A357758 Numbers k such that in the binary expansion of k, the Hamming weight of each block differs by at most 1 from every other block of the same length.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 26, 27, 29, 30, 31, 32, 33, 34, 36, 37, 41, 42, 43, 45, 46, 47, 53, 54, 55, 59, 61, 62, 63, 64, 65, 66, 68, 72, 73, 74, 82, 84, 85, 86, 90, 91, 93, 94, 95, 106, 107, 109, 110, 111

Offset: 1

Author

Rémy Sigrist, Oct 12 2022

Keywords

• nonn
• base

Comments

Leading zeros in binary expansions are ignored.

For any n > 0, there are A005598(n)/2 positive terms with binary length n.

Empirically, if t is a term, then at least one of 2*t or 2*t + 1 is also a term.

If t is a term, then floor(t/2) is also a term.

Examples

			For k = 42:
- the binary expansion of 42 is "101010",
- blocks of length 1 have Hamming weight 0 or 1,
- blocks of length 2 have Hamming weight 1,
- blocks of length 3 have Hamming weight 1 or 2,
- blocks of length 4 have Hamming weight 2,
- blocks of length 5 have Hamming weight 2 or 3,
- so 42 belongs to the sequence.
For k = 44:
- the binary expansion of 44 is "101100",
- blocks of length 2 have Hamming weight 0, 1 or 2,
- so 44 does not belong to the sequence.
		

Links

• Rémy Sigrist, PARI program
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A005598, A274008, A357759.

Programs

PARI

See Links section.

Python

def ok(n):
    b = bin(n)[2:]
    if "00" in b and "11" in b: return False
    for l in range(3, len(b)):
        h = set(b[i:i+l].count("1") for i in range(len(b)-l+1))
        if max(h) - min(h) > 1: return False
    return True
print([k for k in range(112) if ok(k)]) # Michael S. Branicky, Oct 12 2022