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

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Last modified May 28 15:04 EDT 2022. Contains 354115 sequences. (Running on oeis4.)