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 \(\textrm{Difference}(a,s,i)\).


Bad Root

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.

sequence expansion comments
\((1)\) \((1)\) 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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