cp's OEIS Frontend

A082410 a(1)=0. Thereafter, the sequence is constructed using the rule: for any k >= 0, if a(1), a(2), ..., a(2^k+1) are known, the next 2^k terms are given as follows: a(2^k+1+i) = 1 - a(2^k+1-i) for 1 <= i <= 2^k.

%I A082410 #59 Jul 18 2021 20:59:39
%S A082410 0,1,1,0,1,1,0,0,1,1,1,0,0,1,0,0,1,1,1,0,1,1,0,0,0,1,1,0,0,1,0,0,1,1,
%T A082410 1,0,1,1,0,0,1,1,1,0,0,1,0,0,0,1,1,0,1,1,0,0,0,1,1,0,0,1,0,0,1,1,1,0,
%U A082410 1,1,0,0,1,1,1,0,0,1,0,0,1,1,1,0,1,1,0,0,0,1,1,0,0,1,0,0,0,1,1,0,1,1,0,0,1
%N A082410 a(1)=0. Thereafter, the sequence is constructed using the rule: for any k >= 0, if a(1), a(2), ..., a(2^k+1) are known, the next 2^k terms are given as follows: a(2^k+1+i) = 1 - a(2^k+1-i) for 1 <= i <= 2^k.
%C A082410 a(n) is A014577 shifted right twice (the definition here is similar to one of the constructions for A034947). - _N. J. A. Sloane_, Jul 27 2012
%C A082410 Complement of characteristic function of A060833.
%C A082410 From _Tanya Khovanova_, Apr 21 2020: (Start)
%C A082410 Suppose you have a deck of cards face down with 2^n cards such that the color pattern corresponds to this sequence: 0 for one color, 1 for the other. Then you proceed in the following manner: transfer to top card to the bottom of the deck, deal the next card, then repeat. The dealt cards will have alternating colors.
%C A082410 Even terms of this sequence alternate: 1, 0, 1, 0 and so on.
%C A082410 Removing even-indexed terms doesn't change the sequence. (End)
%H A082410 <a href="/index/Fo#fold">Index entries for sequences obtained by enumerating foldings</a>
%F A082410 For n >= 2, Sum_{k=1..n} a(k) = (n + A037834(n-1))/2.
%F A082410 a(1) = 0, a(4*n+2) = 1, a(4*n+4) = 0, a(2*n+1) = a(n+1) for n >= 0. - _A.H.M. Smeets_, Jul 27 2018
%e A082410 First 3 terms are 0,1,1; therefore, a(4) = a(3+1) = 1 - a(3-1) = 1 - a(2) = 0, a(5) = a(3+2) = 1 - a(3-2) = 1 - a(1) = 1 and the sequence begins 0, 1, 1, 0, 1, ...
%o A082410 (Python)
%o A082410 def A082410(n):
%o A082410     if n == 1:
%o A082410         return 0
%o A082410     s = bin(n-1)[2:]
%o A082410     m = len(s)
%o A082410     i = s[::-1].find('1')
%o A082410     return 1-int(s[m-i-2]) if m-i-2 >= 0 else 1 # _Chai Wah Wu_, Apr 08 2021
%Y A082410 The following are all essentially the same sequence: A014577, A014707, A014709, A014710, A034947, A038189, A082410. - _N. J. A. Sloane_, Jul 27 2012
%K A082410 nonn
%O A082410 1,1
%A A082410 _Benoit Cloitre_, Apr 24 2003