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Let p = A295895(n) = parity of the binary weight of A005940(1+n). If A005940(1+n) is a square or twice a square (in A028982) then a(n) = 1 - p, otherwise a(n) = p.
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%I #12 Dec 01 2017 18:51:40

%S 0,0,0,0,0,0,1,0,1,0,0,0,0,1,0,0,1,1,1,0,1,0,0,0,0,0,0,1,0,0,0,0,1,1,

%T 0,1,1,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,1,0,0,0,0,0,0,1,0,1,

%U 0,0,0,1,1,1,0,1,0,0,0,0,1,0,0,1,1,0,0,0,1,0,0,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,1,1,1,0,0,1,0,1,1,1,0,0

%N Let p = A295895(n) = parity of the binary weight of A005940(1+n). If A005940(1+n) is a square or twice a square (in A028982) then a(n) = 1 - p, otherwise a(n) = p.

%H Antti Karttunen, <a href="/A295875/b295875.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) = A295895(n) + A295896(n) (mod 2).

%F a(n) = A295894(n) + A000203(A005940(1+n)) mod 2.

%F a(n) = A295297(A005940(1+n)).

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

%e The first six levels of the binary tree (compare also to the illustrations given at A005940, A295894 and A295895):

%e 0

%e |

%e 0

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

%e 0 0

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

%e 0 0 1 0

%e / \ / \ / \ / \

%e / \ / \ / \ / \

%e 1 0 0 0 0 1 0 0

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

%e 1 1 1 0 1 0 0 0 0 0 0 1 0 0 0 0

%o (Scheme) (define (A295875 n) (A000035 (+ (A295895 n) (A295896 n))))

%Y Cf. A000035, A005940, A295297, A295894, A295895, A295896.

%K nonn

%O 0

%A _Antti Karttunen_, Dec 01 2017