A343030 Number of 1-bits in the binary expansion of n which have an odd number of 0-bits at less significant bit positions.

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

Offset

0,7

Comments

See A343029 for further notes.

Links

Kevin Ryde, Table of n, a(n) for n = 0..8192

Formula

a(n) = A343029(n) - A004718(n).

a(n) = A000120(n) - A343029(n), where A000120 is the number of 1-bits in n (binary weight).

a(2*n) = A000120(n) - a(n).

a(2*n+1) = a(n).

G.f. satisfies g(x) = (x-1)*g(x^2) + A000120(x^2).

G.f.: (1/2)* Sum_{k>=0} x^(2^k)*( (1-x^(2^k))/(1-x) - Prod_{j=0..k-1} x^(2^j)-1 )/( 1-x^(2*2^k ) ).

a(2(2^n - 1)) = n. - Michael S. Branicky, Apr 03 2021

Example

n = 628 = binary 1001110100
                 ^  ^^^      a(n) = 4

Prog

(PARI) a(n) = my(t=0, ret=0); for(i=0, if(n, logint(n, 2)), if(bittest(n, i), ret+=t, t=!t)); ret;
(Python)
def a(n):
  b = bin(n)[2:]
  return sum(bi=='1' and b[i:].count('0')%2==1 for i, bi in enumerate(b))
print([a(n) for n in range(87)]) # Michael S. Branicky, Apr 03 2021

Crossrefs

Cf. A343029, A004718, A000120, A000918 (indices of new highs).

Sequence in context: A136745 A214157 A246720 * A246690 A317748 A090465

Adjacent sequences: A343027 A343028 A343029 * A343031 A343032 A343033

Keyword

nonn,easy

Author

Kevin Ryde, Apr 03 2021

Status

approved