A267584 a(0)=1; thereafter a(n) = 2^(1 + number of zeros in binary expansion of n).

1, 2, 4, 2, 8, 4, 4, 2, 16, 8, 8, 4, 8, 4, 4, 2, 32, 16, 16, 8, 16, 8, 8, 4, 16, 8, 8, 4, 8, 4, 4, 2, 64, 32, 32, 16, 32, 16, 16, 8, 32, 16, 16, 8, 16, 8, 8, 4, 32, 16, 16, 8, 16, 8, 8, 4, 16, 8, 8, 4, 8, 4, 4, 2, 128, 64, 64, 32, 64, 32, 32, 16, 64

Offset: 0

Author

N. J. A. Sloane, Jan 17 2016

Keywords

• nonn

Examples

			12 = 1100 in binary, which contains two 0's, so a(12) = 2^3 = 8.
		

Links

• N. J. A. Sloane, Table of n, a(n) for n = 0..20000

Crossrefs

Partial sums give A064194.

Cf. A023416.

Programs

Mathematica

Join[{1},Table[2^(1+DigitCount[n,2,0]),{n,80}]] (* Harvey P. Dale, Oct 08 2023 *)

Formula

For n >= 1, a(n) = 2^(1+A023416(n)).