A320642 Number of 1's in the base-(-2) expansion of -n.

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

Offset: 1

Author

Jianing Song, Oct 18 2018

Keywords

• nonn
• base

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(k-1).

Examples

			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.
		

Links

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

Crossrefs

Cf. A000120, A005352, A027615, A086893, A212529.

Programs

Mathematica

b[n_] := b[n] = b[Quotient[n - 1, -2]] + Mod[n, 2]; b[0] = 0; a[n_] := b[-n]; Array[a, 100] (* Amiram Eldar, Jul 23 2023 *)

PARI

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

Formula

a(n) == -n (mod 3).

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