

A320642


Number of 1's in the base(2) expansion of n.


2



2, 1, 3, 2, 4, 3, 2, 1, 3, 2, 4, 3, 5, 4, 3, 2, 4, 3, 5, 4, 6, 5, 4, 3, 5, 4, 3, 2, 4, 3, 2, 1, 3, 2, 4, 3, 5, 4, 3, 2, 4, 3, 5, 4, 6, 5, 4, 3, 5, 4, 6, 5, 7, 6, 5, 4, 6, 5, 4, 3, 5, 4, 3, 2, 4, 3, 5, 4, 6, 5, 4, 3, 5, 4, 6, 5, 7, 6, 5, 4, 6, 5, 7, 6, 8, 7, 6
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OFFSET

1,1


COMMENTS

Number of 1's in A212529(n).
Define f(n) as: f(0) = 0, f(2*n) = f(n), f(2*n+1) = f(n) + 1, then a(n) = f(n), n >= 1. See A027615 for the other half of f.
For k > 1, the earliest occurrence of k is n = A086893(k1).


LINKS

Jianing Song, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Negadecimal
Eric Weisstein's World of Mathematics, Negabinary
Wikipedia, Negative base


FORMULA

a(n) == n (mod 3).
a(n) = A000120(A005352(n)).  Michel Marcus, Oct 23 2018


EXAMPLE

A212529(11) = 110101 which has four 1's, so a(11) = 4.
A212529(25) = 111011 which has five 1's, so a(25) = 5.
A212529(51) = 11011101 which has six 1's, so a(51) = 6.


PROG

(PARI) b(n) = if(n==0, 0, b(n\(2))+n%2)
a(n) = b(n)


CROSSREFS

Cf. A000120, A005352, A027615, A086893, A212529.
Sequence in context: A274121 A052306 A322886 * A046823 A224989 A124876
Adjacent sequences: A320639 A320640 A320641 * A320643 A320644 A320645


KEYWORD

nonn,base


AUTHOR

Jianing Song, Oct 18 2018


STATUS

approved



