A327186 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 x OR y (where OR denotes the bitwise OR operator).
0, 1, 1, 1, 1, 1, 3, 3, 1, 1, 2, 3, 3, 3, 3, 3, 1, 1, 2, 3, 5, 5, 6, 7, 3, 3, 3, 3, 7, 7, 7, 7, 1, 1, 2, 3, 4, 5, 6, 7, 5, 5, 7, 7, 5, 5, 7, 7, 3, 3, 3, 3, 6, 7, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 1, 1, 2, 3, 4, 5, 6, 7, 9, 9, 10, 11, 12, 13, 14, 15, 5, 5, 7, 7, 5
Offset: 0
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: 0 OR 42 = 42, - "1" and "01010": x=1 and y=10: 1 OR 10 = 11, - "10" and "1010": x=2 and y=10: 2 OR 10 = 10, - "101" and "010": x=5 and y=2: 5 OR 2 = 7, - "1010" and "10": x=10 and y=2: 10 OR 2 = 10, - "10101" and "0": x=21 and y=0: 21 OR 0 = 21, - "101010" and "": x=42 and y=0: 42 OR 0 = 42, - hence a(42) = 7.
Links
- Rémy Sigrist, Table of n, a(n) for n = 0..8192
Crossrefs
Programs
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PARI
a(n) = my (v=oo, b=binary(n)); for (w=0, #b, v=min(v, bitor(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.