DylanN wrote: For example, if $a_{10} = 11$, then $b_{10} = 11_{11} (= 12_{10})$; $a_{11} = 11_{11} - 1 = 10_{11} (= 11_{10})$; $b_{11} = 10_{12} (= 12_{10})$; $a_{12} = 9_{12} (= 9_{10})$. I dont understand your example : $b_{11}=12$ and so $a_{12}=11$ and not, as you wrote, $9$ Maybe I misunderstood something in the process?

pco wrote: DylanN wrote: For example, if $a_{10} = 11$, then $b_{10} = 11_{11} (= 12_{10})$; $a_{11} = 11_{11} - 1 = 10_{11} (= 11_{10})$; $b_{11} = 10_{12} (= 12_{10})$; $a_{12} = 9_{12} (= 9_{10})$. I dont understand your example : $b_{11}=12$ and so $a_{12}=11$ and not, as you wrote, $9$ Maybe I misunderstood something in the process? Yes, you are correct. There was a mistake in the source that I was copying from. I should have spotted it when posting the problem.

DylanN wrote: Let a sequence $(a_i)_{i=10}^{\infty}$ be defined as follows: $a_{10}$ is some positive integer, which can of course be written in base 10. For $i \geq 10$ if $a_i > 0$, let $b_i$ be the positive integer whose base-$(i + 1)$ representation is the same as $a_i$'s base-$i$ representation. Then let $a_{i + 1} = b_i - 1$. If $a_i = 0$, $a_{i + 1} = 0$. For example, if $a_{10} = 11$, then $b_{10} = 11_{11} (= 12_{10})$; $a_{11} = 11_{11} - 1 = 10_{11} (= 11_{10})$; $b_{11} = 10_{12} (= 12_{10})$; $a_{12} = 11$. Does there exist $a_{10}$ such that $a_i$ is strictly positive for all $i \geq 10$? No. There does not : Writing $a_i=\sum_{k=0}^{n}\alpha_ki^k$, we get $a_{i+1}=\sum_{k=0}^{n}\alpha_k(i+1)^k-1$ Consider the ordered sequence $s_i$ of base_i representation of $a_i$ Using lexicographic order, the sequence $s_{10},s_{11},$ ... is strictly decreasing and so eventually becomes $1$, and then $0$ And from there $a_i=0$ To illustrate : Consider $s_{10}=(1,0,3)$ $s_{11}=(1,0,2)$ $s_{12}=(1,0,1)$ $s_{13}=(1,0,0)$ $s_{14}=(13,13)$ $s_{15}=(13,12)$ ... $s_{27}=(13,0)$ $s_{28}=(12,27)$ $s_{29}=(12,26)$ ... $s_{55}=(12,0)$ $s_{56}=(11,55)$ ... $s_{114687}=(1,0)$ $s_{114688}=(114687)$ ... $s_{229374}=(1)$ And so $a_{229375}=0$