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A195187
a(1)=1 and recursively a(n+1) = R(1+(a(n))^3).
0
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.
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
Sequence in context: A122673 A241454 A139102 * A041515 A010750 A373908
KEYWORD
nonn,base
AUTHOR
Jonathan Vos Post, Sep 12 2011
STATUS
approved