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Taro's multivariable Ackermann function
Based onsuccessor function
Growth rate$$f_{\omega^\omega}(n)$$

Taro's multivariable Ackermann function $$A(x_1,x_2, …, x_n)$$ is a recursive function invented by a Japanese googologist Taro (2007) and described by Fish (2013)

[1]. It is similar to Ackermann function with 2 variables, and with more than 3 variables grows at similar rate with Array notation, despite its simple definition.

## Definition

• X : vector of integers larger than or equal to 0, with the length larger than or equal to 0 (ex., [3, 1, 0, 0])
• Y : vector of 0 with the length larger than or equal to 0 (ex., [0, 0, 0, 0, 0])
• a, b : integer larger than or equal to 0

\begin{eqnarray*} A(Y, a) & = & a+1 \\ A(X, b+1, 0) & = & A(X, b, 1) \\ A(X, b+1, a+1) & = & A( X, b, A(X, b+1, a) ) \\ A(X, b+1, 0, Y, a ) & = & A(X, b, a, Y, a) \end{eqnarray*}

In other words:

• If all but/and the last variable are zeros,the function value is the successor of the last variable.
• If the second to last variable is a non-zero:
• If the last variable is zero:
• Replace the last variable to 1.
• Reduce the second to last variable in 1.
• Else:
• Replace the last variable with value of the function, but with its last variable reduced by 1.
• Else (the second to last entry is a end of a Y vector (which will be named Z):
• Replace the first variable of Z with the last variable.
• Reduce the last variable before Z by 1.

## Calculation

Secret of the fast rate of growing of this function is in the 4th equation. In order to reduce the "b" variable by 1, it diagonizes the right variable from 0 to a. It is enough to make strong recursion. If we count the variables from right to left, 2nd variable recurses the 1st variable (primitive recursion in Ackermann function), 3rd variable recurses the second variable, where recursion of primitive recursion makes double recursion, 4th variable recurses the 3rd variable. As the 'n'th variable recurses the 'n-1'th variable, level of recursion steadily goes up.

Example calculation is shown to compare this with Graham's number with help of Conway's chained arrow notation. With 2 variable, it is similar to standard Ackermann function and can be compared as

\begin{eqnarray*} A(x,y) & \approx & 3 \rightarrow y \rightarrow x-2 \\ \end{eqnarray*}

With 3 variables,

\begin{eqnarray*} A(1,1,0) & = & A(1,0,1) = A(0,1,1) = A(1,1) = 3 \\ A(1,1,1) & = & A(1,0,A(1,1,0)) = A(1,0,3) = A(3,3) = 61 \\ A(1,1,2) & = & A(1,0,61) = A(61,61) > 3 \rightarrow 3 \rightarrow 2 \rightarrow 2 \\ A(1,1,3) & \approx & A(1,0,3 \rightarrow 3 \rightarrow 2 \rightarrow 2) \approx 3 \rightarrow 3 \rightarrow 3 \rightarrow 2 \\ A(1,1,4) & \approx & A(1,0,3 \rightarrow 3 \rightarrow 2 \rightarrow 3) \approx 3 \rightarrow 3 \rightarrow 4 \rightarrow 2 \\ A(1,1,x) & \approx & 3 \rightarrow 3 \rightarrow x \rightarrow 2 \\ A(1,1,65) & \approx & 3 \rightarrow 3 \rightarrow 65 \rightarrow 2 > G \\ A(1,2,0) & = & A(1,1,1) = 61 \\ A(1,2,1) & = & A(1,1,61) > 3 \rightarrow 3 \rightarrow 61 \rightarrow 2 \\ A(1,2,2) & \approx & A(1,1,3 \rightarrow 3 \rightarrow 61 \rightarrow 2) > 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow 61 \rightarrow 2) \rightarrow 2 > A(1,1,65) \end{eqnarray*}

And therefore A(1,2,2) > A(1,1,65) > Graham's number.

The following relationship was calculated:[2]

For $$x=1, y>1$$ or $$x>1, y+z>0$$

$A(x,y,z) < \underbrace{3 \rightarrow3 \rightarrow \cdots \rightarrow 3}_{x+1 \text{reps of} 3s} \rightarrow z+2 \rightarrow y+1 < A(x,y,z+1)$

Therefore function A(x,y,z) has recursive level corresponding to x+3 variables of Conway's chained arrow notation, and A(1,0,1,2) exceeds Conway's chained arrow notation with Graham's numbers length.

\begin{eqnarray*} A(1,0,1,0) & = & A(1,0,0,1) = A(1,0,1) = A(1,1) = 3 \\ A(1,0,1,1) & = & A(1,0,0,A(1,0,1,0)) \\ & = & A(1,0,0,3) = A(3,0,3) = A(2,3,3) \\ A(1,0,1,2) & = & A(1,0,0,A(1,0,1,1)) \\ & = & A(1,0,0,A(2,3,3)) \\ & = & A(A(2,3,3),0,A(2,3,3)) \approx \underbrace{3 \rightarrow 3 \rightarrow 3 ... 3 \rightarrow 3 \rightarrow 3}_{A(2,3,3)} \end{eqnarray*}

The growth rate of this function was compared with fast-growing hierarchy as follows.

\begin{eqnarray*} A(n, n) & \approx & f_{\omega}(n) \\ A(1, 0, n) & \approx & f_{\omega}(n) \\ A(a, 0, n) & \approx & f_{\omega･a}(n) \\ A(n, 0, n) & \approx & f_{\omega^2}(n) \\ A(1, 0, 0, n) & \approx & f_{\omega^2}(n) \\ A(a, 0, 0, n) & \approx & f_{\omega^2･a}(n) \\ A(n, 0, 0, n) & \approx & f_{\omega^3}(n) \\ A(1, 0, 0, 0, n) & \approx & f_{\omega^3}(n) \\ A(a, 0, 0, 0, n) & \approx & f_{\omega^3･a}(n) \\ A(n, 0, 0, 0, n) & \approx & f_{\omega^4}(n) \\ A(..., a3, a2, a1, a0, n) & \approx & f_{... + \omega^3･a3 + \omega^2･a2 + \omega･a1 + a0}(n) \\ A(\underbrace{1,1,...,1}_n) & \approx & f_{\omega^\omega}(n) \\ \end{eqnarray*}

## Approximations in other notations

$$A(..., d, c, b, a, n)$$ is approximated as follows.

Notation Approximation
BEAF $$\lbrace n,2,a+1,b+1,c+1,d+1,... \rbrace$$
Fast-growing hierarchy $$f_{... \omega^3 d + \omega^2 c + \omega b + a}(n)$$
Hardy hierarchy $$H_{\omega^{... \omega^3 d + \omega^2 c + \omega b + a} }(n)$$

## Sources

Fish numbers: Fish number 1 · Fish number 2 · Fish number 3 · Fish number 4 · Fish number 5 · Fish number 6 · Fish number 7
Mapping functions: S map · SS map · S(n) map · M(n) map · M(m,n) map
By Aeton: Okojo numbers · N-growing hierarchy
By BashicuHyudora: Primitive sequence number · Pair sequence number · Bashicu matrix system
By Kanrokoti: KumaKuma ψ function
By 巨大数大好きbot: Flan numbers
By Jason: Irrational arrow notation · δOCF · δφ · ε function
By mrna: 段階配列表記 · 降下段階配列表記 · 多変数段階配列表記 · SSAN · S-σ
By Nayuta Ito: N primitive
By p進大好きbot: Large Number Garden Number
By Yukito: Hyper primitive sequence system · Y sequence · YY sequence · Y function
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Chinese, Japanese and Korean counting system: Wan · Yi · Zhao · Jing · Gai · Zi · Rang · Gou · Jian · Zheng · Zai · Ji · Gougasha · Asougi · Nayuta · Fukashigi · Muryoutaisuu
Other: Taro's multivariable Ackermann function · TR function · Arai's $$\psi$$ · Sushi Kokuu Hen

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