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

Author

Rémy Sigrist, Aug 25 2019

Keywords

• nonn
• look
• base

Examples

			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.
		

Links

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

Crossrefs

See A327186 for other variants.

Programs

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

Formula

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