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:
- Put \(L := \textrm{Lng}(a)\).
- If \(L = 0\), then \(\textrm{Ancestor}(a) := ()\).
- If \(L \neq 0\) and \(a_{L-1} > L-1\), then \(\textrm{Ancestor}(a) := ()\).
- If \(L \neq 0\) and \(a_{L-1} = 0\), then \(\textrm{Ancestor}(a) = (L-1)\).
- Suppose \(L \neq 0\) and \(0 < a_{L-1} \leq L-1\).
- Denote by \(a'\) the initial segment of \(a\) of length \(a_{L-1}\).
- 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:
- Put \(L := \textrm{Lng}(a)\).
- Suppose \(\textrm{Lng}(s) \geq L\)
- 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:
- Denote by \(a' \in \mathbb{N}_+^{< \omega}\) the initial segment of \(a\) of length \(i+1\).
- Denote by \(s' \in \mathbb{N}_+^{< \omega}\) the initial segment of \(s\) of length \(i+1\).
- Suppose that there is an entry \(j\) of \(\textrm{Ancestor}(s')\) satisfying \(0 \neq a_j < a_i\).
- Denote by \(p \in \mathbb{N}\) the maximum of such a \(j\).
- Set \(b_i := a_i - a_p\).
- Set \(c_i := i - p\).
- Suppose that there is no entry \(j\) of \(\textrm{Ancestor}(s')\) satisfying \(0 \neq a_j < a_i\).
- Set \(b_i := 0\).
- Set \(c_i := 0\).
- Set \(\textrm{PreDifference}(a,s) := (b_j)_{j=0}^{L-1}\).
- Set \(\textrm{PreMountain}(a,s) := (c_j)_{j=0}^{L-1}\).
- 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:
- Suppose \(\textrm{Lng}(s) < L\).
- Set \(\textrm{PreDifference}(a,s) := ()\).
- 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:
- Suppose \(i = 0\).
- Set \(\textrm{Difference}(a,s,i) := a\).
- Set \(\textrm{Mountain}(a,s,i) := \textrm{PreMountain}(a,s)\).
- Suppose \(i \neq 0\).
- Set \(\textrm{Difference}(a,s,i) := \textrm{PreDifference}(\textrm{Difference}(a,s,i-1),\textrm{Mountain}(a,s,i-1))\).
- 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:
- Put \(H := \textrm{Height}(a,s)\).
- For each \(h \in \mathbb{N}\) smaller than \(H\), defined a \(b^{(h)} \in \mathbb{N}^{< \omega} \setminus \{()\}\) in the following recursive way:
- Suppose \(h = H-1\).
- Put \(L := \textrm{Lng}(\textrm{Ancestor}(\textrm{Mountain}(a,s,h)))\).
- Set \(b^{(h)} := ((h,\textrm{Ancestor}(\textrm{Mountain}(a,s,h))_i))_{i=0}^{L-1}\).
- Suppose \(h \neq H-1\).
- Denote by \(i\) the second entry of \(b^{(h+1)}_0\).
- Denote by \(c \in \mathbb{N}^{< \omega}\) the initial segment of \(\textrm{Mountain}(a,s,h)\) of length \(i+1\).
- Put \(L := \textrm{Lng}(\textrm{Ancestor}(c))\).
- Set \(b^{(h)} := ((h,\textrm{Ancestor}(c)_i))_{i=0}^{L-1}\).
- \(\textrm{Descent}(a,s)\) 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:
- Put \(L := \textrm{Lng}(\textrm{Descent}(a,s))\).
- For each \(i \in \mathbb{N}\) smaller than \(L\), put \((h_i,j_i) := \textrm{Descent}(a,s)_i\).
- 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:
- Put \(L := \textrm{Lng}(a)\).
- Put \(b := \textrm{Royal}(a,\textrm{Trivial}_L)\).
- If \(b = ()\), then set \(\textrm{BadRoot}(a) := (0,L-1,L-1)\).
- Suppose \(b \neq ()\).
- Put \((h,r,k) := \textrm{BadRoot}(b)\).
- 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:
- Put \(L := \textrm{Lng}(a)\).
- Put \(s := \textrm{Trivial}_L\).
- Put \((h,r,k) := \textrm{BadRoot}(a)\).
- Denote by \(g \in \mathbb{N}_+^{< \omega}\) the initial segment of \(a\) of length \(r\).
- Denote by \(g' \in \mathbb{N}_+^{< \omega}\) the initial segment of \(\textrm{Difference}(a,s,h)\) of length \(r\).
- Denote by \(b \in \mathbb{N}_+^{< \omega}\) the sequence given by removing the first \(r\) entries from \(\textrm{Difference}(a,s,h)\).
- Denote by \(s' \in M\) the sequence given by removing the first \(r\) entries from \(\textrm{Mountain}(a,s,h)\).
- 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:
- If \(a = (1)\), then set \(a[n] := (1)\).
- If \(a \neq (1)\) and \(\textrm{Lng}(\textrm{Ancestor}(a)) = 1\), then \(a[n]\) is the sequence given by removing the rightmost entry from \(a\).
- Suppose \(a \neq (1)\) and \(\textrm{Lng}(\textrm{Ancestor}(a)) \neq 1\).
- Put \((h,r,k) := \textrm{BadRoot}(a)\).
- Put \((g,g',b,s) := \textrm{Prince}(a)\).
- Put \(L := \textrm{Lng}(a)\).
- Put \(H := \textrm{Height}(a,\textrm{Trivial}_L)\).
- If \(H = 0\), denote by \(\beta^{(n)} \in \mathbb{N}_+^{< \omega}\) the concatenation of \(n+1\) copies of \(b\).
- Suppose \(H \neq 0\).
- Put \(L' := \textrm{Lng}(b)-1\).
- Denote by \(c \in \mathbb{N}_+^{< \omega}\) the sequence given by removing the first \(k\) entries from \(\textrm{Royal}(a,\textrm{Trivial}_L)[n]\).
- Denote by \(s^{(n)} \in M\) the unique mountain characterised by the following:
- The equality \(\textrm{Lng}(s^{(n)}) = (n+1)L'\) holds.
- For each \((j,j') \in \mathbb{N}^2\) satisfying \(j < n+1\), \(j' < L'\), and \(s_{j'} \leq j'\), \(s^{(n)}_{jL'+j'} = s_{j'}\).
- For each \((j,j') \in \mathbb{N}^2\) satisfying \(j < n+1\), \(j' < L'\), and \(s_{j'} > j'\), \(s^{(n)}_{jL'+j'}= jL'+s_{j'}\).
- For each \(i \in \mathbb{N}\), denote by \(T_i \in \mathbb{N}_+^{< \omega}\) the sequence given by removing the rightmost entry from \(\textrm{Mountain}(a,\textrm{Trivial}_L,i)\).
- 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 \((i,j) \in \textrm{Descent}(b,s)\).
- 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:
- The inequality \(b^{(n)} <_{\textrm{lex}} b\) holds.
- The inequality \(\textrm{Lng}(b') \leq (n+1)L'\) holds.
- The inequality \(\textrm{Royal}(b',s^{(n)}) \leq_{\textrm{lex}} c\) holds.
- 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
- 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.
- 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:
- \(\textrm{Mountain}(b',s^{(n)},i(H-h)+i')_{r+j(L-r-1)+j'} = 0\)
- \(\textrm{Mountain}(b',s^{(n)},i(H-h)+i')_{r+j(L-r-1)+j'} \geq (T_{i'+h})_{j'} \neq 0\)
- 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.
- 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:
- The inequality \(g \frown b' <_{\textrm{lex}} a\) holds.
- The inequality \(\textrm{Lng}(b') \leq (n+1)L'\) holds.
- The inequality \(\textrm{Difference}(g \frown b',\textrm{Trivial}_{(n+1)L'},h) \leq_{\textrm{lex}} g' \frown b^{(n)}\) holds.
- 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.
- 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:
- If \(A = ()\), then set \(\langle A, n \rangle := n\).
- Suppose \(A = (a)\) for some \(a \in Y\).
- If \(a = (1)\), then set \(\langle A, n \rangle := n+1\).
- If \(a \neq (1)\), then set \(\langle A, n \rangle := \langle ((\underbrace{a[n+1],\ldots,a[n+1]}_{n+1}), n+1 \rangle\).
- Suppose that the length of \(A\) is greater than \(1\).
- Denote by \(B\) the array given by removing the rightmost entry from \(A\).
- Denote by \(a\) the rightmost entry of \(A\).
- 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 \((1,2,4,8,10,7,14,11,17,19,\ldots)\). See Y sequence#(1,2,4,8,10,8) Problem for more details. |
| \((1,2,4,8,15)\) | \((1,2,4,8,14,22,32,44,\ldots)\) | 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. |
| \((1,3,1)\) | \((1,3)\) | |
| \((1,3,2)\) | \((1,3,1,3,\ldots)\) | |
| \((1,3,3)\) | \((1,3,2,5,4,9,8,17,16,33,32,55,\ldots)\) | |
| \((1,3,4)\) | \((1,3,3,3,\ldots)\) | |
| \((1,3,4,1)\) | \((1,3,4)\) | |
| \((1,3,4,2)\) | \((1,3,4,1,3,4,\ldots)\) | |
| \((1,3,4,2,5)\) | \((1,3,4,2,4,9,13,8,16,33,\ldots)\) | |
| \((1,3,4,2,5,6)\) | \((1,3,4,2,5,5,9,4,9,9,18,8,17,17,\ldots)\)
|
|
| \((1,3,4,2,5,6,5)\) | \((1,3,4,2,5,6,4,9,10,8,17,18,16,33,34,32,\ldots)\)
|
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| \((1,3,4,2,5,6,8,5)\) | \((1,3,4,2,5,6,8,4,9,10,12,8,\ldots)\) | |
| \((1,3,4,3)\) | \((1,3,4,2,5,9,4,9,18,8,17,35,16,\ldots)\)
|
|
| \((1,3,4,4)\) | \((1,3,4,3,10,31,83,\ldots)\) | |
| \((1,3,5)\) | \((1,3,4,7,11,18,29,\ldots)\) | |
| \((1,3,5,1)\) | \((1,3,5)\) | |
| \((1,3,5,2)\) | \((1,3,5,1,3,5,1,3,5,\ldots)\) | |
| \((1,3,5,2,1)\) | \((1,3,5,2)\) | |
| \((1,3,5,2,2)\) | \((1,3,5,2,1,3,5,2,\ldots)\) | |
| \((1,3,5,2,3)\) | \((1,3,5,2,2,2,2,\ldots)\) | |
| \((1,3,5,2,4)\) | \((1,3,5,2,3,4,5,6,\ldots)\) | |
| \((1,3,5,2,5)\) | \((1,3,5,2,4,8,16,32,\ldots)\) | 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 \((1,1,2)\). |
| \((1,3,5,2,5,6)\) | \((1,3,5,2,5,5,5,5,\ldots)\) | |
| \((1,3,5,2,5,6,8)\) | \((1,3,5,2,5,6,7,8,9,\ldots)\) | |
| \((1,3,5,2,5,6,8,1)\) | \((1,3,5,2,5,6,8)\) | |
| \((1,3,5,2,5,6,8,2)\) | \((1,3,5,2,5,6,8,1,3,5,2,5,6,8,\ldots)\) | |
| \((1,3,5,2,5,6,8,3)\) | \((1,3,5,2,5,6,8,2,5,6,8,\ldots)\) | |
| \((1,3,5,2,5,6,8,4)\) | \((1,3,5,2,5,6,8,3,6,7,9,4,7,8,10,\ldots)\)
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| \((1,3,5,2,5,6,8,5)\) | \((1,3,5,2,5,6,8,4,9,11,14,8,17,21,26,\ldots)\)
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It is a standard expression, but the computation of its expansion uses the expansion of the non-standard expression \((1,1,2)\). |
| \((1,3,5,2,5,6,8,6)\) | \((1,3,5,2,5,6,8,5,6,8,5,6,8,\ldots)\) | |
| \((1,3,5,2,5,6,8,7)\) | \((1,3,5,2,5,6,8,6,8,6,8,\ldots)\) | |
| \((1,3,5,2,5,6,8,9)\) | \((1,3,5,2,5,6,8,8,8,8,\ldots)\) | |
| \((1,3,5,2,5,6,8,10)\) | \((1,3,5,2,5,6,8,9,11,12,14,15,17,\ldots)\)
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| \((1,3,5,2,5,6,8,11)\) | \((1,3,5,2,5,6,8,10,12,14,16,\ldots)\) | |
| \((1,3,5,2,5,6,8,12)\) | \((1,3,5,2,5,6,8,11,15,20,26,33,\ldots)\) | |
| \((1,3,5,2,5,6,9)\) | \((1,3,5,2,5,6,8,12,20,36,\ldots)\) | |
| \((1,3,5,2,5,7)\) | \((1,3,5,2,5,6,9,7,10,8,11,\ldots)\) | |
| \((1,3,5,2,5,8)\) | \((1,3,5,2,5,7,12,19,31,50,81,131,\ldots)\)
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It is a standard expression, but the computation of its expansion uses the expansion of the non-standard expression \((1,1,2)\). |
| \((1,3,5,2,5,9)\) | \((1,3,5,2,5,8,11,14,17,\ldots)\) | |
| \((1,3,5,2,5,10)\) | \((1,3,5,2,5,9,16,27,45,74,121,197,\ldots)\)
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It is a standard expression, but the computation of its expansion uses the expansion of the non-standard expression \((1,1,2,2)\). |
| \((1,3,5,3)\) | \((1,3,5,2,5,10,4,9,19,8,\ldots)\) | |
| \((1,3,5,4)\) | \((1,3,5,3,5,3,5,\ldots)\) | |
| \((1,3,5,5)\) | \((1,3,5,4,7,12,11,18,30,29,\ldots)\) | |
| \((1,3,5,6)\) | \((1,3,5,5,5,5,\ldots)\) | |
| \((1,3,5,7)\) | \((1,3,5,6,9,14,20,29,43,63,\ldots)\) | |
| \((1,3,6)\) | \((1,3,5,7,9,11,13,15,17,\ldots)\) | |
| \((1,3,6,8)\) | \((1,3,6,7,10,16,23,33,49,71,104,153,\ldots)\)
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| \((1,3,7)\) | \((1,3,6,12,24,48,96,192,384,\ldots)\) | |
| \((1,3,7,6)\) | \((1,3,7,5,12,36,\ldots)\) | |
| \((1,3,7,11)\) | \((1,3,7,10,16,29,52,91,159,279,\ldots)\) | |
| \((1,3,7,12)\) | \((1,3,7,11,19,35,67,\ldots)\) | |
| \((1,3,7,13)\) | \((1,3,7,12,20,35,62,109,191,\ldots)\) | |
| \((1,3,7,14)\) | \((1,3,7,13,21,31,43,\ldots)\) | It is a standard expression, but the computation of its expansion uses the expansion of the non-standard expression \((1,2,2,3)\). |
| \((1,3,8)\) | \((1,3,7,15,31,63,127,255,\ldots)\) | |
| \((1,3,9)\) | \((1,3,8,20,48,112,256,\ldots)\) | |
\((1,3,9,27,81,243,633,1323,2121,2256)\)
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\((1,3,9,27,81,243,633,1323,2121,2255,2545,3135,4345,6899,12225,18171,19353,21723,26481,36051,55425,94935,139257,148799,\ldots)\)
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It is a sequence in Koteitan's test. |
| \((1,4)\) | \((1,3,9,27,81,\ldots)\) | |
| \((1,4,1)\) | \((1,4)\) | |
| \((1,4,2)\) | \((1,4,1,4,\ldots)\) | |
| \((1,4,3)\) | \((1,4,2,6,4,10,8,18,16,\ldots)\) | |
| \((1,4,3,3)\) | \((1,4,3,2,6,5,4,\ldots)\) | |
| \((1,4,4)\) | \((1,4,3,10,9,28,27,82,81,244,243,\ldots)\)
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| \((1,4,13,12,15,12)\) | \((1,4,13,12,15,11,29,78,77,162,76,\ldots)\)
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The corresponding Y-sequence was an unsolved problem in Y-sequence system, but solved in 29/06/2020. |
| \((1,5,3,4)\) | \((1,5,3,3,3,\ldots)\) | |
| \((1,5,4,10,10)\) | \((1,5,4,10,9,22,21,50,49,\ldots)\) | |
| \((1,6,14,29,37,29)\) | \((1,6,14,29,37,28,55,63,54,105,113,104,203,211,202,\ldots)\)
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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. |