

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+1i) for 1 <= i <= 2^k.


11



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, 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, 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
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OFFSET

1,1


COMMENTS

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
Complement of characteristic function of A060833.
From Tanya Khovanova, Apr 21 2020: (Start)
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.
Even terms of this sequence alternate: 1, 0, 1, 0 and so on.
Removing evenindexed terms doesn't change the sequence. (End)


LINKS

Table of n, a(n) for n=1..105.
Index entries for sequences obtained by enumerating foldings


FORMULA

For n >= 2, Sum_{k=1..n} a(k) = (n + A037834(n1))/2.
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


EXAMPLE

3 first terms are 0,1,1; therefore, a(4) = a(3+1) = 1  a(31) = 1  a(2) = 0, a(5) = a(3+2) = 1  a(32) = 1  a(1) = 1 and the sequence begins 0, 1, 1, 0, 1, ...


CROSSREFS

The following are all essentially the same sequence: A014577, A014707, A014709, A014710, A034947, A038189, A082410.  N. J. A. Sloane, Jul 27 2012
Sequence in context: A286400 A288478 A225183 * A189479 A260394 A181932
Adjacent sequences: A082407 A082408 A082409 * A082411 A082412 A082413


KEYWORD

nonn


AUTHOR

Benoit Cloitre, Apr 24 2003


STATUS

approved



