Definition of Dimensional BMS and its correspondence to Y sequence

\( \newcommand{\node}{ {\mathcal N} } \newcommand{\gnd}{ {\mathcal G} } \newcommand{\nat}{ {\mathbb N} } \newcommand{\bm}[1]{\boldsymbol #1} \newcommand{\if}{~{\rm if}~} \newcommand{\lex}{\lt_{\rm lex}} \newcommand{\lexeq}{=_{\rm lex}} \newcommand{\lexlt}{\lt_{\rm lex}} \newcommand{\lexgt}{\gt_{\rm lex}} \newcommand{\lexseq}{\lt_{\rm lexseq}} \newcommand{\lexseqeq}{=_{\rm lexseq}} \newcommand{\lexseqlt}{\lt_{\rm lexseq}} \newcommand{\lexseqgt}{\gt_{\rm lexseq}} \newcommand{\rows}{ {\rm rows}} \newcommand{\cols}{ {\rm cols}} \newcommand{\seq}{ {\rm seq}} \) Japanese translation is available here.

Dimensional BMS (DBMS) is the notation which has quite similar features to Bashicu Matrix System (BMS) and it forms like ((0),(1),(2,1),(3,2,1)) and so on.

Conventional Study

Ecl1psed, the discord user, posted "Dimensional BMS" named notation and its correspondence to some ordinals and the expansion result examples on bashicu-matrix channel in Googology Server of discord on Sep. 14,2019. Though Ecl1psed said that he was "might not have been the first person to come up with this idea", the earliest record of the seemds to be the post above.

On the other hand, Yukito said that he invented a notation named YBM which has the equivalent to this DBMS independently to Ecl1psed, and the expansion rule of YBM and DBMS formed (0)(1)(2,1)(3,2,1)(4,3,2,1)... are identical to the rule of BM4 \({\rm expand(X[n])}\), and he also said YBM and DBMS are correspond to Y sequence under Y(1,3).

AxiomaticSystem, the discord user, created an expansion program of Pointer-BMS, which is the expansion system which is identical to the one of DBMS and posted it to discord on June 3, 2020. In second, AxiomaticSystem also posted the program which translate Pointer-BMS and DBMS and Y sequence each other to the Repl.it, the program posting site and published on July 7, 2020. With that, it is clarified that DBMS and Y sequence can be transpose each other.

Proposition

At first, I'm going to redefine DBMS with mathematical equation in this article.

In second, though that equals to the result by AxiomaticSystem, in the other aspect from that, I try to redefine Y sequence by the correspondence to DBMS.

Notation

I define sets \(T_\seq\) of formal strings consisting of the natural numbers, \((\) and \()\), and also define a function \(\rows(s) \) on \(T_\seq \mapsto \nat \) recursively in the followings:

1. \(() \in T_\seq\).
2. \(\rows(())=0\).
3. \(\left( s\frown (n) \in T_\seq \right) \leftarrow \left( s \in T_\seq \land n \in \nat \right) \).
4. \(\left( \rows(s\frown (n))=\rows(s)+1 \right) \leftarrow \left( s \in T_\seq \land n \in \nat \right) \).

Where \(\frown\) is a binary operator on \( \nat^2 \rightarrow \nat \) and \( (a_0,a_1,\cdots,a_{k_a}) \frown (b_0,b_1,\cdots,b_{k_b}) \) \(= (a_0,a_1,\cdots,a_{k_a},b_0,b_1,\cdots,b_{k_b}) \) when \(a_0,a_1,\cdots,a_{k_a},b_0,b_1,\cdots,b_{k_b} \in \nat\).

I define sets \(T\) of formal strings consisting of \(T_\seq\), \((\) and \()\), and also define a function \(\cols(s) \) on \(T \mapsto \nat \) recursively in the followings:

1. \( (()) \in T \).
2. \( \cols((()))=0 \).
3. \(\left( m\frown (s) \in T \right) \leftarrow \left( m \in T \land s \in T_\seq \right) \).
4. \(\left( \cols(m)=\cols(m)+1 \right) \leftarrow \left( m \in T \land s \in T_\seq \right) \).

Where \(\frown\) is a binary operator on \( T_\seq^2 \rightarrow \nat \) and \( (a_0,a_1,\cdots,a_{k_a}) \frown (b_0,b_1,\cdots,b_{k_b}) \) \(= (a_0,a_1,\cdots,a_{k_a},b_0,b_1,\cdots,b_{k_b}) \) when \(a_0,a_1,\cdots,a_{k_a},b_0,b_1,\cdots,b_{k_b} \in T_\seq\).

-informal plaza-

\(T_\seq\) is the finite sequence and it corresponds to a column of a DBMS, and \(T\) is a finite sequence of the finite sequences and it corresponds to a matrix of a DBMS. By the convention, the trailing zeros in \(T_\seq\) are often omitted, for example, (3,0,0)=(3,0)=(3). By the convention, the commas in \(T\) and the parenthesis which covers whole \(T\) are often ommited, for example, ((0,0,0),(1,1,1))=(0,0,0)(1,1,1). \(\rows(y)\) is the function returning the number of the row(=elements) of \(y\). \(\cols(x)\) is the function \(\cols(x)\) returning the number of the columns of \(x\).

Lexicographical ordered sets

I define \( x\lexseqlt y, ~x\lexseqgt y\) which is 2-arities relation on \(T_\seq^2\) as followings. (\(x,y,z \in T_\seq\), \(a,b \in \nat\), \(x\neq(),y\neq()\),\(z\neq() \))：

1. \( () \lexseq x \).
2. \( x\frown z \lexseq y \leftarrow x \lexseq y \).
3. \( x \lexseq y\frown z \leftarrow x \lexseq y \).
4. \( z\frown (a) \lexseq z\frown (b) \leftrightarrow a \lt b \).
5. \(x \lexseqgt y \leftrightarrow y \lexseqlt x\).

I define \( x\lexlt y, ~x\lexgt y\) which is 2-arities relation on \(T^2\)as followings. (\(x,y,z \in T\),\(a,b \in T_\seq\), \(x\neq(()),y\neq(())\),\(z\neq(()) \)：

1. \( (()) \lex x \).
2. \( x \lex y \rightarrow x\frown z \lex y \).
3. \( x \lex y \rightarrow x \lex y\frown z \).
4. \( z\frown (a) \lex z\frown (b) \leftrightarrow a \lexseq b \).
5. \( y \lexlt x \leftrightarrow x \lexgt y \).

-Informal plaza-

With both \(\lex\) and \(\lexseq\), the rule is correspond to that repeating comparison of the elements 1 by 1 from left-end and going left if they are the same. For example, about the comparison for (4,3,2,0) and (4,3,1,1), 4=4, 3=3, 2>1, so (4,3,2,0) \(\lexseqgt\) (4,3,1,1). about the comparison for (0)(1)(2,1)(3,2,1)(4) and (0)(1)(2,1)(3,2)(4,3), (0)=(0), (1)=(1), (2,1)=(2,1), (3,2,1)\(\lexseqgt\)(3,2)=(3,2,0) then ((0)(1)(2,1)(3,2,1)(4)) \(\lexgt\) ((0)(1)(2,1)(3,2)(4,3)). 1. means that empty sequence is grater than the sequences which has the elements. 2. and 3. means it is less if its head is less. 4. means that comparison is done with the following elements if the first element is equivarent. 5. means that the opposite of \(\lex\) is \(\lexgt\).

Linear order sets

I define \(LT_\seq \subset T_\seq \) as the following:

1. \(() \in LT_\seq \).
2. \((0) \in LT_\seq \).
3. \( (a_0,a_1) \in LT_\seq \leftarrow \left( a_0,a_1 \in \nat \land a_0 \gt a_1 \right) \).
4. \( a \frown (a_0,a_1) \in LT_\seq \\ \leftarrow \left( a, \in LT_\seq \land a_0,a_1 \in \nat \land a_1 \lt a_0 \right) \).

I define \(LT \subset T \) as the following:

1. \((()) \in LT\).
2. \(((0)) \in LT \).
3. \(a\frown ((0)) \in LT \leftarrow a \in LT\).
4. \(a \frown (b_0\frown b)\frown c \frown(d_0\frown d) \leftarrow \left( \\ a \in LT \\ (b\frown b_0) \in T \land \\ c \in T \land \\ (d\frown d_0) \in T \land \\ a \frown (b_0\frown b)\frown c \in LT \land \\ b,d \in LT_\seq \land \\ b_0,d_0 \in \nat \land \\ d_0 \leq b_0+1 \\ \right) \).

-Informal plaza-

\(LT\) is the set which gathers \(T\) accespting only the sequence which meets the following condition. (condition 1.) The value in a row should be reduced. For example, (5,4,3),(5,4,2),(5,4,1),(5,4,0) are allowed but (5,4,5) is increased from 4 to 5 and (5,4,4) is not decrease and it keep 4 to 4, so that they are not allowed. (condition 2.) When you add the new column in the matrix, the first value in a new column should be less than or equal to (the heading number on a row in the previous matrix +1), or 0. For example, the accepted columns to add into the right of (0)(1)(2,1)(2,1) are less than and equal to 0+1=1, 1+1=2, 1+1=2, 2+1=3. In other words, (0)(1)(2,1)(2,1)(0), (0)(1)(2,1)(2,1)(2,1), (0)(1)(2,1)(2,1)(3) and (0)(1)(2,1)(2,1)(3,2,1) can be accepted but (0)(1)(2,1)(2,1)(4) is not allowed. This \(LT\) should be the linear ordered sets because the lexicographical order \(\lex\) has linearity. On the contrary to that, \(LT\) isn't necessary to be well-ordered set because if the expansion rule is harmful and if it has infinite loop for the calculation, it doesn't have the well-ordered feature. After I add proper expansion fundamental sequence, it will gain well-ordered feature.

Fundamental sets

I define \(FT \in LT \) as the followings:

1. \((()) \in FT \).
2. \(((0)) \in FT \).
3. \(x\frown y \in FT \leftarrow \left(\\ x \in FT \land \\ y = \max_{\lex}\{y'| x\frown y' \in LT \land \cols(y')=1 \} \\\right)\)

-Informal plaza-

Fundamental sets \(FT\) are \(LT\) which has the \(\lex\)-maximum column in the matrices whose number of the column is the same. They form (0), (0)(1), (0)(1)(2,1), (0)(1)(2,1)(3,2,1) and \((0)(1)(2,1)\cdots(x,x-1,x-2,\cdots,2,1,0)\cdots\). They are the similar matrices of (0),(0,0)(1,1),(0,0,0)(1,1,1),...,(0…0)(1…1) in Bashicu matrices and (1),(1,2),(1,2,4),(1,2,4,8),...,(1,2,...,2^x,...) in the Y sequnces.

Standard notation

I define Dimensional BMS (DBMS) \(OT \in FT\) as the followings:

1. \(X \in OT \leftarrow X \in FT\).
2. \(\forall k \in \nat~\forall n0\cdots n_k \in \nat~X \in OT \\ \leftarrow {\rm expand}(X)[n0]\cdots[n_k] \in FT\).

-Informal plaza-
The definition of standard notation needs the following definition of the expansion function \(\cdot[\cdot]\). The way to define standard notation can be said that the standard notations are defined as all the fundamental and all the expansion of them. Though There is no proof, \((OT,\lex)\) is expected to be an ordinal notation.

Fundamental sequence

I define the function \(\cdot[\cdot]\) on \(OT \times \nat \mapsto OT \) as the followings.

\begin{eqnarray*} S[n]&=&\left\{\begin{array}{ll} (()) & {\rm if}~S=(()) \\ S_0 \frown S_1 \frown \cdots \frown S_{\cols(S)-2} & {\rm if~}\forall yS_{\cols(S)-1y}=0\\ G \frown B^{(0)} \frown B^{(1)} \frown \cdots \frown B^{(n)} & {\rm otherwise} \end{array}\right.\\ \mathrm{Matrix:}~S&=&(S_0,S_1,\cdots, S_{\cols(S)-1})\in OT\\ \mathrm{Column:}~S_x&=&(S_{x0},S_{x1},\cdots,S_{x(\rows(S_x)-1)})\in T_\seq\\ \mathrm{Good part:}~G&=&(S_0,S_1,\cdots,S_{r-1})\\ \mathrm{Bad part:}~B^{(a)}&=&(B_0^{(a)},B_1^{(a)},\cdots,B_{\cols(S)-2-r}^{(a)})\in T\\ \mathrm{Bad part columns:}~B_x^{(a)}&=&(B_{x0}^{(a)},B_{x1}^{(a)},\cdots,B_{x(\rows(S_x)-1)}^{(a)})\in T_\seq\\ \mathrm{Bad part elements:}~B_{xy}^{(a)}&=&S_{(r+x)y}+a\Delta_{y}A_{xy}\in \nat\\ \mathrm{Ascension amount:}~\Delta_{y}&=&\left\{\begin{array}{ll} S_{(\cols(S)-1)y}-S_{ry}&(\mathrm{if}~y\lt t)\\ 0 &(\mathrm{if}~y\geq t) \end{array}\right.\in \nat\\ \mathrm{Ascension matrix:}~A_{xy}&=&\left\{\begin{array}{ll} 1 &(\mathrm{if}~ \exists a( r=(P_{y})^a(r+x)))\\ 0 &(\mathrm{otherwise}) \end{array}\right.\in \nat\\ \mathrm{Lowermost nonzero:}~t&=&\max\{y|S_{(\cols(S)-1)y}\gt 0\}\in \nat\\ \mathrm{bad root:}~r &=& P_t(\cols(S)-1)\in \nat\\ \mathrm{parent of~}S_{xy}:~P_{y}(x)&=&\left\{\begin{array}{ll} \max\{p|p\lt x \land S_{py} \lt S_{xy} \land \exists a( p=(P_{y-1})^a(x))\} & (\mathrm{if}~y\gt 0)\\ \max\{p|p\lt x \land S_{py} \lt S_{xy} \} & (\mathrm{if}~y=0)\\ \end{array}\right.\in \nat\\ \end{eqnarray*}

-Informal plaza-
Where \(S, S_x, S_{xy}, G, B^{(k)}, B_x^{(k)}, B_{xy}^{(k)}, \Delta_y, A_{xy}, t, r, P_y(x)\) are the similar definition as the article about BM4. In \(X'[b]={\rm expand}(X[b])\) of BMS, its output is a natural number. On the contrary to that, \(\cdot[\cdot]\) of DBMS, its output is a matrix. \(X\) and \(Y\) are replaced into \(\cols(S)\) and \(\rows(S)\).

Fast Growing Hierarchy

I defined the function \(f_X(n)\) on \(OT \times \nat \mapsto \nat \) as the following:

1. \(f_{X\frown((0))}(n)=g(n)\)
2. \(f_{X}(n)=f_{X[g(n)]}(g(n)) \leftarrow \not\exists X'~(X=X' \frown (0)) \)
3. \(g(n)=n^2\)

Large number

Using DBMS, its estimated the following large numbers:

1. [Primitive_Sequence_Number] is equal to \(f^{10}_{(0)(1)(2,1)}(9)\).
2. [Pair_Sequence_Number] is equal to \(f^{10}_{(0)(1)(2,1)(3,2,1)}(9)\).
3. [Bashicu_Matrix_System|Trio Sequence Number] is equal to \(f^{10}_{(0)(1)(2,1)(3,2,1)(4,3,2,1)}(9)\).
4. [Bashicu_Matrix_System|Bashicu Matrix System] is represented by using the function \(\bm{BM}(n)\) on \(\nat\mapsto\nat\) as \({\bm BM}^{10}(9)\).
   1. \(\bm{BM}(n)=f_{X}(n) \leftarrow X=\max_{\lex}\{X'|\cols(X)\leq n\}\)

The correspondence to Y-Sequence

In the followings, I denote Y sequence as \(Y(s_0,s_1,s_2,\cdots,s_k)\) and denote DBMS X as DBMS(X), and made 1-to-1 correspondence of them and redefine Y sequence from that:

1. \({\rm Y}()={\rm DBMS}(((())))\)
2. \({\rm Y}(1)={\rm DBMS}(((0)))\)
3. \({\rm Y}(Y\frown (y))=({\rm DBMS}(X\frown (x))\)<br\>\(\leftarrow (x=\) (the \(y\) th smallest elements which meets the condition that \( (y\in \nat \land x \in T_\seq \land\) \(Y \in T_\seq \land\) <br\>\(X\in OT\land\)<br\> \({\rm Y}(Y)={\rm DBMS}(X) \land\)<br\> \(\rows(X\frown x)=\rows({\rm DBMS}(X))+1\)).)

-Informal plaza-

There is no proof which prove it correspond to existing rules. with the DBMS defined here and the definition of BM4 which defined by the programs and mathematical equation, it can be said that it gives a one of definition of Y sequence under Y(1,3). On the contrary to that Axiomatic System gives the new value of the column of the new added child by the differential seuqnce with its parents arr[i][-1]-arr[i-nex[i][-1]][-2], it equals to that, on my proposal, the fact that DBMS is created by adding k-th smallest column in the columns which can put in the right of the original matrices corresponding to the original Y sequence can be the one of new discovery. You can understand clearly if you see the following correspondence results.

Examples of the correspondence to Y-Sequences

I show the correspondence example based on the construction of Y(1,2,4,8,9,8):

1. Y()=DBMS(()).
2. Y((1)) = DBMS((0)).
3. Y(1,1) = (1st DBMS made by putting in the right of DBMS(0)=Y(1)) = DBMS(0)(0).
4. Y(1,2) = (2nd DBMS made by putting in the right of DBMS(0)=Y(1)) = DBMS(0)(1).
5. Y(1,2,1) = (1st DBMS made by putting in the right of DBMS(0)(1)=Y(1,2)) = DBMS(0)(1)(0).
6. Y(1,2,2) = (2nd DBMS made by putting in the right of DBMS(0)(1)=Y(1,2)) = DBMS(0)(1)(1).
7. Y(1,2,3) = (3rd DBMS made by putting in the right of DBMS(0)(1)=Y(1,2)) = DBMS(0)(1)(2).
8. Y(1,2,4) = (4th DBMS made by putting in the right of DBMS(0)(1)=Y(1,2)) = DBMS(0)(1)(2,1).
9. Y(1,2,4,1) = (1st DBMS made by putting in the right of DBMS(0)(1)(2,1)=Y(1,2,4)) = DBMS(0)(1)(2,1)(0).
10. Y(1,2,4,2) = (2nd DBMS made by putting in the right of DBMS(0)(1)(2,1)=Y(1,2,4)) = DBMS(0)(1)(2,1)(1).
11. Y(1,2,4,3) = (3rd DBMS made by putting in the right of DBMS(0)(1)(2,1)=Y(1,2,4)) = DBMS(0)(1)(2,1)(2).
12. Y(1,2,4,4) = (4th DBMS made by putting in the right of DBMS(0)(1)(2,1)=Y(1,2,4)) = DBMS(0)(1)(2,1)(2,1).
13. Y(1,2,4,5) = (5th DBMS made by putting in the right of DBMS(0)(1)(2,1)=Y(1,2,4)) = DBMS(0)(1)(2,1)(3).
14. Y(1,2,4,6) = (6th DBMS made by putting in the right of DBMS(0)(1)(2,1)=Y(1,2,4)) = DBMS(0)(1)(2,1)(3,1).
15. Y(1,2,4,7) = (7th DBMS made by putting in the right of DBMS(0)(1)(2,1)=Y(1,2,4)) = DBMS(0)(1)(2,1)(3,2).
16. Y(1,2,4,8) = (8th DBMS made by putting in the right of DBMS(0)(1)(2,1)=Y(1,2,4)) = DBMS(0)(1)(2,1)(3,2,1).
17. Y(1,2,4,8,1) = (1st DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)=Y(1,2,4,8)) = DBMS(0)(1)(2,1)(3,2,1)(0).
18. Y(1,2,4,8,2) = (2nd DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)=Y(1,2,4,8)) = DBMS(0)(1)(2,1)(3,2,1)(1).
19. Y(1,2,4,8,3) = (3rd DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)=Y(1,2,4,8)) = DBMS(0)(1)(2,1)(3,2,1)(2).
20. Y(1,2,4,8,4) = (4th DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)=Y(1,2,4,8)) = DBMS(0)(1)(2,1)(3,2,1)(2,1).
21. Y(1,2,4,8,5) = (5th DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)=Y(1,2,4,8)) = DBMS(0)(1)(2,1)(3,2,1)(3).
22. Y(1,2,4,8,6) = (6th DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)=Y(1,2,4,8)) = DBMS(0)(1)(2,1)(3,2,1)(3,1).
23. Y(1,2,4,8,7) = (7th DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)=Y(1,2,4,8)) = DBMS(0)(1)(2,1)(3,2,1)(3,2).
24. Y(1,2,4,8,8) = (8th DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)=Y(1,2,4,8)) = DBMS(0)(1)(2,1)(3,2,1)(3,2,1).
25. Y(1,2,4,8,9) = (9th DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)=Y(1,2,4,8)) = DBMS(0)(1)(2,1)(3,2,1)(4).
26. Y(1,2,4,8,9,1) = (1th DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)(4)=Y(1,2,4,8,9)) = DBMS(0)(1)(2,1)(3,2,1)(4)(0).
27. Y(1,2,4,8,9,2) = (2th DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)(4)=Y(1,2,4,8,9)) = DBMS(0)(1)(2,1)(3,2,1)(4)(1).
28. Y(1,2,4,8,9,3) = (3th DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)(4)=Y(1,2,4,8,9)) = DBMS(0)(1)(2,1)(3,2,1)(4)(2).
29. Y(1,2,4,8,9,4) = (4th DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)(4)=Y(1,2,4,8,9)) = DBMS(0)(1)(2,1)(3,2,1)(4)(2,1).
30. Y(1,2,4,8,9,5) = (5th DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)(4)=Y(1,2,4,8,9)) = DBMS(0)(1)(2,1)(3,2,1)(4)(3).
31. Y(1,2,4,8,9,6) = (6th DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)(4)=Y(1,2,4,8,9)) = DBMS(0)(1)(2,1)(3,2,1)(4)(3,1).
32. Y(1,2,4,8,9,7) = (7th DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)(4)=Y(1,2,4,8,9)) = DBMS(0)(1)(2,1)(3,2,1)(4)(3,2).
33. Y(1,2,4,8,9,8) = (8th DBMS made by putting in the right of DBMS(0)(1)(2,1)(3,2,1)(4)=Y(1,2,4,8,9)) = DBMS(0)(1)(2,1)(3,2,1)(4)(3,2,1).

Others

DBMS is, originally, made to construct the Bashicu matrix system with the elements in the multiple dimensional space with the expansion using "doubled comma" and "tripled comma" like (0)(1)(2,1,,1) and said that it can be extended on the limit of (0)(1)(2,1,,1,,,1,,,,1,,,,,1...)。