User blog comment:P進大好きbot/Elementary Large Number/@comment-11227630-20190604040637/@comment-11227630-20190604053233

Inferred from your ordinal table: Then \(2\text{Code}(\alpha)+1\) fits your left column.
 * 1) \(\text{Code}(0)=0\)
 * 2) If \(\alpha=\omega^\beta+\gamma\) with \(\alpha>\gamma\), then \(\text{Code}(\alpha)=2^{\text{Code}(\beta)}\cdot(2\text{Code}(\gamma)+1)\)
 * 3) *Written in CNF, \(\alpha=\omega^{\beta_1}+\omega^{\beta_2}+\cdots\omega^{\beta_{n-1}}+\omega^{\beta_n}\) with \(\beta_1\ge\beta_2\ge\cdots\beta_{n-1}\ge\beta_n\), then \(\text{Code}(\alpha)=2^{\text{Code}(\beta_1)}+2^{\text{Code}(\beta_1)+\text{Code}(\beta_2)+1}+2^{\text{Code}(\beta_1)+\text{Code}(\beta_2)+\text{Code}(\beta_3)+2}+\cdots2^{\text{Code}(\beta_1)+\text{Code}(\beta_2)+\cdots\text{Code}(\beta_n)+n-1}\)