A108737 Start with S = {}. For m = 0, 1, 2, 3, ... let u be the binary expansion of m. If u is not a substring of S, append the minimal number of 0's and 1's to S to remedy this. Sequence gives S.

0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1

Offset

1,1

Links

Michael S. Branicky, Table of n, a(n) for n = 1..10000

Example

We construct S as follows, starting with S = {}.
0 is missing, so S = {0};
1 is missing, so S = {0,1};
10 is missing, so S = {0,1,0};
11 is missing, so S = {0,1,0,1,1};
100 is missing, so S = {0,1,0,1,1,0,0};
101 and 110 are visible, but 111 is missing, so S = {0,1,0,1,1,0,0,1,1,1}; etc.

Prog

(Python)
from itertools import count, islice, product
def a(): # generator of terms
    S = ""
    for m in count(0):
        Sm = bin(m)[2:]
        if Sm in S: continue
        for i in range(1, len(Sm)+1):
            v = Sm[-i:]
            t = "" if len(v) == len(Sm) else S[-len(Sm)+i:]
            if t+v == Sm: break
        S += v
        yield from list(map(int, v))
print(list(islice(a(), 105))) # Michael S. Branicky, Oct 27 2023

Crossrefs

Cf. A108736.

Sequence in context: A189203 A296028 A165263 * A165221 A295891 A093879

Adjacent sequences: A108734 A108735 A108736 * A108738 A108739 A108740

Keyword

nonn,easy

Author

N. J. A. Sloane, Jun 23 2005

Extensions

More terms from John W. Layman, Jun 24 2005

Status

approved