A357449 a(0) = 0; for n > 0, a(n) is the smallest positive number not occurring earlier such that the binary string of a(n) plus the largest previous term does not appear in the binary string concatenation of a(0)..a(n-1).

0, 1, 2, 3, 4, 5, 10, 6, 7, 9, 14, 15, 16, 17, 18, 20, 12, 24, 8, 28, 26, 30, 22, 33, 11, 21, 31, 32, 36, 37, 27, 35, 41, 13, 23, 40, 44, 38, 62, 46, 66, 19, 42, 63, 65, 69, 39, 59, 60, 68, 72, 56, 57, 71, 76, 52, 53, 80, 48, 49, 55, 58, 61, 64, 83, 45, 73, 77, 81, 82, 85, 43, 50, 75, 79, 87, 51

Offset: 0

Author

Scott R. Shannon, Sep 29 2022

Keywords

• nonn
• base
• look

Comments

The main concentration of terms lies near the line a(n) = n; there are 26 fixed points in the first 100000 terms. The sequence is conjectured to be a permutation of the positive integers.

Examples

			a(9) = 9 as the concatenation of a(0)..a(8) in binary is "0110111001011010110111" and 9 plus the largest previous term = 9 + 10 = 19 = 10011_2 which does not appear in the concatenated string. Since 10 + 8 = 18 = 10010_2 appears in the concatenated string, a(9) cannot be 8.
		

Links

• Scott R. Shannon, Image for n = 0..100000.

Crossrefs

Cf. A357082, A355611, A007088, A030302, A118248, A341766.

Programs

Python

from itertools import islice
def agen():
    aset, astr, an, mink = {0}, "0", 0, 1
    while True:
        yield an; k, m = mink, max(aset)
        while k in aset or bin(m+k)[2:] in astr: k += 1
        while mink in aset: mink += 1
        an = k; aset.add(an); astr += bin(an)[2:]
print(list(islice(agen(), 77))) # Michael S. Branicky, Sep 29 2022