A089224 In binary representation: number of zeros of number of zeros of n.

0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 2, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 2, 2, 0, 2, 0, 0, 1, 2, 0, 0, 1, 0, 1, 1, 0, 2, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 2, 1, 2, 2, 0, 1, 2, 2, 0, 2, 0, 0, 1, 1, 2, 2, 0, 2, 0, 0, 1, 2, 0, 0, 1, 0, 1, 1, 0, 1, 2, 2, 0, 2, 0

Offset: 0

Author

Reinhard Zumkeller, Dec 10 2003

Keywords

• nonn
• base

Examples

			a(0) = 0; a(1) = 1; a(16) = 2; a(256) = 3; a(65536) = 4.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
• Index entries for sequences related to binary expansion of n.

Crossrefs

Cf. A054868, A007088.

Cf. A023416.

Programs

Haskell

a089224 = a023416 . a023416  -- Reinhard Zumkeller, Mar 31 2015

Maple

a:= n-> (z-> z(z(n)))(k-> `if`(k=0, 1, add(1-i, i=Bits[Split](k)))):
seq(a(n), n=0..100);  # Alois P. Heinz, Jul 04 2022

Mathematica

a[n_] := DigitCount[DigitCount[n, 2, 0], 2, 0]; Array[a, 100, 0] (* Amiram Eldar, Jul 24 2023 *)

Python

def a(n): return bin(bin(n)[2:].count("0"))[2:].count("0")
print([a(n) for n in range(102)]) # Michael S. Branicky, Jul 04 2022

Formula

a(n) = A023416(A023416(n)).