

A014577


The regular paperfolding sequence (or dragon curve sequence).


34



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,1


COMMENTS

a(n) is the complement of the bit to the left of the least significant "1" in the binary expansion of n. E.g., n = 4 = 100, so a(4) = (complement of bit to left of 1) = 1.  Robert L. Brown, Nov 28 2001
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
A014577 is a generator for A088748: begin A088748 with "1", then add 1 if A014577: (1, 1, 0, 1, 1,...) = 1; subtract 1 otherwise, getting (1, 2, 3, 2,...).  Gary W. Adamson, Aug 30 2009
The characteristic function is 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 Lsystem with rules L>L1R, R>L0R, 1>1, 0>0, starting with L, and dropping all L and R (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 onetoone 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 SternBrocot 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, else 0; getting A014557, the HarterHeighway dragon curve: (1, 1, 0, 1, 1, 0, 0, 1, 1,...). (End)
The paperfolding 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


REFERENCES

J.P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, pp. 155, 182.
Danielle Cox and K. McLellan, A problem on generation sets containing Fibonacci numbers, Fib. Quart., 55 (No. 2, 2017), 105113.
M. Gardner, Mathematical Magic Show. New York: Vintage, pp. 207209 and 215220, 1978.
G. Melancon, Factorizing infinite words using Maple, MapleTech journal, vol. 4, no. 1, 1997, pp. 3442, esp. p. 36.
Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence  see "List of Sequences" in Vol. 2.


LINKS

Ivan Panchenko, Table of n, a(n) for n = 0..10000
Ibrahim M. Alabdulmohsin, "Analytic Summability Theory", in Summability Calculus: A Comprehensive Theory of Fractional Finite Sums, Springer, Cham, pp 6591.
J.P. Allouche and M. Mendes France, Automata and Automatic Sequences.
Joerg Arndt, Matters Computational (The Fxtbook), pp. 8892; image of the dragon curve on p. 89.
Michael Coons, An Irrationality Measure for Regular Paperfolding Numbers, Journal of Integer Sequences, Vol. 15 (2012), Article #12.1.6.
Alexey Garber, On triangular paperfolding patterns, arXiv:1807.05627 [math.CO], 2018.
Franz Gahler and Johan Nilsson, Substitution rules for higherdimensional paperfolding structures, arXiv:1408.4997 [math.DS], 2014.
A. M. Hinz, S. Klavžar, U. Milutinović, C. Petr, The Tower of Hanoi  Myths and Maths, Birkhäuser 2013. See page 63. Book's website
Luke Schaeffer, Jeffrey Shallit, Closed, Palindromic, Rich, Privileged, Trapezoidal, and Balanced Words in Automatic Sequences, Electronic Journal of Combinatorics 23(1) (2016), #P1.25.
J. E. S. Socolar and J. M. Taylor, An aperiodic hexagonal tile, arXiv:1003.4279 [math.CO], 2010.
Eric Weisstein's World of Mathematics, Dragon curve.
Index entries for 2automatic sequences.
Index entries for sequences related to binary expansion of n
Index entries for sequences obtained by enumerating foldings


FORMULA

a(n) = (1+jacobi(1,n))/2 (cf. A034947).  N. J. A. Sloane, Jul 27 2012
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).
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.  Ralf Stephan, Jul 03 2003
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
Can be generated directly from A005811:
1,...2,...1,...2,...3,...2,...1,...2,...3,... = A005811.
1,...1,...0,...1,...1,...0,...0,...1,...1,... = A014577.
By inspection, A014577 = 1 if the corresponding term in A005811 is greater than the previous A005811 term; else 0.  Gary W. Adamson, Jun 20 2012
a((2*n+1)*2^p1) = (n+1) mod 2, p >= 0.  Johannes W. Meijer, Jan 28 2013
G.f. g(x) satisfies g(x) = x*g(x^2) + 1/(1x^4).  Robert Israel, Jan 06 2015


EXAMPLE

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)


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^p1) := (n+1) mod 2 od: od: seq(a(n), n=0..nmax); # Johannes W. Meijer, Jan 28 2013


MATHEMATICA

a[n_] := Boole[EvenQ[((n+1)/2^IntegerExponent[n+1, 2]1)/2]]; Table[a[n], {n, 0, 98}] (* JeanFrançois 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 *)


PROG

(C++) /* code from the fxt library, about 5 CPU cycles per computation */
bool bit_paper_fold(ulong k)
{
ulong h = k & k; /* == lowest_one(k) */
k &= (h<<1);
return ( k==0 );
} /* Joerg Arndt, Apr 15 2010 */
(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
(MAGMA) [(1+KroneckerSymbol(1, n))/2 : n in [1..100]] /* or */ [Floor(1/2*(1+(1)^(1/2*((n+1)/2^Valuation(n+1, 2)1)))): n in [0..100]]; // Vincenzo Librandi, Aug 05 2015


CROSSREFS

The following are all essentially the same sequence: A014577, A014707, A014709, A014710, A034947, A038189, A082410.  N. J. A. Sloane, Jul 27 2012
Cf. A038189, A059125, A065339, A005811, A220466.
A082410(n+2)=a(n).
Cf. A088748, A091072.
Sequence in context: A079559 A175480 A229062 * A157926 A131377 A077049
Adjacent sequences: A014574 A014575 A014576 * A014578 A014579 A014580


KEYWORD

nonn,easy,nice


AUTHOR

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


EXTENSIONS

More terms from Ralf Stephan, Jul 03 2003


STATUS

approved



