cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

Original entry on oeis.org

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
Offset: 1

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Author

Benoit Cloitre, Apr 24 2003

Keywords

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 even-indexed terms doesn't change the sequence. (End)

Examples

			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, ...
		

Crossrefs

The following are all essentially the same sequence: A014577, A014707, A014709, A014710, A034947, A038189, A082410. - N. J. A. Sloane, Jul 27 2012

Programs

  • Python
    def A082410(n):
        if n == 1:
            return 0
        s = bin(n-1)[2:]
        m = len(s)
        i = s[::-1].find('1')
        return 1-int(s[m-i-2]) if m-i-2 >= 0 else 1 # Chai Wah Wu, Apr 08 2021

Formula

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