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A296615 Lexicographically earliest sequence of distinct positive terms such that, for any n > 0, a(n) XOR a(n+1) is a cube (where XOR denotes the XOR binary operator).

%I A296615 #11 Dec 20 2017 12:13:24
%S A296615 1,9,8,19,18,26,27,91,38,39,47,46,53,52,60,61,64,65,73,72,83,82,90,
%T A296615 130,131,139,138,145,144,152,153,217,164,124,103,102,110,111,116,117,
%U A296615 125,165,173,172,183,182,190,191,194,195,203,202,209,208,216,399,343
%N A296615 Lexicographically earliest sequence of distinct positive terms such that, for any n > 0, a(n) XOR a(n+1) is a cube (where XOR denotes the XOR binary operator).
%C A296615 This sequence has similarities with A175428: here a(n) XOR a(n+1) is a cube, there a(n) + a(n+1) is a cube.
%C A296615 This sequence is conjectured to the a permutation of the natural numbers.
%C A296615 The first fixed points are: 1, 5676, 5677, 5698, 11677, 13226, 26943, 26946, 27575, 28039, 28569, 28625, 30127, 30162, 37660, 37661, 44672, 44673, 45934.
%C A296615 The scatterplot of the first terms of the sequence shows hatches (see Links section).
%H A296615 Rémy Sigrist, <a href="/A296615/b296615.txt">Table of n, a(n) for n = 1..10000</a>
%H A296615 Rémy Sigrist, <a href="/A296615/a296615.png">Scatterplot of the first 50000 terms</a>
%e A296615 The first terms, alongside a(n) XOR a(n+1), are:
%e A296615   n     a(n)    a(n) XOR a(n+1)
%e A296615   --    ----    ---------------
%e A296615    1       1    2^3
%e A296615    2       9    1^3
%e A296615    3       8    3^3
%e A296615    4      19    1^3
%e A296615    5      18    2^3
%e A296615    6      26    1^3
%e A296615    7      27    4^3
%e A296615    8      91    5^3
%e A296615    9      38    1^3
%e A296615   10      39    2^3
%e A296615   11      47    1^3
%e A296615   12      46    3^3
%e A296615   13      53    1^3
%e A296615   14      52    2^3
%e A296615   15      60    1^3
%e A296615   16      61    5^3
%e A296615   17      64    1^3
%e A296615   18      65    2^3
%e A296615   19      73    1^3
%e A296615   20      72    3^3
%o A296615 (PARI) seen = 0; unseen = 1
%o A296615 other(p) = seen += 2^p; while (bittest(seen, unseen), unseen++); \
%o A296615 for (v=unseen, oo, if (!bittest(seen, v) && ispower(bitxor(p,v),3), return (v)))
%o A296615 for (n=1, 57, v=if (n==1, 1, other(v)); print1 (v ", "))
%Y A296615 Cf. A000578, A175428.
%K A296615 nonn,base
%O A296615 1,2
%A A296615 _Rémy Sigrist_, Dec 17 2017