OFFSET
1,3
COMMENTS
A simple heuristic argument suggests that this sequence (albeit rather sparse) is infinite. The numbers of terms of k digits, for k=1..14, are 8, 4, 8, 6, 10, 14, 20, 18, 33, 23, 42, 37, 46, 77, respectively. The 5 numbers obtained multiplying the first h=1..5 terms of (1+10^2, 1+10^8, 1+10^32, 1+10^128, 1+10^512), are all member of the sequence. The largest one is a number of 683 digits whose alternating cube has 2047 digits. - Giovanni Resta, Aug 16 2018
LINKS
Giovanni Resta, Table of n, a(n) for n = 1..350
MATHEMATICA
n3pdaQ[n_]:=Module[{pty=Boole[EvenQ/@IntegerDigits[n^3]], len= IntegerLength[ n^3]}, pty== PadRight[{}, len, {1, 0}]||pty==PadRight[ {}, len, {0, 1}]]; Join[{0}, Select[Range[450000], n3pdaQ]] (* Harvey P. Dale, Mar 26 2018 *)
CROSSREFS
KEYWORD
nonn,base
AUTHOR
STATUS
approved