A274009 1's distance from a number in its binary expansion.

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

Offset: 0

Author

William K. Grannis, Jun 06 2016

Keywords

• nonn
• base

Comments

To generate the value for n, write out n's decimal expansion. Then, write out 1's decimal expansion (0000000000....001). Compute how many times you need to change 0 to a 1 or a 1 to a 0 in order to switch from one number to the other.

The value for 2^x is always 2. The value for 2^x +1 is always 1. The value for 2^x -1 is always x-1 when x > 0. To get to 2^x, you need to drop the 1 at the beginning and add the 1 in the 2^x place value.

For 2^x + 1, you need to add the 1 in the 2^n place value, but you keep the 1 in the 1s place value. Thus you are only adding or getting rid of 1 digit.

For 2^x -1, it will have x digits, and all of them will be 1's. You already have 1 in the 1's place value, so there are n-1 digits left over.

Crossrefs

Cf. A000120.

Programs

Mathematica

Table[If[OddQ@ n, # - 1, # + 1] &@ DigitCount[n, 2, 1], {n, 0, 120}] (* Michael De Vlieger, Jul 13 2016 *)

PARI

a(n) = hammingweight(n) + (-1)^n; \\ Michel Marcus, Jul 14 2016

Formula

a(n) = A000120(n) + (-1)^n. - Michel Marcus, Jul 14 2016

Extensions

More terms from Michel Marcus, Jul 13 2016