User blog:P進大好きbot/Elementary Large Number beyond PTO(Π11-CA0)

This is an English translation of my Japanese blog post submitted to a Japanese googological event.

I created a \(1\)-ary function \(\textrm{萃}(n)\) only using elementary arithmetic such as \(+\), \(-\), \(\times\), and \(/\). This is a portion of the contiuation of my attempt to create an elementary large number. By this restriction, the definition is very complicated. If you want to understand how it works, compare it with the previous attempt.

= Large Function =

The digit of a positive integer is the digit of its decimal expression. For positive integers \(a\) and \(b\), the quotient of \(a\) by \(b\) is the greatest natural number \(q\) satisfying \(q \geq qb\).

For a positive integer \(n\), I define a natural number \(\textrm{萃}(n)\) in the following recursive way:
 * 1) Denote by \(n_0\) how many times \(x\) is divisible by \(10\), and by \(a_0\) the quotient of \(x\) by \(10^{n_0}\).
 * 2) Denote by \(b_0+1\) the digit of \(a_0\), and by \(c_0\) the quotient of \(a_0\) by \(10^{b_0}\). Put \(d_0 = a_0 - c_0 10^{b_0}\).
 * 3) Suppose \(d_0 = 0\).
 * 4) If \(n_0 = 0\), then set \(\textrm{萃}(x) = 0\).
 * 5) If \(n_0 > 0\), then set \(\textrm{萃}(x) = 10^x\).
 * 6) Suppose \(0 < d_0 < 10\).
 * 7) Denote by \(n_1\) how many times \(b_0\) is divisible by \(10\), and by \(a_1\) the quotient of \(b_0\) by \(10^{n_1}\).
 * 8) Suppose \(c_0 = 9\).
 * 9) Suppose \(\textrm{萃}(a_1) = 0\).
 * 10) Suppose \(\textrm{萃}(1+10n_1) = 0\).
 * 11) If \(n_0 = 0\), then set \(\textrm{萃}(x) = 1\).
 * 12) If \(n_0 = 1\), then set \(\textrm{萃}(x) = 1\).
 * 13) If \(n_0 > 1\), then set \(\textrm{萃}(x) = 10^x\).
 * 14) Suppose \(\textrm{萃}(1+10n_1) = 1\).
 * 15) If \(n_0 = 0\), then set \(\textrm{萃}(x) = 9\).
 * 16) If \(n_0 = 1\), then set \(\textrm{萃}(x) = 10^x\).
 * 17) Suppose \(n_0 > 1\).
 * 18) Put \(y = \textrm{萃}(10(1+10n_1))\). Denote by \(n_2\) the quotient of \(y\) by \(10\).
 * 19) Put \(z = \textrm{萃}(10^{n_0-1}a_0)\). Denote by \(n_3\) how many times \(z\) is divisible by \(10\), and by \(a_2\) the quotient of \(z\) by \(10^{n_3}\).
 * 20) Put \(b_1 = 10^{n_2}a_2\), and \(a_3 = 1 + 10^{b_1}9\).
 * 21) Set \(\textrm{萃}(x) = 10^z a_3\).
 * 22) Suppose \(\textrm{萃}(1+10n_1) > 1\).
 * 23) If \(n_0 = 0\), then set \(\textrm{萃}(x) = \textrm{萃}(1+10n_1)\).
 * 24) Suppose \(n_0 > 0\).
 * 25) Put \(y = \textrm{萃}(10^{n_0}(1+10n_1))\). Denote by \(n_2\) how many times \(y\) is divisible by \(10\), by \(n_3\) the quotient of \(y\) by \(10^{n_2+1}\).
 * 26) Put \(b_1 = 10^{n_3}\), and \(a_2 = 1 + 10^{b_1}9\).
 * 27) Set \(\textrm{萃}(x) = 10^y a_2\).
 * 28) Suppose \(\textrm{萃}(a_1) = 1\).
 * 29) Suppose \(\textrm{萃}(1+10n_1) = 0\).
 * 30) If \(n_0 = 0\), then set \(\textrm{萃}(x) = 9\).
 * 31) Suppose \(n_0 = 1\).
 * 32) Put \(b_2 = \textrm{萃}(10a_1)\), and \(a_2 = 1+10^{b_2}9\).
 * 33) Set \(\textrm{萃}(x) = 10^x a_2\).
 * 34) Suppose \(n_0 > 1\).
 * 35) Put \(y = \textrm{萃}(10^{n_0-1}a_0)\). Denote by \(n_2\) how many times \(y\) is divisible by \(10\), and by \(d_1\) the quotient of \(y\) by \(10^{n_2}\), and by \(b_1+1\) the digit of \(d_1\).
 * 36) Put \(z = \textrm{萃}(10a_0)\). Denote by \(n_3\) how many times \(z\) is divisible by \(10\), and by \(a_2\) the quotient of \(z\) by \(10^{n_3+1}\). Put \(a_3 = d_1 + 10^{b_1+1}a_2\).
 * 37) Set \(\textrm{萃}(x) = 10^y a_3\).
 * 38) Suppose \(\textrm{萃}(1+10n_1) = 1\).
 * 39) If \(n_0 = 0\), then set \(\textrm{萃}(x) = 9\).
 * 40) Suppose \(n_0 = 1\).
 * 41) Put \(a_2 = \textrm{萃}(10a_1)\), \(b_1 = 10^{n_1}a_2\), \(d_1 = 1 + 10^{b_1}9\), and \(a_3 = d_1 + 10^{b_1+1}9\).
 * 42) Set \(\textrm{萃}(x) = 10^x a_3\).
 * 43) Suppose \(n_0 > 1\).
 * 44) Put \(y = \textrm{萃}(10^{n_0-1}a_0)\).  Denote by \(n_2\) how many times \(y\) is divisible by \(10\), and by \(a_2\) the quotient of \(y\) by \(10^{n_2}\).
 * 45) Put \(z = \textrm{萃}(10(1+10n_1))\). Denote by \(n_3\) the quotient of \(z\) by \(10\).
 * 46) Put \(b_1 = 10^{n_3}a_2\), and \(a_3 = 1 + 10^{b_1}9\).
 * 47) Set \(\textrm{萃}(x) = 10^y a_3\).
 * 48) Suppose \(\textrm{萃}(1+10n_1) > 1\).
 * 49) If \(n_0 = 0\), then set \(\textrm{萃}(x) = \textrm{萃}(1+10n_1)\).
 * 50) Suppose \(n_0 > 0\).
 * 51) Put \(y = \textrm{萃}(10^{n_0}(1+10n_1))\). Denote by \(n_2\) how many times \(y\) is divisible by \(10\), and by \(n_3\) the quotient of \(y\) by \(10^{n_2+1}\).
 * 52) Put \(a_2 = \textrm{萃}(10a_1)\), \(b_1 = 10^{n_1}a_2\), \(d_1 = 1 + 10^{b_1}9\), and \(a_3 = d_1 + 10^{b_1+1}9\).
 * 53) Put \(b_2 = 10^{n_3}a_3\), and \(a_4 = 1 + 10^{b_2}9\).
 * 54) Set \(\textrm{萃}(x) = 10^y a_4\).
 * 55) Suppose \(\textrm{萃}(a_1) > 1\).
 * 56) If \(n_0 = 0\), then set \(\textrm{萃}(x) = \textrm{萃}(a_1)\).
 * 57) Suppose \(n_0 > 0\).
 * 58) Put \(y = \textrm{萃}(10^{n_0}a_1)\). Denote by \(n_2\)  how many times \(y\) is divisible by \(10\), and by \(a_2\) the quotient of \(y\) by \(10^{n_2}\).
 * 59) Put \(b_1 = 10^{n_1}a_2\), and \(a_3 = 1 + 10^{b_1}9\).
 * 60) Set \(\textrm{萃}(x) = 10^y a_3\).
 * 61) Suppose \(c_0 \neq 9\).
 * 62) Suppose \(\textrm{萃}(a_1) = 0\).
 * 63) Suppose \(n_1 = 0\).
 * 64) If \(n_0 = 0\), then set \(\textrm{萃}(x) = 9\).
 * 65) If \(n_0 = 1\), then set \(\textrm{萃}(x) = 10^x\).
 * 66) Suppose \(n_0 > 1\).
 * 67) Put \(y = \textrm{萃}(10^{n_0-1}a_0)\). Denote by \(n_2\) how many times \(y\) is divisible by \(10\), and by \(n_3\) the quotient of \(y\) by \(10^{n_2}\).
 * 68) Put \(b_1 = 10^{n_3}\), and \(a_3 = 1 + 10^{b_1}9\).
 * 69) Set \(\textrm{萃}(x) = 10^y a_3\).
 * 70) Suppose \(n_1 > 0\).
 * 71) If \(n_0 = 0\), then set \(\textrm{萃}(x) = 9 + 10n_1\).
 * 72) If \(n_0 > 0\), then set \(\textrm{萃}(x) = 1+10n_0\).
 * 73) Suppose \(\textrm{萃}(a_1) = 1\).
 * 74) If \(n_0 = 0\), then set \(\textrm{萃}(x) = 9\).
 * 75) Suppose \(n_0 = 1\).
 * 76) Put \(a_2 = \textrm{萃}(10a_1)\), \(b_1 = 10^{n_1}a_2\), \(d_1 = 1 + 10^{b_1}\), and \(a_3 = d_1 + 10^{b_1+1} 9\).
 * 77) Set \(\textrm{萃}(x) = 10^x a_3\).
 * 78) Suppose \(n_0 > 1\).
 * 79) Put \(y = \textrm{萃}(10^{n_0-1}a_0)\). Denote by \(n_2\) how many times \(y\) is divisible by \(10\), and by \(n_3\) the quotient of \(y\) by \(10^{n_2}\).
 * 80) Put \(b_1 = 10^{n_3}\), and \(a_3 = 1 + 10^{b_1}9\).
 * 81) Set \(\textrm{萃}(x) = 10^y a_3\).
 * 82) Suppose \(1 < \textrm{萃}(a_1) < 10\).
 * 83) If \(n_0 = 0\), then set \(\textrm{萃}(x) = 9\).
 * 84) Suppose \(n_0 > 0\).
 * 85) Put \(y = \textrm{萃}(10^{n_0}a_1)\). Denote by \(n_2\) how many times \(y\) is divisible by \(10\), and by \(a_2\) the quotient of \(y\) by \(10^{n_2}\).
 * 86) Put \(b_1 = 10^{n_1}a_2\), and \(a_3 = 1 + 10^{b_1}\).
 * 87) Set \(\textrm{萃}(x) = 10^y a_3\).
 * 88) Suppose \(\textrm{萃}(a_1) \geq 10\).
 * 89) Denote by \(n_2\) the quotient of \(\textrm{萃}(a_1)\) by \(10\).
 * 90) Suppose \(n_2 \leq n_1\).
 * 91) If \(n_0 = 0\), then set \(\textrm{萃}(x) = \textrm{萃}(a_1)\).
 * 92) Suppose \(n_0 > 0\).
 * 93) Put \(y = \textrm{萃}(10^{n_0}a_1)\). Denote by \(n_3\) how many times \(y\) is divisible by \(10\), and by \(a_2\) the quotient of \(y\) by \(10^{n_3}\).
 * 94) Put \(b_1 = 10^{n_1}a_2\), and \(a_3 = 1 + 10^{b_1}\).
 * 95) Set \(\textrm{萃}(x) = 10^y a_3\).
 * 96) Suppose \(n_2 > n_1\).
 * 97) If \(n_0 = 0\), then set \(\textrm{萃}(x) = 9\).
 * 98) Suppose \(n_0 = 1\).
 * 99) Put \(y = \textrm{萃}(10a_1)\). Denote by \(n_3\) how many times \(y\) is divisible by \(10\), and by \(a_2\) the quotient of \(y\) by \(10^{n_3}\).
 * 100) Put \(b_1 = 10^{n_1}a_2\), and \(a_3 = 1 + 10^{b_1}\).
 * 101) Set \(\textrm{萃}(x) = 10^y a_3\).
 * 102) Suppose \(n_0 > 1\).
 * 103) Put \(y = \textrm{萃}(10^{n_0-1}a_0)\). Denote by \(n_3\) how many times \(y\) is divisible by \(10\), and by \(a_2\) the quotient of \(y\)を\(10^{n_3}\) by \(a_2\).
 * 104) Denote by \(b_1+1\) the digit of \(a_2\), by \(n_3\) how many times \(b_1\) is divisible by \(10\), and by \(n_4\) the quotient of \(b_1\) by \(10^{n_3}\).
 * 105) Put \(z = \textrm{萃}(10^{n_4}a_1)\). Denote by \(n_5\) how many times \(z\) is divisible by \(10\), and by \(a_3\) the quotient of \(z\) by \(10^{n_5}\).
 * 106) Put \(b_1 = 10^{n_1}a_3\), and \(a_3 = 1 + 10^{b_1}\).
 * 107) Set \(\textrm{萃}(x) = 10^y a_3\).
 * 108) Suppose \(d_0 \geq 10\).
 * 109) Denote by \(b_1+1\) the digit of \(d_0\). Put \(d_1 = 10^{b_0-b_1}c_0+1\).
 * 110) Suppose \(\textrm{萃}(d_1) \leq 1\).
 * 111) If \(n_0 = 0\), then set \(\textrm{萃}(x) = 1\).
 * 112) If \(n_0 = 1\), then set \(\textrm{萃}(x) = d_0\).
 * 113) If \(n_0 > 1\), then set \(\textrm{萃}(x) = 10^x d_0\).
 * 114) Suppose \(\textrm{萃}(d_1) > 1\).
 * 115) Put \(y = \textrm{萃}(10^{n_0} d_1)\).
 * 116) If \(n_0 = 0\) or \(y = 0\), then set \(\textrm{萃}(x) = y\).
 * 117) Suppose \(n_0 > 0\) and \(y > 0\).
 * 118) Denote by \(n_1\) how many times \(y\) is divisible by \(10\), and by \(a_1\) the quotient of \(y\) by \(10^{n_1+1}\).
 * 119) Put \(a_2 = d_0+10^{b_1+1}a_1\).
 * 120) Set \(\textrm{萃}(x) = 10^y a_2\).

= Large Number =

I name \(\textrm{萃} \left( 10^{10^{10^{10^{10}}}+1}+1 \right)\) "\(\textrm{十}^{\textrm{十}^{\textrm{十}^{\textrm{十}^{\textrm{十}}}}}\textrm{鬼夜行}\)".

= Analysis =

I lazily display the result of my analysis in the following table before showing the definition. Here, \(\psi\) denotes Buchholz's OCF, and \(f\) denotes FGH with respect to a suitable recursive system of fundamental sequences. \begin{eqnarray*} \textrm{萃}(1) & = & 0 \\ \textrm{萃}(2) & = & 0 \\ \textrm{萃}(3) & = & 0 \\ \textrm{萃}(11) & = & 9 \\ \textrm{萃}(12) & = & 9 \\ \textrm{萃}(13) & = & 9 \\ \textrm{萃}(100) & = & 10^{100} \\ \textrm{萃}(900) & = & 10^{900} \\ \textrm{萃}(1000) & = & 10^{1000} \\ \textrm{萃}(10000) & = & 10^{10000} \\ \textrm{萃}(11000) & > & f_{\Gamma_0}(11000) \\ \textrm{萃} \left( 10^{10^{100}+1}+1 \right) & > & f_{\psi_0(\Omega_2)}(10^{10^{100}+1}+1) \\ \end{eqnarray*} \(\textrm{十}^{\textrm{十}^{\textrm{十}^{\textrm{十}^{\textrm{十}}}}}\textrm{鬼夜行}\) is greater than \(f_{\psi_0(\Omega_{\omega})} \left( 10^{10^{10^{10^{10}}}+1}+1 \right)\). Namely, it is a large number beyond \(\textrm{PTO}(\Pi_1^1{-}\textrm{CA}_0)\)-level.