A327188 For any n >= 0: consider the different ways to split the binary representation of n into two (possibly empty) parts, say with value x and y; a(n) is the greatest possible value of x AND y (where AND denotes the bitwise AND operator).

0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 2, 2, 0, 1, 2, 3, 0, 1, 2, 2, 0, 1, 2, 2, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 2, 4, 4, 4, 4, 0, 1, 2, 2, 4, 5, 4, 5, 0, 1, 2, 3, 4, 4, 6, 6, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 2, 4, 4, 4, 4, 0, 1, 2, 2, 4, 4, 4, 4, 0, 1, 2, 2, 4, 5, 4

Offset

0,11

Comments

The first 10000 positive integers where the sequence equals zero match the first 10000 terms of A082662; is that true forever?

Links

Rémy Sigrist, Table of n, a(n) for n = 0..8192

Example

For n=42:
- the binary representation of 42 is "101010",
- there are 7 ways to split it:
   - "" and "101010": x=0 and y=42: 0 AND 42 = 0,
   - "1" and "01010": x=1 and y=10: 1 AND 10 = 0,
   - "10" and "1010": x=2 and y=10: 2 AND 10 = 2,
   - "101" and "010": x=5 and y=2: 5 AND 2 = 0,
   - "1010" and "10": x=10 and y=2: 10 AND 2 = 2,
   - "10101" and "0": x=21 and y=0: 21 AND 0 = 0,
   - "101010" and "": x=42 and y=0: 42 AND 0 = 0,
- hence a(42) = 2.

Prog

(PARI) a(n) = my (v=-oo, b=binary(n)); for (w=0, #b, v=max(v, bitand(fromdigits(b[1..w], 2), fromdigits(b[w+1..#b], 2)))); v

Crossrefs

See A327186 for other variants.

Cf. A082662.

Sequence in context: A363623 A101565 A292944 * A330270 A029341 A240181

Adjacent sequences: A327185 A327186 A327187 * A327189 A327190 A327191

Keyword

nonn,look,base

Author

Rémy Sigrist, Aug 25 2019

Status

approved