%I #12 Aug 17 2018 10:58:59
%S 0,1,2,3,5,6,7,9,18,23,85,87,101,103,168,206,301,303,363,725,1683,
%T 2461,2788,7921,9563,9668,20606,28443,29501,45168,46701,49501,63556,
%U 78206,80901,90009,167861,168069,208288,278636,331841,375121,440468
%N Numbers n such that in n^3 the parity of digits alternates.
%C 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
%H Giovanni Resta, <a href="/A030159/b030159.txt">Table of n, a(n) for n = 1..350</a>
%t 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 *)
%Y Cf. A030160, A030161, A030162, A030151.
%K nonn,base
%O 1,3
%A _Patrick De Geest_
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