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Define Z(1) = {1}, and Z(n+1) = Z(n) (+) { x+y, with x and y in Z(n) } for any n>0 (where (+) stands for the symmetric difference of two sets). Then a(n) gives the number of elements in Z(n).
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%I #15 Apr 25 2016 12:00:16

%S 1,2,3,7,10,22,42,87,170,342,686,1365,2727,5468,10919,21857,43680,

%T 87389,174756,349539,699039,1398115,2796191,5592422,11184795,22369639,

%U 44739229,89478503,178956950,357913967,715827858,1431655793,2863311503,5726623097,11453246088

%N Define Z(1) = {1}, and Z(n+1) = Z(n) (+) { x+y, with x and y in Z(n) } for any n>0 (where (+) stands for the symmetric difference of two sets). Then a(n) gives the number of elements in Z(n).

%C a(n) can also be interpreted as the number of ON cells at the n-th stage of the following automaton:

%C - At first stage, we have only one ON cell at position 1,

%C - An ON cell appears at position x+y if the cells at positions x and y are ON,

%C - An ON cell dies at position x+y if the cells at positions x and y are ON.

%C a(n) <= 2^(n-1) for any n>0.

%H Paul Tek, <a href="/A263402/b263402.txt">Table of n, a(n) for n = 1..250</a>

%H Paul Tek, <a href="/A263402/a263402.pl.txt">PERL program for this sequence</a>

%H Paul Tek, <a href="/A263402/a263402.png">Graphic representation of the elements in the range [1..300] of the first 300 sets Z(n)</a>

%F a(n) = A000120(z(n)) for any n>0

%F where z(n) is a binary encoding of Z(n), defined as follows:

%F - z(1) = 2^1,

%F - z(n+1) = z(n) XOR A067398(z(n)) for any n>0 (where XOR stands for the binary XOR operator).

%e Z(1) = {1};

%e Z(2) = {1} (+) {2} = {1,2};

%e Z(3) = {1,2} (+) {2,3,4} = {1,3,4};

%e Z(4) = {1,3,4} (+) {2,4,5,6,7,8} = {1,2,3,5,6,7,8};

%e Hence: a(1) = 1, a(2) = 2, a(3) = 3 and a(4) = 7.

%o (Perl) See Links section.

%o (PARI) lista(nn) = {zprec = Set([1]); print1(#zprec, ", "); for (n=2, nn, zs = setbinop((x,y)->x+y, zprec); zn = setminus(setunion(zprec, zs), setintersect(zprec, zs)); print1(#zn, ", "); zprec = zn;);} \\ _Michel Marcus_, Oct 20 2015

%Y Cf. A067398.

%K nonn

%O 1,2

%A _Paul Tek_, Oct 17 2015