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

%I #9 Oct 08 2023 13:41:33

%S 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,

%T 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,

%U 8,8,4,8,4,4,2,128,64,64,32,64,32,32,16,64

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

%H N. J. A. Sloane, <a href="/A267584/b267584.txt">Table of n, a(n) for n = 0..20000</a>

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

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

%t Join[{1},Table[2^(1+DigitCount[n,2,0]),{n,80}]] (* _Harvey P. Dale_, Oct 08 2023 *)

%Y Partial sums give A064194.

%Y Cf. A023416.

%K nonn

%O 0,2

%A _N. J. A. Sloane_, Jan 17 2016