A195187 a(1)=1 and recursively a(n+1) = R(1+(a(n))^3).

1, 2, 9, 37, 45605, 62102890005849, 50579867441435760410772290593896794615932

Offset

1,2

Comments

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.

The number of digits is 1, 1, 1, 2, 5, 14, 41, 123, 369, 1105, ... in the n=1st, 2nd, 3rd etc. term.

Links

Table of n, a(n) for n=1..7.

Formula

a(n) = A004086(1+a(n-1)^3).

Mathematica

Nest[Append[#, IntegerReverse[1 + #[[-1]]^3]] &, {1}, 6] (* Michael De Vlieger, Feb 15 2020 *)

Crossrefs

Cf. A000578, A004086.

Sequence in context: A122673 A241454 A139102 * A041515 A010750 A373908

Adjacent sequences: A195184 A195185 A195186 * A195188 A195189 A195190

Keyword

nonn,base

Author

Jonathan Vos Post, Sep 12 2011

Status

approved