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A328979
Trajectory of 0 under repeated application of the morphism 0 -> 0010, 1 -> 1010.
6
0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0
OFFSET
1
COMMENTS
The morphism in the definition, 0 -> 0010, 1 -> 1010, is the square of the morphism tau: 0 -> 10, 1 -> 00.
This sequence is also the 0-limiting word of tau (see A284948).
It is also the image of A080426 under the morphism 1 -> 0,0,1,0; 3 -> 0,0,1,0,1,0,1,0.
This sequence underlies all of A297469, A298468, A328190, and A328196.
Theorem: a(n) = 1 iff the binary expansion of n-1 ends in an odd number of 0's (cf. A036554, A072939). For proof see comments by Michel Dekking in A284948.
Is this A096268 with an additional 0 added in front? - R. J. Mathar, Nov 13 2019 [Yes: it follows, e.g., from the above theorem. - Andrey Zabolotskiy, Jan 12 2020]
MAPLE
F(0):= (0, 0, 1, 0): F(1):= (1, 0, 1, 0):
B:= [0]: # if start at 0 get the present sequence, if start at 1 get A284948
for i from 1 to 4 do B:= map(F, B) od:
B;
# Or, construction via A080426:
f(1):= (1, 3, 1): f(3):= (1, 3, 3, 3, 1):
A:= [1]:
for i from 1 to 5 do A:= map(f, A) od:
A;
g(1):= (0, 0, 1, 0); g(3):= (0, 0, 1, 0, 1, 0, 1, 0);
map(g, A):
MATHEMATICA
SubstitutionSystem[{0 -> {0, 0, 1, 0}, 1 -> {1, 0, 1, 0}}, 0, 4] // Last (* Jean-François Alcover, Apr 06 2020 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Nov 04 2019. Extensively revised Nov 05 2019 thanks to comments from R. J. Mathar.
STATUS
approved