%I #14 Jul 11 2023 08:16:27
%S 10,101,1211,14321,150145,1163512,4340370,8144773,8144773
%N Successive base conversions of prior base representations.
%C Breaks down when 8144773 (base 11) = 466331A.
%F a(1) = 10; for n>1: a(n) = the base (10) interpretation of a(n-1), represented in base (n+1).
%e a(2) = 101 because a(1) = 10 and 10 (base 3) = 101.
%e a(3) = 1211 because a(2) = 101 and 101 (base 4) = 1211.
%e a(4) = 14321 because a(3) = 1211 and 1211 (base 5) = 14321.
%e a(5) = 150145 because a(4) = 14321 and 14321 (base 6) = 150145.
%e a(6) = 1163512 because a(5) = 150145 and 150145 (base 7) = 1163512.
%p a:= proc(n) option remember; `if`(n=1, 10, parse(cat(
%p ListTools[Reverse](convert(a(n-1), base, n+1))[])))
%p end:
%p seq(a(n), n=1..9); # _Alois P. Heinz_, Jul 10 2023
%K nonn,base,easy,fini,full,less
%O 1,1
%A _Jonathan Vos Post_, Jul 14 2005
%E Corrected (a(6) and onward) by _Daniel T. Martin_, Jul 10 2023