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A117964 a(n) = A117963(n) mod 2.

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%I A117964 #11 Jan 01 2023 12:35:55
%S A117964 1,1,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,1,1,0,0,0,0,0,0,0,0,
%T A117964 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,1,1,0,0,0,0,0,0,
%U A117964 0,0,0,0,1,1,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
%N A117964 a(n) = A117963(n) mod 2.
%C A117964 a(3n+2) = 0, a(3n) = a(3n+1). a(3n) may be equal to A088917(n).
%H A117964 Antti Karttunen, <a href="/A117964/b117964.txt">Table of n, a(n) for n = 0..16384</a>
%H A117964 <a href="/index/Ch#char_fns">Index entries for characteristic functions</a>
%F A117964 a(n)=sum{k=0..floor(n/2), L(C(n-k,k)/3)} mod 2 where L(j/p) is the Legendre symbol of j and p.
%F A117964 a(2*A081601(n)) = a(1+2*A081601(n)) = 1. [Conjectured, also these two formulas together seem to give the positions of all 1's] - _Antti Karttunen_, Jan 01 2023
%Y A117964 Cf. A000035, A005836, A117963, A081601, A088917.
%K A117964 easy,nonn
%O A117964 0,1
%A A117964 _Paul Barry_, Apr 05 2006