User blog:P進大好きbot/Elementary Large Number

I introduce an elementary large function whose growth rate is greater than \(f_{\varepsilon_0}\) in FGH. I guess that the growth rate is \(f_{\varepsilon_0+2}\) or \(f_{\varepsilon_0+3}\) (or is perhaps bounded by \(f_{\varepsilon_0+5}\)). Could someone estimate an upper bound in a precise (not intuition-based) way?

= Rule =

This is an entry of a Japanese googology event. The aim of the event is "to make a well-defined analogue of Sam's number". Participants submit large numbers, and create a new large number as the sum of them.

Unlike other existing analogues of Sam's number, participants are allowed only to use \({+}, {-}, {\times}, {/}\), and functions whose definitions are precisely explicitly written by them, in order to avoid intuition-based cheating like submitting "the natural number defined as the resulting large number of this event plus \(1\)" or an ill-defined number whose creator is not able to understand the reason why it is ill-defined even though problems are pointed out because of the lack of knowledge on notions appearing in the description.

Therefore this blogpost is quite elementary. Indeed, I only used \({+}, {-}, {\times}. {/}\), power, and composition. Well, in order to use power and composition, I should write down the precise definition in the next section. Readers can skip it, because I think that almost all of us know what power \(x^n\) means.

Please be patient if you do not like saladiblity derived from such restrictions. I like such rules as well as restricting large numbers to computable ones ;D

= Preparation =

I denote by \(\mathbb{N}_{+}\) the set of positive integers.

For a positive integer \(x\) and an integer \(n\), I define a positive rational number \(x^n\) in the following recursive way:
 * 1) If \(n = 0\), then \(x^n = 1\).
 * 2) If \(n > 0\), then \(x^n = x^{n-1} \times x\).
 * 3) If \(n < 0\), then \(x^n = \frac{1}{x^{-n}}\).

For a function \(f \colon \mathbb{N}_{+} \to \mathbb{N}_{+}\) and a natural number \(n\), I define a function \begin{eqnarray*} f^n \colon \mathbb{N}_{+} & \to & \mathbb{N}_{+} \\ x & \mapsto & f^n(x) \end{eqnarray*} in the following recursive way:
 * 1) If \(n = 0\), then \(f^n(x) = x\).
 * 2) If \(n > 0\), then \(f^n(x) = f(f^{n-1}(x))\).

= Elementary Large Function =

I define a function \(\textrm{AC}_1(x)\). For this purpose, I introduce two other functions \(\downarrow(x)\) and \(\uparrow(x)\).

I define a function \begin{eqnarray*} \downarrow \colon \mathbb{N}_{+} & \to & \mathbb{N}_{+} \\ x & \mapsto & \downarrow(x) \end{eqnarray*} in the following way:
 * 1) Put \(n_0 := \max \{i \in \mathbb{N} \mid 2^{-i} x \in \mathbb{N}_{+}\} \in \mathbb{N}\).
 * 2) Put \(a_0 := 2^{-n_0} x \in \mathbb{N}_{+}\).
 * 3) If \(a_0 = 1\), then \(\downarrow(x) = 2^x\).
 * 4) Suppose \(a_0 > 1\).
 * 5) Put \(m_0 := \max \{i \in \mathbb{N} \mid 2^i \leq a_0\} \in \mathbb{N}_{+}\).
 * 6) Put \(b_0 := a_0-2^{m_0} \in \mathbb{N}_{+}\).
 * 7) If \(2^{m_0-1} \leq b_0\), then \(\downarrow(x) = 2^x b_0\).
 * 8) Suppose \(2^{m_0-1} > b_0\).
 * 9) Put \(c := \max \{i \in \mathbb{N} \mid 2^i \leq b_0\} \in \mathbb{N}_{+}\).
 * 10) Put \(a_1 := 2(m_0-c)-1 \in \mathbb{N}_{+}\).
 * 11) Put \(m_1 := \max \{i \in \mathbb{N} \mid 2^i \leq a_1\} \in \mathbb{N}\).
 * 12) Put \(b_1 = a_1-2^{m_1} \in \mathbb{N}_{+}\).
 * 13) Suppose \(2^{m_1-1} \leq b_1\).
 * 14) Put \(d := 2^{-1}(b_1+1) \in \mathbb{N}_{+}\).
 * 15) Then \(\downarrow(x) = 2^x(b_0+2^{c+d}(2^{dx}-1))\).
 * 16) Suppose \(2^{m_1-1} > b_1\).
 * 17) Put \(y := \downarrow(2^x a_1) \in \mathbb{N}_{+}\).
 * 18) Put \(n_1 := \max \{i \in \mathbb{N} \mid 2^{-i} y \in \mathbb{N}_{+}\} \in \mathbb{N}\).
 * 19) Put \(a_2 := 2^{-n_1} y \in \mathbb{N}_{+}\).
 * 20) Put \(d := 2^{-1}(a_2+1) \in \mathbb{N}_{+}\).
 * 21) Then \(\downarrow(x) = 2^y(b_0+2^{c+d})\).

I define a function \begin{eqnarray*} \uparrow \colon \mathbb{N}_{+} & \to & \mathbb{N}_{+} \\ x & \mapsto & \uparrow(x) \end{eqnarray*} in the following recursive way:
 * 1) Put \(n_0 := \max \{i \in \mathbb{N} \mid 2^{-i} x \in \mathbb{N}_{+}\} \in \mathbb{N}\).
 * 2) Put \(a_0 := 2^{-n_0} x \in \mathbb{N}_{+}\).
 * 3) If \(a_0 = 1\), then \(\uparrow(x) = \downarrow(x)\).
 * 4) If \(a_0 > 1\), then \(\uparrow(x) = 2^{\uparrow^x(\downarrow(x))} x\).

I define a function \begin{eqnarray*} \textrm{AC}_1 \colon \mathbb{N}_{+} & \to & \mathbb{N}_{+} \\ x & \mapsto & \textrm{AC}_1(x) \end{eqnarray*} in the following recursive way:
 * 1) If \(x = 1\), then \(\textrm{AC}_1(x) = \uparrow(x)\).
 * 2) If \(x > 1\), then \(\textrm{AC}_1(x) = \uparrow \left( \textrm{AC}_1(x-1) \left( 1 + 2^{\textrm{AC}_1(x-1)} \right) \right)\).

= Elementary Large Number =

I submit \(\textrm{AC}_1(10)\) to the event.

= Example =

For convenience for displaying the examples, I define a positive integer \(x \otimes^m n\) for positive integers \(x\), \(n\), and \(m\) in the following recursive way:
 * 1) If \(m = 1\), then \(x \otimes^m n = x^n\).
 * 2) If \(m > 1\), then \(x \otimes^m n = x^{x \otimes^{m-1} n}\).

\begin{eqnarray*} \downarrow(1) & = & 2^1 = 2 \\ \downarrow(2) & = & 2^2 = 4 \\ \downarrow(3) & = & 2^3 \times 1 = 2^3 = 8 \\ \downarrow(4) & = & 2^4 = 16 \\ \downarrow(5) & = & 2^8(1+2^{0+1}) = 768 \\ \downarrow(6) & = & 2^6 \times 1 = 64 \\ \downarrow(7) & = & 2^7 \times 3 = 384 \\ \downarrow(8) & = & 2^8 = 256 \\ \downarrow(9) & = & 2^{2^{2^{2^{2560}}} \times 3} \left( 1+2^{0+2^{0+2}} \right)=2^{2^{2^{2^{2560}}} \times 3} \times 17 \\ \downarrow(10) & = & 2^{2^{3072}}(1+2^{0+1}) = 2^{2^{3072}} \times 3 \\ \downarrow(11) & = & 2^{2^{6144}}(3+2^{1+1}) = 2^{2^{6144}} \times 7 \\ \downarrow(12) & = & 2^{12} \times 1 = 4048 \\ \downarrow(13) & = & 2^{13} \times 5 = 40480 \\ \downarrow(14) & = & 2^{14} \times 3 = 48576 \\ \downarrow(15) & = & 2^{15} \times 7 = 226688 \\ \downarrow(16) & = & 2^{16} \\ \downarrow(17) & = & 2^{2^{2^{17} \times 7} \times 3}(1+2^{0+2}) = 2^{2^{2^{17} \times 7} \times 3} \times 5 \\ \downarrow(18) & = & 2^{2^{2^{18} \times 5} \times 3}(1+2^{0+2}) = 2^{2^{2^{18} \times 5} \times 3} \times 5 \\ \downarrow(19) & = & 2^{2^{2^{19} \times 5} \times 3}(3+2^{1+2}) = 2^{2^{2^{19} \times 5} \times 3} \times 11 \\ \downarrow(20) & = & 2^{2^{2^{20} \times 3}}(1+2^{0+1}) = 2^{2^{2^{20} \times 3}} \times 3 \\ & \vdots & \\ \downarrow(2560) & = & 2^{2^{2^{2560}}}(1+2^{0+1}) = 2^{2^{2^{2560}}} \times 3 \\ \downarrow(3072) & = & 2^{3072} \times 1 = 2^{3072} \\ \downarrow(6144) & = & 2^{6144} \times 1 = 2^{6144} \\ & \vdots & \\ \downarrow(2^{17} \times 7) & = & 2^{2^{17} \times 7} \times 3 \\ \downarrow(2^{18} \times 5) & = & 2^{2^{18} \times 5}(1+2^{0+1}) = 2^{2^{18} \times 5} \times 3 \\ \downarrow(2^{19} \times 5) & = & 2^{2^{19} \times 5}(1+2^{0+1}) = 2^{2^{19} \times 5} \times 3 \\ \downarrow(2^{20} \times 3) & = & 2^{2^{20} \times 3} \times 1 = 2^{2^{20} \times 3} \\ & \vdots & \\ \downarrow(2^{2560} \times 3) & = & 2^{2^{2560}} \times 1 = 2^{2^{2560}} \\ & \vdots & \\ \downarrow \left( 2^{2^{3072}} \times 3 \right) & = & 2^{2^{2^{3072}} \times 3} \times 1 = 2^{2^{2^{3072}} \times 3} \\ \downarrow \left( 2^{2^{2^{3072}} \times 3} \right) & = & 2^{2^{2^{2^{3072}} \times 3}} \\ \downarrow \left( 2^{2^{2^{2^{3072}} \times 3}} \right) & = & 2^{2^{2^{2^{2^{3072}} \times 3}}} \\ & \vdots & \end{eqnarray*}

\begin{eqnarray*} \uparrow(1) & = & \downarrow(1) = 2 \\ \uparrow(2) & = & \downarrow(2) = 4 \\ \uparrow(10) & = & 2^{\uparrow^{10}(\downarrow(10))} \times 10 = 2^{\uparrow^{10} \left( 2^{2^{3072}} \times 3 \right)} \times 10 \\ & = & 2^{\uparrow^9 \left( 2^{\left( 2 \otimes^{2^{2^{3072}} \times 3+1} \left( 2^{2^{3072}} \times 3 \right) \right)+2^{3072}} \times 3 \right)} \times 10 \\ & = & 2^{\uparrow^8 \left( 2^{2^{\left( 2 \otimes^{2^{2^{3072}} \times 3+1} \left( 2^{2^{3072}} \times 3 \right) \right)+2^{3072}} \times 3} \right)} \times 10 \\ & \vdots & \\ & = & \left( 2 \otimes^{10} 2^{\left( 2 \otimes^{2^{2^{3072}} \times 3+1} \left( 2^{2^{3072}} \times 3 \right) \right)+2^{3072}} \times 3 \right) \times 10 \\ \uparrow \left( 2^{2^{3072}} \times 3 \right) & = & 2^{\uparrow^{2^{2^{3072}} \times 3} \left( \downarrow \left( 2^{2^{3072}} \times 3 \right) \right)} \times 2^{2^{3072}} \times 3 \\ & = & 2^{\uparrow^{2^{2^{3072}} \times 3} \left( 2^{2^{2^{3072}} \times 3} \right)+2^{3072}} \times 3 \\ & = & 2^{\uparrow^{2^{2^{3072}} \times 3-1} \left( 2^{2^{2^{2^{3072}} \times 3}} \right)+2^{3072}} \times 3 \\ & = & 2^{\uparrow^{2^{2^{3072}} \times 3-2} \left( 2^{2^{2^{2^{2^{3072}} \times 3}}} \right)+2^{3072}} \times 3 \\ & \vdots & \\ & = & 2^{\left( 2 \otimes^{2^{2^{3072}} \times 3+1} 2^{2^{3072}} \times 3 \right)+2^{3072}} \times 3 \\ \uparrow \left( 2^{2^{2^{3072}} \times 3} \right) & = & \downarrow \left( 2^{2^{2^{3072}} \times 3} \right) = 2^{2^{2^{2^{3072}} \times 3}} \\ \uparrow \left( 2^{\left( 2 \otimes^{2^{2^{3072}} \times 3+1} \left( 2^{2^{3072}} \times 3 \right) \right)+2^{3072}} \times 3 \right) & = & 2^{2^{\left( 2 \otimes^{2^{2^{3072}} \times 3+1} \left( 2^{2^{3072}} \times 3 \right) \right)+2^{3072}} \times 3} \times 1 = 2^{2^{\left( 2 \otimes^{2^{2^{3072}} \times 3+1} \left( 2^{2^{3072}} \times 3 \right) \right)+2^{3072}} \times 3} \\ \uparrow \left( 2^{2^{\left( 2 \otimes^{2^{2^{3072}} \times 3+1} \left( 2^{2^{3072}} \times 3 \right) \right)+2^{3072}} \times 3} \right) & = & \downarrow \left( 2^{2^{\left( 2 \otimes^{2^{2^{3072}} \times 3+1} \left( 2^{2^{3072}} \times 3 \right) \right)+2^{3072}} \times 3} \right) = 2^{2^{2^{\left( 2 \otimes^{2^{2^{3072}} \times 3+1} \left( 2^{2^{3072}} \times 3 \right) \right)+2^{3072}} \times 3}} \\ \uparrow \left( 2^{2^{2^{\left( 2 \otimes^{2^{2^{3072}} \times 3+1} \left( 2^{2^{3072}} \times 3 \right) \right)+2^{3072}} \times 3}} \right) & = & \downarrow \left( 2^{2^{2^{\left( 2 \otimes^{2^{2^{3072}} \times 3+1} \left( 2^{2^{3072}} \times 3 \right) \right)+2^{3072}} \times 3}} \right) = 2^{2^{2^{2^{\left( 2 \otimes^{2^{2^{3072}} \times 3+1} \left( 2^{2^{3072}} \times 3 \right) \right)+2^{3072}} \times 3}}} \\ & \vdots & \end{eqnarray*}

\begin{eqnarray*} \textrm{AC}_1(1) & = & \uparrow(1) = 2 \\ \textrm{AC}_1(2) & = & \uparrow \left( \textrm{AC}_1(1) \left( 1+2^{\textrm{AC}_1(1)} \right) \right) = \uparrow(10) = \left( 2 \otimes^{10} \left( 2^{\left( 2 \otimes^{2^{2^{3072}} \times 3+1} \left( 2^{2^{3072}} \times 3 \right) \right)+2^{3072}} \times 3 \right) \right) \times 10 \\ & \vdots & \end{eqnarray*}

= Analysis =

It is not so difficult to verify \(\textrm{AC}_1(x) > f_{\varepsilon_0}(x-1)\) with respect to FGH. Intuitively, the growth rate of \((\textrm{AC}_1(x)\) seems to be \(f_{\varepsilon_0+2}(x-1)\) or \(f_{\varepsilon_0+3}(x-1)\). At least, it might be bounded by \(f_{\varepsilon_0+5}(x-1)\). I appreciate anyone's precise estimation!

ja:User_blog:p進大好きbot/%E5%B7%A8%E5%A4%A7%E6%95%B0%E3%81%9F%E3%82%93%E3%82%A2%E3%83%89%E3%83%99%E3%83%B3%E3%83%88%E3%82%AB%E3%83%AC%E3%83%B3%E3%83%80%E3%83%BC%E5%88%9D%E6%97%A5%E3%82%A8%E3%83%B3%E3%83%88%E3%83%AA%E3%83%BC