A030159 Numbers n such that in n^3 the parity of digits alternates.

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

Author

Patrick De Geest

Keywords

• nonn
• base

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

Crossrefs

Cf. A030160, A030161, A030162, A030151.

Programs

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 *)