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.

Showing 1-3 of 3 results.

A038219 The Ehrenfeucht-Mycielski sequence (0,1-version): a maximally unpredictable sequence.

Original entry on oeis.org

0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1
Offset: 1

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Comments

The sequence starts 0,1,0 and continues according to the following rule: find the longest suffix that has occurred at least once previously. If there is more than one previous occurrences select the most recent one. The next digit of the sequence is the opposite of the one following the previous occurrence. - Christopher Carl Heckman, Feb 10 2005

Examples

			We start with a(1)=0, a(2)=1, a(3)=0.
The longest suffix we have seen before is "0", when it occurred at the start, followed by 1. So a(4) = 0. We now have 0100.
The longest suffix we have seen before is again "0", when it occurred at a(3), followed by a(4)=0. So a(5) = 1. We now have 01001.
The longest suffix we have seen before is "01", when it occurred at a(1),a(2), followed by a(3)=0. So a(6) = 1. We now have 010011.
And so on.
For further illustrations of calculating these terms, see A308174 and A308175. - _N. J. A. Sloane_, May 21 2019
		

Crossrefs

Cf. A007061 (1, 2 version).
Cf. A201881 (run lengths).
Cf. also A253059, A253060, A253061.
For first appearance of subwords see A308173.
A308174, A308175 are used in the calculation of a(n).

Programs

  • Haskell
    a038219 n = a038219_list !! n
    a038219_list = 0 : f [0] where
       f us = a' : f (us ++ [a']) where
            a' = b $ reverse $ map (`splitAt` us) [0..length us - 1] where
               b ((xs,ys):xyss) | vs `isSuffixOf` xs = 1 - head ys
                                | otherwise          = b xyss
            vs = fromJust $ find (`isInfixOf` init us) $ tails us
    -- Reinhard Zumkeller, Dec 05 2011
    
  • Maple
    See Lunnon link.
  • Perl
    See Links section.
    
  • Python
    from itertools import count, islice
    def agen():
        astr, preval = "010", 1
        yield from [0, 1, 0]
        while True:
            an = 1 - preval
            yield an
            astr += str(an)
            for l in range(len(astr)-1, 0, -1):
                idx = astr.rfind(astr[-l:], 0, len(astr)-1)
                if idx >= 0: preval = int(astr[idx+l]); break
    print(list(islice(agen(), 105))) # Michael S. Branicky, Aug 03 2022

Extensions

More terms from Joshua Zucker, Aug 11 2006
Offset changed by Reinhard Zumkeller, Dec 11 2011
Edited by N. J. A. Sloane, May 12 2019

A007061 The Ehrenfeucht-Mycielski sequence (1,2-version): a maximally unpredictable sequence.

Original entry on oeis.org

1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 2, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 2, 2, 1, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 2
Offset: 1

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Comments

Klaus Sutner remarks (Jun 26 2006) that this sequence is very similar to the Kimberling sequence A079101. Both sequences have every finite binary word as a factor; in fact, essentially the same proof works for both sequences.
Sutner continues: All words of length k seem to appear in the first 2^{k+2} bits. This is true for the first billion bits of the sequence, but no proof is known. The main open problem is whether the limiting density of 0's is 1/2. It seems to require a large amount of effort just to show that it is bounded away from 0, never mind some of the more exotic properties of the sequence (see the Sutner reference).
Start with a single bit 0. If the first n bits U(n) = a(1)a(2)...a(n) have already been chosen, let v be the longest suffix of U(n) that already appears in U(n-1). Find the last occurrence of v in U(n-1) and let b the bit that occurs immediately after. Then a(n+1) is the complement of b. (The entry gives the bits as 1's and 2s instead of 0's and 1's - compare A038219) - Joshua Zucker, Aug 11 2006

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A038219 (0-1 version), A079101.
Cf. A201881 (run lengths).
Cf. also A253059, A253060, A253061.

Programs

  • Haskell
    a007061 n = a007061_list !! (n-1)
    a007061_list = 1 : f [1] where
       f us = a' : f (us ++ [a']) where
         a' = b $ reverse $ map (`splitAt` us) [0..length us - 1] where
            b ((xs,ys):xyss) | vs `isSuffixOf` xs = 3 - head ys
                             | otherwise          = b xyss
         vs = fromJust $ find (`isInfixOf` init us) $ tails us
    -- Reinhard Zumkeller, Dec 05 2011
    
  • Python
    from itertools import count, islice
    def agen():
        astr, preval = "121", 2
        yield from [1, 2, 1]
        while True:
            an = 3 - preval
            yield an
            astr += str(an)
            for l in range(len(astr)-1, 0, -1):
                idx = astr.rfind(astr[-l:], 0, len(astr)-1)
                if idx >= 0: preval = int(astr[idx+l]); break
    print(list(islice(agen(), 105))) # Michael S. Branicky, Aug 03 2022

Extensions

More terms from Joshua Zucker, Aug 11 2006
Offset changed from 0 to 1, Aug 18 2006

A201882 Where the first run of length n occurs in maximally unpredictable sequences A007061, A038219.

Original entry on oeis.org

1, 3, 8, 11, 27, 36, 84, 231, 349, 535, 1267, 2916, 4114, 14349, 27045, 35059, 89723, 234443, 408129, 799350, 1926026, 2170589, 4291892, 10758318, 21141201, 57927399, 122141530, 138265841
Offset: 1

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Author

Reinhard Zumkeller, Dec 11 2011

Keywords

Comments

A201881(a(n)) = n and A201881(m) < n for m < a(n).
Since every substring appears in A007061 (and A038219) this sequence is infinite. - N. J. A. Sloane, May 17 2019

Crossrefs

Programs

  • C
    // See Links section
  • Haskell
    a201882 = ((+ 1) . fromJust . (`elemIndex` a201881_list))
    

Extensions

a(11)-a(28) from Rémy Sigrist, Jul 31 2022
Showing 1-3 of 3 results.