A327192 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 least possible value of max(x, y).

0, 1, 1, 1, 1, 1, 2, 3, 1, 1, 2, 3, 3, 3, 3, 3, 1, 1, 2, 3, 4, 5, 5, 5, 3, 3, 3, 3, 4, 5, 6, 7, 1, 1, 2, 3, 4, 5, 6, 7, 5, 5, 5, 5, 5, 5, 6, 7, 3, 3, 3, 3, 4, 5, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 9, 9, 9, 9, 9, 5, 5, 5, 5, 5, 5, 6

Offset

0,7

Links

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

Formula

a(n) = 1 iff n = 2^k or n = 2^k + 1 for some k >= 0.

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: max(0, 42) = 42,
   - "1" and "01010": x=1 and y=10: max(1, 10) = 10,
   - "10" and "1010": x=2 and y=10: max(2, 10) = 10,
   - "101" and "010": x=5 and y=2: max(5, 2) = 5,
   - "1010" and "10": x=10 and y=2: max(10, 2) = 10,
   - "10101" and "0": x=21 and y=0: max(21, 0) = 21,
   - "101010" and "": x=42 and y=0: max(42, 0) = 42,
- hence a(42) = 5.

Prog

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

Crossrefs

See A327186 for other variants.

Sequence in context: A338854 A330957 A352924 * A351651 A157813 A111879

Adjacent sequences: A327189 A327190 A327191 * A327193 A327194 A327195

Keyword

nonn,look,base

Author

Rémy Sigrist, Aug 25 2019

Status

approved