The OEIS is supported by the many generous donors to the OEIS Foundation.
%I #28 Feb 15 2020 23:57:00
%S 1,2,9,37,45605,62102890005849,
%T 50579867441435760410772290593896794615932
%N a(1)=1 and recursively a(n+1) = R(1+(a(n))^3).
%C Add one to the cube of the previous term, and reverse the digits in base 10 to obtain the next term. a(1)= 1, a(n+1) = A004086(1+A000578(a(n))). This is a flawed try to make the analog to the exponent 3 as A193914 is to the exponent 2.
%C The number of digits is 1, 1, 1, 2, 5, 14, 41, 123, 369, 1105, ... in the n=1st, 2nd, 3rd etc. term.
%F a(n) = A004086(1+a(n-1)^3).
%t Nest[Append[#, IntegerReverse[1 + #[[-1]]^3]] &, {1}, 6] (* _Michael De Vlieger_, Feb 15 2020 *)
%Y Cf. A000578, A004086.
%K nonn,base
%O 1,2
%A _Jonathan Vos Post_, Sep 12 2011