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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). 3
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 (list; graph; refs; listen; history; text; internal format)
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: A108483 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

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Last modified January 26 07:25 EST 2022. Contains 350573 sequences. (Running on oeis4.)