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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.

A014577 The regular paper-folding sequence (or dragon curve sequence). Alphabet {1,0}.

Original entry on oeis.org

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
Offset: 0

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Author

N. J. A. Sloane, Eric W. Weisstein

Keywords

Comments

a(n) is the complement of the bit to the left of the least significant "1" in the binary expansion of n+1. E.g., n = 3, n+1 = 4 = 100_2, so a(3) = (complement of bit to left of 1) = 1. - Robert L. Brown, Nov 28 2001 [Adjusted to match offset by N. J. A. Sloane, Apr 15 2021]
To construct the sequence: start from 1,(..),0,(..),1,(..),0,(..),1,(..),0,(..),1,(..),0,... and fill undefined places with the sequence itself. - Benoit Cloitre, Jul 08 2007
This is the characteristic function of A091072 - 1. - Gary W. Adamson, Apr 11 2010
Turns (by 90 degrees) of the Heighway dragon which can be rendered as follows: [Init] Set n=0 and direction=0. [Draw] Draw a unit line (in the current direction). Turn left/right if a(n) is zero/nonzero respectively. [Next] Set n=n+1 and goto (draw). See fxtbook link below. - Joerg Arndt, Apr 15 2010
Sequence can be obtained by the L-system with rules L->L1R, R->L0R, 1->1, 0->0, starting with L, and then dropping all L's and R's (see example). - Joerg Arndt, Aug 28 2011
From Gary W. Adamson, Jun 20 2012: (Start)
One half of the infinite Farey tree can be mapped one-to-one onto A014577 since both sequences can be derived directly from the binary. First few terms are
1, 1, 0, 1, 1, 0, 0, 1, 1, 1, ...
1/2 2/3 1/3 3/4 3/5 2/5 1/4 4/5 5/7 5/8, ...
Infinite Farey tree fractions can be derived from the binary by appending a repeat of rightmost binary term to the right, then recording the number of runs to obtain the continued fraction representation. Example: 9 = 1001 which becomes 10011 which becomes [1,2,2] = 5/7. (End)
The sequence can be considered as a binomial transform operator for a target sequence S(n). Replace the first 1 in A014577 with the first term in S(n), then for successive "1" term in A014577, map the next higher term in S(n). If "0" in A014577, map the next lower term in S(n). Using the sequence S(n) = (1, 3, 5, 7, ...), we obtain (1), (3, 1), (3, 5, 3, 1), (3, 5, 7, 5, 3, 5, 3, 1), .... Then parse the terms into subsequences of 2^k terms, adding the terms in each string. We obtain (1, 4, 12, 32, 80, ...), the binomial transform of (1, 3, 5, 7, ...). The 8-bit string has one 1, three 5's, three 7's and one 1) as expected, or (1, 3, 3, 1) dot (1, 3, 5, 7). - Gary W. Adamson, Jun 24 2012
From Gary W. Adamson, May 29 2013: (Start)
The sequence can be generated directly from the lengths of continued fraction representations of fractions in one half of the Stern-Brocot tree (fractions between 0 and 1):
1/2
1/3 2/3
1/4 2/5 3/5 3/4
1/5 2/7 3/8 3/7 4/7 5/8 5/7 4/5
...
and their corresponding continued fraction representations are:
[2]
[3] [1,2]
[4] [2,2] [1,1,2] [1,3]
[5] [3,2] [2,1,2] [2,3] [1,1,3] [1,1,1,2] [1,2,2] [1,4]
...
Record the lengths by rows then reverse rows, getting:
1,
2, 1,
2, 3, 2, 1,
2, 3, 4, 3, 2, 3, 2, 1,
...
start with "1" and if the next term is greater than the current term, record a 1, otherwise 0; getting the present sequence, the Harter-Heighway dragon curve. (End)
The paper-folding word "110110011100100111011000..." can be created by concatenating the terms of a fixed point of the morphism or string substitution rule: 00 -> 1000, 01 -> 1001, 10 -> 1100 & 11 -> 1101, beginning with "11". - Robert G. Wilson v, Jun 11 2015
From Gary W. Adamson, Jun 04 2021: (Start)
Since the Heighway dragon is composed of right angles, it can be mapped with unit trajectories (Right = 1, Left = (-1), Up = i and Down = -i) on the complex plane where i = sqrt(-1). The initial (0th) iterate is chosen in this case to be the unit line from (0,0) to (-1,0). Then follow the directions below as indicated, resulting in a reflected variant of the dragon curve shown at the Eric Weisstein link. The conjectured system of complex plane trajectories is:
0 -1
1 -1, i
2 -1, i, 1, i
3 -1, i, 1, i, 1, -i, 1, i
4 -1, i, 1, i, 1, -i, 1, i, 1, -i, -1, -i, 1, -i, 1, i
...
The conjecture succeeds through the 4th iterate. It appears that to generate the (n+1)-th row, bring down the n-th row as the left half of row (n+1). For the right half of row (n+1), bring down the n-th row but change the signs of the first half of row n. For example, to get the complex plane instructions for the third iterate of the dragon curve, bring down (-1, i, 1, i) as the left half, and the right half is (1, -i, 1, i). (End)
From Gary W. Adamson, Jun 09 2021: (Start)
Partial sums of the iterate trajectories produce a sequence of complex addresses for unit segments. Partial sums of row 4 are: -1, (-1+i), i, 2i, (1+2i), (1+i), (2+i), (2+2i), (3+2i), (3+i), (2+i), 2, 3, (3-i), (4-i), 4. (zeros are omitted with terms of the form a+0i). The reflected variant of the dragon curve has the 0th iterate from (0,0) to (1,0), and the respective addresses simply change the signs of the real terms. (End)

Examples

			1 + x + x^3 + x^4 + x^7 + x^8 + x^9 + x^12 + x^15 + x^16 + x^17 + x^19 + ...
From _Joerg Arndt_, Aug 28 2011: (Start)
Generation via string substitution:
Start: L
Rules:
  L --> L1R
  R --> L0R
  0 --> 0
  1 --> 1
-------------
0:   (#=1)
  L
1:   (#=3)
  L1R
2:   (#=7)
  L1R1L0R
3:   (#=15)
  L1R1L0R1L1R0L0R
4:   (#=31)
  L1R1L0R1L1R0L0R1L1R1L0R0L1R0L0R
5:   (#=63)
  L1R1L0R1L1R0L0R1L1R1L0R0L1R0L0R1L1R1L0R1L1R0L0R0L1R1L0R0L1R0L0R
Drop all L and R to obtain 1101100111001001110110001100100
(End)

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, pp. 155, 182.
  • Chandler Davis and Donald E. Knuth, Number Representations and Dragon Curves -- I and II, Journal of Recreational Mathematics, volume 3, number 2, April 1970, pages 66-81, and number 3, July 1970, pages 133-149. Reprinted in Donald E. Knuth, Selected Papers on Fun and Games, CSLI Publications, 2010, pages 571-614.
  • Michel Dekking, Michel Mendes France, and Alf van der Poorten, "Folds", The Mathematical Intelligencer, 4.3 (1982): 130-138 & front cover, and 4:4 (1982): 173-181 (printed in two parts).
  • Martin Gardner, Mathematical Magic Show, New York: Vintage, pp. 207-209 and 215-220, 1978.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Links

Crossrefs

Essentially the same: A014707, A014709, A014710, A034947, A038189, A082410, A089013, A099545, A112347, A121238, A317335, A317336.
Cf. A059125, A065339, A005811, A220466, A088748, A091072, A343173 (first differences), A343174 (partial sums).
The two bisections are A000035 and the sequence itself.
See A343181 for prefixes of length 2^k-1.

Programs

  • Magma
    [(1+KroneckerSymbol(-1,n))/2: n in [1..100]]; // Vincenzo Librandi, Aug 05 2015
  • Magma
    [Floor(1/2*(1+(-1)^(1/2*((n+1)/2^Valuation(n+1, 2)-1)))): n in [0..100]]; // Vincenzo Librandi, Aug 05 2015
  • Maple
    nmax:=98: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 0 to ceil(nmax/(p+2))+1 do a((2*n+1)*2^p-1) := (n+1) mod 2 od: od: seq(a(n), n=0..nmax); # Johannes W. Meijer, Jan 28 2013
a014577 := proc(n) local p,s1,s2,i;
if n=0 then return(1); fi;
s1:=convert(n,base,2); s2:=nops(s1);
for i from 1 to s2 do if s1[i]=1 then p:=i; break; fi; od:
if p <= s2-1 then 1-s1[p+1]; else 1; fi; end;
[seq(a014577(i),i=1..120)]; # N. J. A. Sloane, Apr 08 2021
# third Maple program:
a:= n-> 1-irem(iquo((n+1)/2^padic[ordp](n+1, 2), 2), 2):
seq(a(n), n=0..120);  # Alois P. Heinz, Apr 08 2021
  • Mathematica
    a[n_] := Boole[EvenQ[((n+1)/2^IntegerExponent[n+1, 2]-1)/2]]; Table[a[n], {n, 0, 98}] (* Jean-François Alcover, Feb 16 2012, after Gary W. Adamson, updated Nov 21 2014 *)
Table[1-(((Mod[#1,2^(#2+2)]/2^#2)&[n,IntegerExponent[n,2]])-1)/2,{n,1,100,1}] (* WolframAlpha compatible code; Robert L. Brown, Jan 06 2015 *)
MapThread[(a[x_/;IntegerQ[(x-#1)/4]]:= #2)&,{{1,3},{1,0}}];a[x_/;IntegerQ[x/2]]:=a[x/2];a/@ Range[100] (* Bradley Klee, Aug 04 2015 *)
(1 + JacobiSymbol[-1, Range[100]])/2 (* Paolo Xausa, May 22 2024 *)
Array[Boole[BitAnd[#, BitAnd[#, -#]*2] == 0] &, 100] (* Paolo Xausa, May 22 2024, after Joerg Arndt C++ code *)
  • PARI
    {a(n) = if( n%2, a(n\2), 1 - (n/2%2))} /* Michael Somos, Feb 05 2012 */
  • PARI
    a(n)=1/2*(1+(-1)^(1/2*((n+1)/2^valuation(n+1,2)-1))) \\ Ralf Stephan, Sep 02 2013
  • PARI
    a(n)=!bittest(n+1,valuation(n+1,2)+1); \\ Robert L. Brown, Jul 07 2025
  • Python
    def A014577(n):
    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

a(n) = (1+Jacobi(-1,n+1))/2 (cf. A034947). - N. J. A. Sloane, Jul 27 2012 [Adjusted to match offset by Peter Munn, Jul 01 2022]
Set a=1, b=0, S(0)=a, S(n+1) = S(n), a, F(S(n)), where F(x) reverses x and then interchanges a and b; sequence is limit S(infinity).
From Ralf Stephan, Jul 03 2003: (Start)
a(4*n) = 1, a(4*n+2) = 0, a(2*n+1) = a(n).
a(n) = 1 - A014707(n) = 2 - A014709(n) = A014710(n) - 1. (End)
Set a=1, b=0, S(0)=a, S(n+1)=S(n), a, M(S(n)), where M(S) is S but the bit in the middle position flipped. (Proof via isomorphism of both formulas to a modified string substitution.) - Benjamin Heiland, Dec 11 2011
a(n) = 1 if A005811(n+1) > A005811(n), otherwise a(n) = 0. - Gary W. Adamson, Jun 20 2012
a((2*n+1)*2^p-1) = (n+1) mod 2, p >= 0. - Johannes W. Meijer, Jan 28 2013
G.f. g(x) satisfies g(x) = x*g(x^2) + 1/(1-x^4). - Robert Israel, Jan 06 2015
a(n) = 1 - A065339(n+1) mod 2. - Peter Munn, Jun 29 2022
From A.H.M. Smeets, Mar 19 2023: (Start)
a(n) = 1 - A038189(n+1).
a(n) = A082410(n+2).
a(n) = 1 - A089013(n+1)
a(n) = (3 - A099545(n+1))/2.
a(n) = (A112347(n+1) + 1)/2.
a(n) = (A121238(n+1) + 1)/2.
a(n) = (A317335(n) + 2)/3.
a(n) = (A317336(n) + 10)/3. (End)
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/2. - Amiram Eldar, Sep 14 2024

Extensions

More terms from Ralf Stephan, Jul 03 2003

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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