< \omega} \times \mathbb{N}^{< \omega} \times \mathbb{N} & \to & \mathbb{N}^{< \omega} \\ & & (a,s,i) & \mapsto & \textrm{Difference}(a,s,i) \\ \textrm{Mountain} & \colon & \mathbb{N}^{< \omega} \times \mathbb{N}^{< \omega} \times \mathbb{N} & \to & M \\ & & (a,s,i) & \mapsto & \textrm{Mountain}(a,s,i) \end{array} \end{eqnarray*} simultaneously in the following recursive way:

1. Suppose $$i = 0$$.
1. Set $$\textrm{Difference}(a,s,i) := a$$.
2. Set $$\textrm{Mountain}(a,s,i) := \textrm{PreMountain}(a,s)$$.
2. Suppose $$i \neq 0$$.
1. Set $$\textrm{Difference}(a,s,i) := \textrm{PreDifference}(\textrm{Difference}(a,s,i-1),\textrm{Mountain}(a,s,i-1))$$.
2. Set $$\textrm{Mountain}(a,s,i) := \textrm{PreMountain}(\textrm{Difference}(a,s,i),\textrm{Mountain}(a,s,i-1))$$.

Roughly speaking, $$\textrm{Difference}(a,s,i)$$ is the hydra corresponding to the $$i$$-th difference sequence of $$(a,s)$$ with respect to the upper-branch-ignoring model, and $$\textrm{Mountain}(a,s,i)$$ is the hydra indicating the ancestor relation for $$\textrm{Difference}(a,s,i)$$.

For an $$(a,s) \in Y \times \mathbb{N}^{< \omega}$$, I denote by $$\textrm{Height}(a,s) \in \mathbb{N}$$ the minimum of a $$j \in \mathbb{N}$$ satisfying that the rightmost entry of $$\textrm{Difference}(a,s,j)$$ is smaller than or equal to $$1$$. Roughly speaking, $$\textrm{Hight}(a,s)$$ is the height of the tower of difference sequences of $$(a,s)$$.

I define a total computable map \begin{eqnarray*} \textrm{Descent} \colon Y \times \mathbb{N}^{< \omega} & \to & (\mathbb{N} \times \mathbb{N})^{< \omega} \setminus \{()\} \\ (a,s) & \mapsto & \textrm{Descent}(a,s) \end{eqnarray*} in the following recursive way:

1. Put $$H := \textrm{Height}(a,s)$$.
2. For each $$h \in \mathbb{N}$$ smaller than $$H$$, defined a $$b^{(h)} \in \mathbb{N}^{< \omega} \setminus \{()\}$$ in the following recursive way:
3. Suppose $$h = H-1$$.
1. Put $$L := \textrm{Lng}(\textrm{Ancestor}(\textrm{Mountain}(a,s,h)))$$.
2. Set $$b^{(h)} := ((h,\textrm{Ancestor}(\textrm{Mountain}(a,s,h))_i))_{i=0}^{L-1}$$.
4. Suppose $$h \neq H-1$$.
1. Denote by $$i$$ the second entry of $$b^{(h+1)}_0$$.
2. Denote by $$c \in \mathbb{N}^{< \omega}$$ the initial segment of $$\textrm{Mountain}(a,s,h)$$ of length $$i+1$$.
3. Put $$L := \textrm{Lng}(\textrm{Ancestor}(c))$$.
4. Set $$b^{(h)} := ((h,\textrm{Ancestor}(c)_i))_{i=0}^{L-1}$$.
5. $$\textrm{Descent}(a,s)$$ is the concatenation of $$(b^{(h)})_{h=0}^{H-1}$$.

Roughly speaking, $$\textrm{Descent}(a,s)$$ exhibits the addresses in the path to go down mountains, which will be used in order to find the bad root.

I define a total computable map \begin{eqnarray*} \textrm{Royal} \colon \mathbb{N}_+^{< \omega} \times \mathbb{N}^{< \omega} & \to & \mathbb{N}_+^{< \omega} \\ (a,s) & \mapsto & \textrm{Royal}(a,s) \end{eqnarray*} in the following recursive way:

1. Put $$L := \textrm{Lng}(\textrm{Descent}(a,s))$$.
2. For each $$i \in \mathbb{N}$$ smaller than $$L$$, put $$(h_i,j_i) := \textrm{Descent}(a,s)_i$$.
3. Set $$\textrm{Royal}(a,s) := (\textrm{Difference}(a,s,h_i)_{j_i})_{i=0}^{L-1}$$.

Namely, $$\textrm{Royal}(a,s)$$ is the sequence of the values of difference sequences of $$(a,s)$$ corresponding to the addresses in $$\textrm{Descent}(a,s)$$.

I define a total computable map \begin{eqnarray*} \textrm{BadRoot} \colon Y & \to & \mathbb{N} \times \mathbb{N} \times \mathbb{N} \\ a & \mapsto & \textrm{BadRoot}(a) \end{eqnarray*} in the following recursive way:

1. Put $$L := \textrm{Lng}(a)$$.
2. Put $$b := \textrm{Royal}(a,\textrm{Trivial}_L)$$.
3. If $$b = ()$$, then set $$\textrm{BadRoot}(a) := (0,L-1,L-1)$$.
4. Suppose $$b \neq ()$$.
1. Put $$(h,r,k) := \textrm{BadRoot}(b)$$.
2. Set $$\textrm{BadRoot}(a) := \textrm{Descent}(a,\textrm{Trivial}_L)_r \frown (r)$$.

I intend that $$\textrm{BadRoot}(a)$$ plays the same role as the bad root of $$(a,s)$$ with respect to the bad root searching rule for Y-sequence system.

I define a total computable map \begin{eqnarray*} \textrm{Prince} \colon Y & \to & \mathbb{N}_+^{< \omega} \times \mathbb{N}_+^{< \omega} \times \mathbb{N}_+^{< \omega} \times M \\ a & \mapsto & \textrm{Prince}(a) \end{eqnarray*} in the following recursive way:

1. Put $$L := \textrm{Lng}(a)$$.
2. Put $$s := \textrm{Trivial}_L$$.
3. Put $$(h,r,k) := \textrm{BadRoot}(a)$$.
4. Denote by $$g \in \mathbb{N}_+^{< \omega}$$ the initial segment of $$a$$ of length $$r$$.
5. Denote by $$g' \in \mathbb{N}_+^{< \omega}$$ the initial segment of $$\textrm{Difference}(a,s,h)$$ of length $$r$$.
6. Denote by $$b \in \mathbb{N}_+^{< \omega}$$ the sequence given by removing the first $$r$$ entries from $$\textrm{Difference}(a,s,h)$$.
7. Denote by $$s' \in M$$ the sequence given by removing the first $$r$$ entries from $$\textrm{Mountain}(a,s,h)$$.
8. Set $$\textrm{Prince}(a) := (g,g',b,s')$$ .

Namely, $$\textrm{Prince}(a)$$ is the data of the good parts, the bad part, and the ancestor relation of the bad part.

# Expansion

I define a partial computable map \begin{eqnarray*} [ \ ] \colon Y \times \mathbb{N} & \to & Y \\ (a,n) & \mapsto & a[n] \end{eqnarray*} in the following recursive way:

1. If $$a = (1)$$, then set $$a[n] := (1)$$.
2. If $$a \neq (1)$$ and $$\textrm{Lng}(\textrm{Ancestor}(a)) = 1$$, then $$a[n]$$ is the sequence given by removing the rightmost entry from $$a$$.
3. Suppose $$a \neq (1)$$ and $$\textrm{Lng}(\textrm{Ancestor}(a)) \neq 1$$.
1. Put $$(h,r,k) := \textrm{BadRoot}(a)$$.
2. Put $$(g,g',b,s) := \textrm{Prince}(a)$$.
3. Put $$L := \textrm{Lng}(a)$$.
4. Put $$H := \textrm{Height}(a,\textrm{Trivial}_L)$$.
5. If $$H = 0$$, denote by $$\beta^{(n)} \in \mathbb{N}_+^{< \omega}$$ the concatenation of $$n+1$$ copies of $$b$$.
6. Suppose $$H \neq 0$$.
1. Put $$L' := \textrm{Lng}(b)-1$$.
2. Denote by $$c \in \mathbb{N}_+^{< \omega}$$ the sequence given by removing the first $$k$$ entries from $$\textrm{Royal}(a,\textrm{Trivial}_L)[n]$$.
3. Denote by $$s^{(n)} \in M$$ the unique mountain characterised by the following:
1. The equality $$\textrm{Lng}(s^{(n)}) = (n+1)L'$$ holds.
2. For each $$(j,j') \in \mathbb{N}^2$$ satisfying $$j < n+1$$, $$j' < L'$$, and $$s_{j'} \leq j'$$, $$s^{(n)}_{jL'+j'} = s_{j'}$$.
3. For each $$(j,j') \in \mathbb{N}^2$$ satisfying $$j < n+1$$, $$j' < L'$$, and $$s_{j'} > j'$$, $$s^{(n)}_{jL'+j'}= jL'+s_{j'}$$.
4. For each $$i \in \mathbb{N}$$, denote by $$T_i \in \mathbb{N}_+^{< \omega}$$ the sequence given by removing the rightmost entry from $$\textrm{Mountain}(a,\textrm{Trivial}_L,i)$$.
5. For each $$i \in \mathbb{N}$$ smaller than $$H$$, denote by $$m_i \in \mathbb{N}$$ the minimum of a $$j \in \mathbb{N}$$ satisfying $$(i,j) \in \textrm{Descent}(b,s)$$.
6. Denote by $$b^{(n)} \in \mathbb{N}^{\omega}$$ the maximum with respect to $$\leq_{\textrm{lex}}$$ of a $$b' \in \mathbb{N}^{< \omega}$$ satisfying the following:
1. The inequality $$b^{(n)} <_{\textrm{lex}} b$$ holds.
2. The inequality $$\textrm{Lng}(b') \leq (n+1)L'$$ holds.
3. The inequality $$\textrm{Royal}(b',s^{(n)}) \leq_{\textrm{lex}} c$$ holds.
4. For any $$i \in \mathbb{N}$$ smaller than $$H$$, the inequality $$\textrm{Difference}(b',s^{(n)},i) <_{\textrm{lex}} \textrm{Difference}(b,s,i)$$ holds
5. For any $$(i,i',j,j') \in \mathbb{N} \times \mathbb{N} \times \mathbb{N}$$ satisfying that $$i'+h < H$$, $$j < n+1$$, $$j' < L'$$, $$\textrm{Mountain}(b',s^{(n)},i(H-h)+i') = 0$$, and $$(i(H-h)+i',jL'+j')$$ is not an entry of $$\textrm{Descent}(b',s^{(n)})$$, the inequality \begin{eqnarray*} & & \textrm{Difference}(b',s^{(n)},i(H-h)+i')_{jL'+j'} - \textrm{Difference}(b',s^{(n)},i(H-h)+i')_{jL'+m_{i'}} \\ & \leq & \textrm{Difference}(b',s^{(n)},i')_{j'} - \textrm{Difference}(b',s^{(n)},i')_{m_{i'}} \end{eqnarray*} holds.
6. For any $$(i,i',j,j') \in \mathbb{N} \times \mathbb{N} \times \mathbb{N} \times \mathbb{N}$$ satisfying $$i'+h < H$$, $$i \leq j$$, and $$j' < L-r-1), either one of the following holds: 1. \(\textrm{Mountain}(b',s^{(n)},i(H-h)+i')_{r+j(L-r-1)+j'} = 0$$
2. $$\textrm{Mountain}(b',s^{(n)},i(H-h)+i')_{r+j(L-r-1)+j'} \geq (T_{i'+h})_{j'} \neq 0$$
7. For any $$(j,j') \in \mathbb{N}^2$$ satisfying $$j < n+1$$, $$j' < L'$$, and $$s^{(n)}_{jL'+j'} > jL'+j'$$, the inequality $$b'_{jL'+j'} \leq b_{j'}$$ holds.
7. Denote by $$\beta^{(n)} \in \mathbb{N}_+^{< \omega}$$ the maximum with respect to $$\leq_{\textrm{lex}}$$ of a $$b' \in \mathbb{N}^{< \omega}$$ satisfying the following:
1. The inequality $$g \frown b' <_{\textrm{lex}} a$$ holds.
2. The inequality $$\textrm{Lng}(b') \leq (n+1)L'$$ holds.
3. The inequality $$\textrm{Difference}(g \frown b',\textrm{Trivial}_{(n+1)L'},h) \leq_{\textrm{lex}} g' \frown b^{(n)}$$ holds.
4. For any $$i \in \mathbb{N}$$ smaller than $$h$$, The inequality $$\textrm{Mountain}(g \frown b',\textrm{Trivial}_{(n+1)L'},i) \leq_{\textrm{lex}} T_i \frown s^{(n)}$$ holds.
7. Set $$a[n] := g \frown \beta^{(n)}$$.

Since I used the expansion itself to search $$b^{(n)}$$, it is not trivial from the definition whether $$(1,m)[n]$$ is defined for any $$(m,n) \in \mathbb{N} \times \mathbb{N}$$ or not. We actually have $$((1,m),n) \in \textrm{dom}([ \ ])$$ by induction on $$m$$, and hence $$(1,m)[n]$$ makes sense. The maximisation steps 3.6.7 and 3.6.8 actually terminate because the restrictions ensure that the length of $$b^{(n)}$$ and $$\beta^{(n)}$$ are bounded by $$(n+1)L'$$ and entries of them is bounded by the maximum of the entry of $$a$$ multiplied by $$2^{((n+1)L)^2}L^2$$. (This upperbound is just a safe one, and the actual supremum is much smaller than it.)

# Large Function

I define a partial computable map \begin{eqnarray*} \langle \cdot, \cdot \rangle \colon Y^{< \omega} \times \mathbb{N} & \to & \mathbb{N} \\ (A,n) & \mapsto & \langle A, n \rangle \end{eqnarray*} in the following recursive way:

1. If $$A = ()$$, then set $$\langle A, n \rangle := n$$.
2. Suppose $$A = (a)$$ for some $$a \in Y$$.
1. If $$a = (1)$$, then set $$\langle A, n \rangle := n+1$$.
2. If $$a \neq (1)$$, then set $$\langle A, n \rangle := \langle ((\underbrace{a[n+1],\ldots,a[n+1]}_{n+1}), n+1 \rangle$$.
3. Suppose that the length of $$A$$ is greater than $$1$$.
1. Denote by $$B$$ the array given by removing the rightmost entry from $$A$$.
2. Denote by $$a$$ the rightmost entry of $$A$$.
3. Set $$\langle A, n \rangle := \langle B, \langle (a), n \rangle \rangle$$.

This is just a computable variant of FGH. Although I do not know whether it is total or not, $$\langle ((1,10)), 10 \rangle$$ will be a computable large number if it actually terminates.

# Example

Here is a table of examples of expansions under the assumption that the programme works as I intend. I do not know whether the expansions in the table coincide with the intended expansions of Y-sequence systems. At least, expansions in the table are consistent with known expansions in Y-sequence system with respect to the version in 02/06/2020.

$$(1)$$ $$(1)$$ The corresponding Y-sequence is believed to have the same strength as $$(0)$$ in BM2.3 = BM4.
$$(1,1)$$ $$(1)$$
$$(1,1,1)$$ $$(1,1)$$
$$(1,1,1,1)$$ $$(1,1,1)$$
$$(1,2)$$ $$(1,1,1,1,\ldots)$$
$$(1,2,1)$$ $$(1,2)$$
$$(1,2,1,1)$$ $$(1,2,1)$$
$$(1,2,1,2)$$ $$(1,2,1,1,\ldots)$$
$$(1,2,2)$$ $$(1,2,1,2,\ldots)$$
$$(1,2,2,1)$$ $$(1,2,2)$$
$$(1,2,2,2)$$ $$(1,2,2,1,2,2,\ldots)$$
$$(1,2,3)$$ $$(1,2,2,2,\ldots)$$
$$(1,2,3,1)$$ $$(1,2,3)$$
$$(1,2,3,2)$$ $$(1,2,3,1,2,3,\ldots)$$
$$(1,2,3,3)$$ $$(1,2,3,2,3,\ldots)$$
$$(1,2,3,4)$$ $$(1,2,3,3,\ldots)$$
$$(1,2,4)$$ $$(1,2,3,4,\ldots)$$ The corresponding Y-sequence is believed to be the limit of PrSS with respect to BM2.3 = BM4.
$$(1,2,4,1)$$ $$(1,2,4)$$
$$(1,2,4,2)$$ $$(1,2,4,1,2,4,\ldots)$$
$$(1,2,4,3)$$ $$(1,2,4,2,4,\ldots)$$
$$(1,2,4,4)$$ $$(1,2,4,3,5,\ldots)$$
$$(1,2,4,5)$$ $$(1,2,4,4,\ldots)$$
$$(1,2,4,5,4)$$ $$(1,2,4,5,3,5,6,4,6,7,5,7,8,\ldots)$$
$$(1,2,4,6)$$ $$(1,2,4,5,7,8,10,11,13,14,\ldots)$$
$$(1,2,4,6,1)$$ $$(1,2,4,6)$$
$$(1,2,4,6,2)$$ $$(1,2,4,6,1,2,4,6,\ldots)$$
$$(1,2,4,6,3)$$ $$(1,2,4,6,2,4,6,\ldots)$$
$$(1,2,4,6,4)$$ $$(1,2,4,6,3,5,7,\ldots)$$
$$(1,2,4,6,5)$$ $$(1,2,4,6,4,6,\ldots)$$
$$(1,2,4,6,6)$$ $$(1,2,4,6,5,7,9,\ldots)$$
$$(1,2,4,6,7)$$ $$(1,2,4,6,6,\ldots)$$
$$(1,2,4,6,8)$$ $$(1,2,4,6,7,9,11,\ldots)$$
$$(1,2,4,7)$$ $$(1,2,4,6,8,10,12,\ldots)$$
$$(1,2,4,8)$$ $$(1,2,4,7,11,\ldots)$$ The corresponding Y-sequence is believed to be the limit of PSS with respect to BM2.3 = BM4.
$$(1,2,4,8,10)$$
$$(1,2,4,8,9,11,15,16,18,22,23,\ldots)$$
$$(1,2,4,8,10,7)$$
$$(1,2,4,8,10,6,10,12,8,12,14,10,14,16,\ldots)$$
$$(1,2,4,8,10,8)$$
$$(1,2,4,8,10,7,12,15,11,17,21,16,23,28,22,\ldots)$$
The expansion is consistent with the version of Y-sequence in 02/06/2020, but the expansion with respect to the current version of Y-sequence seems to be changed into $$(1,2,4,8,10,7,14,11,17,19,\ldots)$$.
$$(1,2,4,8,15)$$ $$(1,2,4,8,14,22,32,44,\ldots)$$ The corresponding Y-sequence is believed to have the same strength as $$(0,0,0)(1,1,1)(2,2,2)$$ in BM2.3 = BM4.
$$(1,3)$$ $$(1,2,4,8,16,32,64,128,256,\ldots)$$ The corresponding Y-sequence is believed to be the limit of BM2.3 = BM4.
$$(1,3,1)$$ $$(1,3)$$
$$(1,3,2)$$ $$(1,3,1,3,\ldots)$$
$$(1,3,3)$$ $$(1,3,2,5,4,9,8,17,16,33,32,55,\ldots)$$
$$(1,3,4)$$ $$(1,3,3,3,\ldots)$$
$$(1,3,4,1)$$ $$(1,3,4)$$
$$(1,3,4,2)$$ $$(1,3,4,1,3,4,\ldots)$$
$$(1,3,4,2,5)$$ $$(1,3,4,2,4,9,13,8,16,33,\ldots)$$
$$(1,3,4,2,5,6)$$
$$(1,3,4,2,5,5,9,4,9,9,18,8,17,17,\ldots)$$
$$(1,3,4,2,5,6,5)$$
$$(1,3,4,2,5,6,4,9,10,8,17,18,16,33,34,32,\ldots)$$
$$(1,3,4,2,5,6,8,5)$$ $$(1,3,4,2,5,6,8,4,9,10,12,8,\ldots)$$
$$(1,3,4,3)$$
$$(1,3,4,2,5,9,4,9,18,8,17,35,16,\ldots)$$
$$(1,3,4,4)$$ $$(1,3,4,3,10,31,83,\ldots)$$
$$(1,3,5)$$ $$(1,3,4,7,11,18,29,\ldots)$$
$$(1,3,5,1)$$ $$(1,3,5)$$
$$(1,3,5,2)$$ $$(1,3,5,1,3,5,1,3,5,\ldots)$$
$$(1,3,5,2,1)$$ $$(1,3,5,2)$$
$$(1,3,5,2,2)$$ $$(1,3,5,2,1,3,5,2,\ldots)$$
$$(1,3,5,2,3)$$ $$(1,3,5,2,2,2,2,\ldots)$$
$$(1,3,5,2,4)$$ $$(1,3,5,2,3,4,5,6,\ldots)$$
$$(1,3,5,2,5)$$ $$(1,3,5,2,4,8,16,32,\ldots)$$ It is a standard expression, but the computation of its expansion uses the expansion of the non-standard expression $$(1,1,2)$$.
$$(1,3,5,2,5,1)$$ $$(1,3,5,2,5)$$
$$(1,3,5,2,5,2)$$ $$(1,3,5,2,5,1,3,5,2,5,\ldots)$$
$$(1,3,5,2,5,3)$$ $$(1,3,5,2,5,2,5,2,5,\ldots)$$
$$(1,3,5,2,5,4)$$ $$(1,3,5,2,5,3,6,4,7,\ldots)$$ It is a standard expression, but the computation of its expansion uses the expansion of the non-standard expression $$(1,1,2)$$.
$$(1,3,5,2,5,5)$$ $$(1,3,5,2,5,4,9,8,17,16,\ldots)$$ It is a standard expression, but the computation of its expansion uses the expansion of the non-standard expression $$(1,1,2)$$.
$$(1,3,5,2,5,6)$$ $$(1,3,5,2,5,5,5,5,\ldots)$$
$$(1,3,5,2,5,6,8)$$ $$(1,3,5,2,5,6,7,8,9,\ldots)$$
$$(1,3,5,2,5,6,8,1)$$ $$(1,3,5,2,5,6,8)$$
$$(1,3,5,2,5,6,8,2)$$ $$(1,3,5,2,5,6,8,1,3,5,2,5,6,8,\ldots)$$
$$(1,3,5,2,5,6,8,3)$$ $$(1,3,5,2,5,6,8,2,5,6,8,\ldots)$$
$$(1,3,5,2,5,6,8,4)$$
$$(1,3,5,2,5,6,8,3,6,7,9,4,7,8,10,\ldots)$$
$$(1,3,5,2,5,6,8,5)$$
$$(1,3,5,2,5,6,8,4,9,11,14,8,17,21,26,\ldots)$$
It is a standard expression, but the computation of its expansion uses the expansion of the non-standard expression $$(1,1,2)$$.
$$(1,3,5,2,5,6,8,6)$$ $$(1,3,5,2,5,6,8,5,6,8,5,6,8,\ldots)$$
$$(1,3,5,2,5,6,8,7)$$ $$(1,3,5,2,5,6,8,6,8,6,8,\ldots)$$
$$(1,3,5,2,5,6,8,9)$$ $$(1,3,5,2,5,6,8,8,8,8,\ldots)$$
$$(1,3,5,2,5,6,8,10)$$
$$(1,3,5,2,5,6,8,9,11,12,14,15,17,\ldots)$$
$$(1,3,5,2,5,6,8,11)$$ $$(1,3,5,2,5,6,8,10,12,14,16,\ldots)$$
$$(1,3,5,2,5,6,8,12)$$ $$(1,3,5,2,5,6,8,11,15,20,26,33,\ldots)$$
$$(1,3,5,2,5,6,9)$$ $$(1,3,5,2,5,6,8,12,20,36,\ldots)$$
$$(1,3,5,2,5,7)$$ $$(1,3,5,2,5,6,9,7,10,8,11,\ldots)$$
$$(1,3,5,2,5,8)$$
$$(1,3,5,2,5,7,12,19,31,50,81,131,\ldots)$$
It is a standard expression, but the computation of its expansion uses the expansion of the non-standard expression $$(1,1,2)$$.
$$(1,3,5,2,5,9)$$ $$(1,3,5,2,5,8,11,14,17,\ldots)$$
$$(1,3,5,2,5,10)$$
$$(1,3,5,2,5,9,16,27,45,74,121,197,\ldots)$$
It is a standard expression, but the computation of its expansion uses the expansion of the non-standard expression $$(1,1,2,2)$$.
$$(1,3,5,3)$$ $$(1,3,5,2,5,10,4,9,19,8,\ldots)$$
$$(1,3,5,4)$$ $$(1,3,5,3,5,3,5,\ldots)$$
$$(1,3,5,5)$$ $$(1,3,5,4,7,12,11,18,30,29,\ldots)$$
$$(1,3,5,6)$$ $$(1,3,5,5,5,5,\ldots)$$
$$(1,3,5,7)$$ $$(1,3,5,6,9,14,20,29,43,63,\ldots)$$
$$(1,3,6)$$ $$(1,3,5,7,9,11,13,15,17,\ldots)$$
$$(1,3,6,8)$$
$$(1,3,6,7,10,16,23,33,49,71,104,153,\ldots)$$
$$(1,3,7)$$ $$(1,3,6,12,24,48,96,192,384,\ldots)$$
$$(1,3,7,6)$$ $$(1,3,7,5,12,36,\ldots)$$
$$(1,3,7,11)$$ $$(1,3,7,10,16,29,52,91,159,279,\ldots)$$
$$(1,3,7,12)$$ $$(1,3,7,11,19,35,67,\ldots)$$
$$(1,3,7,13)$$ $$(1,3,7,12,20,35,62,109,191,\ldots)$$
$$(1,3,7,14)$$ $$(1,3,7,13,21,31,43,\ldots)$$ It is a standard expression, but the computation of its expansion uses the expansion of the non-standard expression $$(1,2,2,3)$$.
$$(1,3,8)$$ $$(1,3,7,15,31,63,127,255,\ldots)$$
$$(1,3,9)$$ $$(1,3,8,20,48,112,256,\ldots)$$
$$(1,3,9,27,81,243,633,1323,2121,2256)$$
$$(1,3,9,27,81,243,633,1323,2121,2255,2545,3135,4345,6899,12225,18171,19353,21723,26481,36051,55425,94935,139257,148799,\ldots)$$
It is a sequence in Koteitan's test.
$$(1,4)$$ $$(1,3,9,27,81,\ldots)$$
$$(1,4,1)$$ $$(1,4)$$
$$(1,4,2)$$ $$(1,4,1,4,\ldots)$$
$$(1,4,3)$$ $$(1,4,2,6,4,10,8,18,16,\ldots)$$
$$(1,4,3,3)$$ $$(1,4,3,2,6,5,4,\ldots)$$
$$(1,4,4)$$
$$(1,4,3,10,9,28,27,82,81,244,243,\ldots)$$
$$(1,4,13,12,15,12)$$
$$(1,4,13,12,15,11,29,78,77,162,76,\ldots)$$
The corresponding Y-sequence was an unsolved problem in Y-sequence system, but solved in 29/06/2020.
$$(1,5,3,4)$$ $$(1,5,3,3,3,\ldots)$$
$$(1,5,4,10,10)$$ $$(1,5,4,10,9,22,21,50,49,\ldots)$$
$$(1,6,14,29,37,29)$$
$$(1,6,14,29,37,28,55,63,54,105,113,104,203,211,202,\ldots)$$
The corresponding Y-sequence was an unsolved problem in Y-sequence system, but solved in 29/06/2020.
$$(1,10)$$ $$(1,9,81,729,\ldots)$$
$$(1,\textrm{Rayo}(10^{100})+1)$$ $$(1,\textrm{Rayo}(10^{100}),\textrm{Rayo}(10^{100})^2,\ldots)$$ The corresponding Y-sequence is believed to be greater than Rayo's number.
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