

A274009


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


0



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
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OFFSET

0,3


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 x1 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 n1 digits left over.


LINKS

Table of n, a(n) for n=0..89.


FORMULA

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


MATHEMATICA

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


PROG

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


CROSSREFS

Cf. A000120.
Sequence in context: A331600 A039637 A194548 * A069157 A294894 A076526
Adjacent sequences: A274006 A274007 A274008 * A274010 A274011 A274012


KEYWORD

nonn,base


AUTHOR

William K. Grannis, Jun 06 2016


EXTENSIONS

More terms from Michel Marcus, Jul 13 2016


STATUS

approved



