%I #8 Dec 01 2017 18:52:28

%S 1,1,2,1,2,2,2,1,3,2,4,2,3,2,4,1,3,3,3,2,3,4,4,2,3,3,4,2,6,4,3,1,3,3,

%T 2,3,5,3,6,2,4,3,4,4,6,4,4,2,5,3,4,3,6,4,4,2,6,6,7,4,5,3,6,1,2,3,4,3,

%U 2,2,4,3,5,5,4,3,4,6,6,2,5,4,6,3,3,4,6,4,5,6,4,4,7,4,5,2,4,5,6,3,6,4,6,3,7,6,8,4,5,4,5,2,6,6,3,6,7,7,5,4,5

%N Binary weight of the contents of node n in Doudna-tree (A005940).

%H Antti Karttunen, <a href="/A295894/b295894.txt">Table of n, a(n) for n = 0..16383</a>

%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>

%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>

%F a(n) = A000120(A005940(1+n)).

%F a(2n+1) = a(n).

%F A000035(a(n)) = A295895(n).

%e The first six levels of the binary tree (compare also to the illustration given at A005940):

%e 1

%e |

%e 1

%e ............../ \..............

%e 2 1

%e ....../ \...... ....../ \......

%e 2 2 2 1

%e / \ / \ / \ / \

%e / \ / \ / \ / \

%e 3 2 4 2 3 2 4 1

%e / \ / \ / \ / \ / \ / \ / \ / \

%e 3 3 3 2 3 4 4 2 3 3 4 2 6 4 3 1

%e For n=0, the corresponding node in A005940(0+1) is 1, in binary also 1, thus a(0) = A000120(1) = 1.

%e For n=1, the corresponding node in A005940(1+1) is 2, in binary "10", thus a(1) = A000120(2) = 1.

%e For n=2, the corresponding node in A005940(1+2) is 3, in binary "11", thus a(2) = A000120(3) = 2.

%e For n=3, the corresponding node in A005940(1+3) is 4, in binary "100", thus a(3) = A000120(4) = 1.

%o (Scheme) (define (A295894 n) (A000120 (A005940 (+ 1 n))))

%Y Cf. A000120, A003961, A005940, A295891, A295893, A295895.

%Y Cf. A000225 (the positions of ones).

%K nonn

%O 0,3

%A _Antti Karttunen_, Nov 30 2017