$f(b)$ is the least $n$ such that either $b=0$ , or $b$ is odd, or $R$ : $p_{\mbox{ln}b-1}^{n+2}$ does not divide $b$ . Here, $<b,n>\in R$ iff for all $l<b$ and $p\le b$ , if $l=\mbox{lh}b-1$ and $p=p_{l}$ , then $<p^{n+2},b>\notin\mathcal{D}$ , where $\mathcal{D}$ is the divisibility relation. This works, because $\mbox{ln}b=0$ iff $b=0$ or $b$ is odd (in which case the function is undefined, so we just return $0$ ), otherwise, $b\ge2$ is even, $\mbox{ln}b\ge1$ , and $p_{\mbox{ln}b-1}^{n+2}$ does not divide $b$ for some large $n$ , moreover, for $b=\langle a_{0},\ldots,a_{m}\rangle$ , $\mbox{ln}b=m+1$ , and $p_{m}^{k}$ divides $b$ iff $k\le a_{m}+1$ , i.e. $k=a_{m}+2$ is the least power such that $p_{m}^{k}$ does not divide $b$ .