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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).
2
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
0,4
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).
EXAMPLE
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.
Sequence in context: A237453 A265754 A089309 * A038374 A284569 A272604
KEYWORD
easy,nonn
AUTHOR
STATUS
approved