A126387 Read binary expansion of n from the left; keep track of the excess of 1's over 0's that have been seen so far; sequence gives maximum(excess of 1's over 0's).

0, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 2, 3, 4, 1, 1, 1, 1, 1, 1, 2, 3, 2, 2, 2, 3, 3, 3, 4, 5, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 3, 4, 2, 2, 2, 2, 2, 2, 3, 4, 3, 3, 3, 4, 4, 4, 5, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 2, 3, 2, 2, 2, 3, 3, 3, 4, 5, 2, 2, 2, 2, 2, 2, 2, 3, 2

Offset: 0

Author

Franklin T. Adams-Watters, Dec 26 2006

Keywords

• easy
• nonn

Examples

			59 in binary is 111011, excess from left to right is 1,2,3,2,3,4, maximum is 4, so a(59) = 4.
		

Crossrefs

Cf. A036989.

Formula

a(0) = 0, a(2^i) = 1, if n = 2^i + 2^j + m with j < i and 0 <= m < 2^j, then a(n) = max(a(2^j+m) + j + 2 - i, 1).