A126387 - OEIS

%I #2 Mar 30 2012 17:35:17

%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,

%T 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,

%U 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

%N 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).

%F 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).

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

%Y Cf. A036989.

%K easy,nonn

%O 0,4

%A _Franklin T. Adams-Watters_, Dec 26 2006