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A030159
Numbers n such that in n^3 the parity of digits alternates.
4
0, 1, 2, 3, 5, 6, 7, 9, 18, 23, 85, 87, 101, 103, 168, 206, 301, 303, 363, 725, 1683, 2461, 2788, 7921, 9563, 9668, 20606, 28443, 29501, 45168, 46701, 49501, 63556, 78206, 80901, 90009, 167861, 168069, 208288, 278636, 331841, 375121, 440468
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
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
STATUS
approved