S(n) map

s(n) map is a, which is a function of "function to function", defined by Japanese googologist Fish in 2002 to define Fish number 3. The alphabet s in \(s(n)\) map was taken from a Japanese word shazou, which means mapping.

\begin{eqnarray*} s(1)f & := & g; g(x)=f^x(x) \\ s(n)f & := & g; g(x)=[s(n-1)^x]f(x) (if n>1) \end{eqnarray*}

Let \(f(x) = x+1\), and the growth rate can be calculated as:

\begin{eqnarray*} f^2(x) & = & x+2 \\ f^3(x) & = & x+3 \\ s(1)f(x) & = & f^x(x) = x+x = 2x \\ s(1)^2f(x) & = & g^x(x) = 2^x x \approx 2^x where g(x)=2x \\ s(1)^3f(x) & \approx & h^{x}(x) \approx 2 \uparrow ^2 x where h(x)=2^x \\ s(2)f(x) & = & s(1)^{x}f(x) \approx 2\uparrow ^x x \approx A(x,x) = A(1,0,x) \approx f_{\omega}(x) \end{eqnarray*}

Here, \(A\) is Taro's multivariable Ackermann function, where the growth rate in FGH is \begin{eqnarray*} A(..., a3, a2, a1, a0, n) \approx f_{... + \omega^3･a3 + \omega^2･a2 + \omega･a1 + a0}(n) \end{eqnarray*}

Let \(f(n) = A(X, b, n)\) (X is a vector in any length), and \begin{eqnarray*} A(X, b+1, n) & = & A(X, b, A(X, b+1, n-1)) \\ & = & f(A(X, b+1, n-1)) \\ & = & f^2(A(X, b+1, n-2)) \\ & = & … = f^n(A(X, b+1, 0)) \\ & \approx & f^n(n) \end{eqnarray*}

Therefore, comparing the 3 functions, they all have similar growth rate. \(s(1)\) map has the same effect of adding 1 to the ordinal in FGH and adding 1 to the second parameter from right in the Ackermann function. This results in
 * \(s(1)f(x) = f^x(x)\)
 * \(f_{\alpha+1}(n) = f^n_\alpha(n)\)
 * \(A(X, b+1, n) = f^n(n)\) where \(f(n) = A(X, b, n)\)

\begin{eqnarray*} s(1)s(2)f(x) & \approx & A(1,1,x) \approx f_{\omega + 1}(x) \\ s(1)^2 s(2)f(x) & \approx & A(1,2,x) \approx f_{\omega + 2}(x) \\ s(1)^n s(2)f(x) & \approx & A(1,n,x) \approx f_{\omega + n}(x) \end{eqnarray*}

and by diagonizing \(s(1)\) again,

\begin{eqnarray*} s(2)^2 f(x) = s(1)^x s(2)f(x) \approx A(1,x,x) = A(2,0,x) \approx f_{\omega \times 2}(x) \end{eqnarray*}

Calculation goes in the same way

\begin{eqnarray*} s(2)^n f(x) & \approx & A(n,0,x) \approx f_{\omega \times n}(x) \\ s(3)f(x) & = & s(2)^{x}f(x) \approx A(x,0,x) = A(1,0,0,x) \approx f_{\omega^2}(x) \\ s(3)^2 f(x) & \approx & A(2,0,0,x) \approx f_{\omega^2 \times 2}(x) \\ s(3)^n f(x) & \approx & A(n,0,0,x) \approx f_{\omega^2 \times n}(x) \\ s(4)f(x) & = & s(3)^{x}f(x) \approx A(x,0,0,x) = A(1,0,0,0,x) \approx f_{\omega^3}(x) \\ s(1)^4 s(2)^3 s(3)^2s(4)f(x) & \approx & A(1,2,3,4,x) \approx f_{\omega^3+\omega^2 \times 2+\omega \times 3 + 4}(x) \\ s(5)f(x) & \approx & f_{\omega^4}(x) \\ s(6)f(x) & \approx & f_{\omega^5}(x) \\ s(n)f(x) & \approx & f_{\omega^{n-1}}(x) \\ s(x)f(x) & \approx & f_{\omega^\omega}(x) \end{eqnarray*}

Therefore, by applying \(s(x)\) map, which diagonizes \(s(n)\) map, to the function \(f(x)=x+1\), the growth rate is \(f_{\omega^\omega}(x)\), similar to array notation and Taro's multivariable Ackermann function.