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%I #9 Dec 01 2017 18:52:21
%S 0,1,0,1,1,0,1,1,0,1,1,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,1,0,1,1,1,
%T 0,1,0,1,0,0,1,1,1,0,1,0,1,1,1,1,0,1,1,0,1,1,0,1,0,1,0,0,0,1,1,1,0,0,
%U 0,0,0,1,1,0,0,1,1,0,0,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,1,1,0,0,1,1,1,1,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0
%N a(n) = 1 if in Doudna-tree (A005940) the contents of the node n and its left-hand child have binary weights with different parity, 0 otherwise.
%C Like A005940 and A252743 this can be viewed as a binary tree. See illustration below.
%H Antti Karttunen, <a href="/A295893/b295893.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/Ch#char_fns">Index entries for characteristic functions</a>
%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>
%F a(n) = A295891(A005940(1+n)).
%F a(n) = A000035(A295894(n) + A295894(2n)) = A295895(n) XOR A295895(2n).
%e The first six levels of the binary tree (compare also to the illustrations given at A005940, A295894, A295895):
%e 0
%e |
%e 1
%e ............../ \..............
%e 0 1
%e ....../ \...... ....../ \......
%e 1 0 1 1
%e / \ / \ / \ / \
%e / \ / \ / \ / \
%e 0 1 1 0 0 0 0 0
%e / \ / \ / \ / \ / \ / \ / \ / \
%e 0 1 0 0 1 0 0 0 0 1 0 0 0 1 0 1
%e For n=0, the corresponding node in A005940(0+1) is 1, which has just one child 2, thus we set a(0) = 0 by convention.
%e For n=1, the corresponding node in A005940(1+1) is 2, which has 3 as its left-hand child, and in binary 2 is "10", and 3 is "11", so the other one has an odd number of 1-bits, while the other has an even number of 1's, thus their parities differ and a(1) = 1.
%e For n=2, the corresponding node in A005940(1+2) is 3 (in binary "11"), which has 5 (in binary "101") as its left-hand child, and they have the same parity, thus a(2) = 0.
%e For n=3, the corresponding node in A005940(1+3) is 4 (in binary "100"), which has 9 (in binary "1001") as its left hand child, and here the parities differ, thus a(3) = 1.
%o (Scheme) (define (A295893 n) (A295891 (A005940 (+ 1 n))))
%Y Cf. A003961, A005940, A295891, A295894, A295895.
%Y Cf. also A252743.
%K nonn,tabf
%O 0
%A _Antti Karttunen_, Nov 30 2017