0,1

The sequence is cube-free (does not contain three consecutive identical blocks) and is overlap-free (does not contain XYXYX where X is 0 or 1 and Y is any string of 0's and 1's).

a(n) = "parity sequence" = parity of number of 1's in binary representation of n.

To construct the sequence: alternate blocks of 0's and 1's of successive lengths A003159(k)-A003159(k-1), k=1,2,3,... (A003159(0)=0). Example: since the first seven differences of A003159 are 1,2,1,1,2,2,2, the sequence starts with 0,1,1,0,1,0,0,1,1,0,0. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Jan 10 2003

Characteristic function of A000069 (odious numbers). a(n) = 1-A010059(n) = A001285(n)-1. - Ralf Stephan (ralf(AT)ark.in-berlin.de), Jun 20 2003

a(n)=S2(n)mod 2, where S2(n) = sum of digits of n, n in base-2 notation. There is a class of generalized Thue-Morse sequences : Let Sk(n) = sum of digits of n; n in base-k notation. Let F(t) be some arithmetic function. Then a(n)= F(Sk(n)) mod m is a generalised Thue-Morse sequence. The classical Thue-Morse sequence is the case k=2, m=2, F(t)= 1. - Ctibor O. Zizka (ctibor.zizka(AT)seznam.cz), Feb 12 2008

More generally, the partial sums of the generalized Thue-Morse sequences a(n)=F(Sk(n)) mod m are fractal, where Sk(n) is sum of digits of n, n in base k; F(t) is an arithmetic function; m integer. - Ctibor O. ZIZKA (ctibor.zizka(AT)seznam.cz), Feb 25 2008

Starting with offset 1, = running sums mod 2 of the kneading sequence (A035263, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1,...); also parity of A005187: (1, 3, 4, 7, 8, 10, 11, 15, 16, 18, 19,...). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 15 2008

Generalized Thue-Morse sequences mod n (n>1) = the array shown in A141803. As n -> Inf. the sequences -> (1, 2, 3,...). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jul 10 2008

The Thue-Morse sequence for N = 3 = A053838, (sum of digits of n in base 3, mod 3): (0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2,...) = A004128 mod 3. [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 24 2008]

For all positive integers k, the subsequence a(0) to a(2^k-1) is identical to the subsequence a(2^k+2^(k-1)) to a(2^(k+1)+2^(k-1)-1). That is to say, the first half of A_k is identical to the second half of B_k, and the second half of A_k is identical to the first quarter of B_{k+1}, which consists of the k/2 terms immediately following B_k.

Proof: The subsequence a(2^k+2^(k-1)) to a(2^(k+1)-1), the second half of B_k, is by definition formed from the subsequence a(2^(k-1)) to a(2^k-1), the second half of A_k, by interchanging its 0s and 1s. In turn, the subsequence a(2^(k-1)) to a(2^k-1), the second half of A_k, which is by definition also B_{k-1}, is by definition formed from the subsequence a(0) to a(2^(k-1)-1), the first half of A_k, which is by definition also A_{k-1}, by interchanging its 0s and 1s. Interchanging the 0s and 1s of a subsequence twice leaves it unchanged, so the subsequence a(2^k+2^(k-1)) to a(2^(k+1)-1), the second half of B_k, must be identical to the subsequence a(0) to a(2^(k-1)-1), the first half of A_k.

Also, the subsequence a(2^(k+1)) to a(2^(k+1)+2^(k-1)-1), the first quarter of B_{k+1}, is by definition formed from the subsequence a(0) to a(2^(k-1)-1), the first quarter of A_{k+1}, by interchanging its 0s and 1s. As noted above, the subsequence a(2^(k-1)) to a(2^k-1), the second half of A_k, which is by definition also B_{k-1}, is by definition formed from the subsequence a(0) to a(2^(k-1)-1), which is by definition A_{k-1}, by interchanging its 0s and 1s, as well. If two subsequences are formed from the same subsequence by interchanging its 0s and 1s then they must be identical, so the subsequence a(2^(k+1)) to a(2^(k+1)+2^(k-1)-1), the first quarter of B_{k+1}, must be identical to the subsequence a(2^(k-1)) to a(2^k-1), the second half of A_k.

Therefore the subsequence a(0),..., a(2^(k-1)-1), a(2^(k-1)),..., a(2^k-1) is identical to the subsequence a(2^k+2^(k-1)),..., a(2^(k+1)-1), a(2^(k+1)),..., a(2^(k+1)+2^(k-1)-1), QED.

J.-P. Allouche and J. Shallit, The ring of k-regular sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.

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Index entries for "core" sequences

Index entries for characteristic functions

a(2n)=a(n), a(2n+1)=1-a(n), a(0)=0. Also, a(k+2^m)=1-a(k) if 0<=k<2^m.

Let S(0) = 0 and for k >=1, construct S(k) from S(k-1) by mapping 0 -> 01 and 1 -> 10; sequence is S(infinity).

G.f.: (1/(1-x) - product_{k>=0} 1-x^(2^k))/2. - Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 23 2003

a(0)=0, a(n)=(n+a(floor(n/2))) mod 2; also a(0)=0, a(n)=(n-a(floor(n/2))) mod 2 - Benoit Cloitre (benoit7848c(AT)orange.fr), Dec 10 2003

a(n)=-1+sum(k=0, n, binomial(n, k){mod 2}) {mod 3}=-1+A001316(n) {mod 3} - Benoit Cloitre (benoit7848c(AT)orange.fr), May 09 2004

Let b(1)=1 and b(n)=b(ceil(n/2))-b(floor(n/2)) then a(n-1)=(1/2)*(1-b(2n-1)) - Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 26 2005

a(n) = A001969(n) - 2n. - Frank Adams-Watters (FrankTAW(AT)Netscape.net), Aug 28 2006

a(n) = A115384(n) - A115384(n-1) for n>0. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Aug 26 2007

For n>=0, a(A004760(n+1))=1-a(n). [From Vladimir Shevelev (shevelev(AT)bgu.ac.il), Apr 25 2009]

a(A160217(n))=1-a(n). [From Vladimir Shevelev (shevelev(AT)bgu.ac.il), May 05 2009]

The evolution starting at 0 is:

.0

.0, 1

.0, 1, 1, 0

.0, 1, 1, 0, 1, 0, 0, 1

.0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0

.0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1

.......

A_2 = 0 1 1 0, so B_2 = 1 0 0 1 and A_3 = A_2 B_2 = 0 1 1 0 1 0 0 1.

s := proc(k) local i, ans; ans := [ 0, 1 ]; for i from 0 to k do ans := [ op(ans), op(map(n->(n+1) mod 2, ans)) ] od; RETURN(ans); end; t1 := s(6); A010060 := n->t1[n]; # s(k) gives first 2^(k+2) terms.

a := proc(k) b := [0]: for n from 1 to k do b := subs({0=[0, 1], 1=[1, 0]}, b) od: b; end; # a(k), after the removal of the brackets, gives the first 2^k terms. # Example: a(3); gives [[[[0, 1], [1, 0]], [[1, 0], [0, 1]]]]

a:=proc(n) local n2: n2:=convert(n, base, 2): sum(n2[j], j=1..nops(n2)) mod 2; end: seq(a(n), n=0..104); - Emeric Deutsch (deutsch(AT)duke.poly.edu), Mar 19 2005

Table[ If[ OddQ[ Count[ IntegerDigits[n, 2], 1]], 1, 0], {n, 0, 100}];

mt = 0; Do[ mt = ToString[mt] <> ToString[(10^(2^n) - 1)/9 - ToExpression[mt] ], {n, 0, 6} ]; Prepend[ RealDigits[ N[ ToExpression[mt], 2^7] ] [ [1] ], 0]

Mod[ Count[ #, 1 ]& /@Table[ IntegerDigits[ i, 2 ], {i, 0, 2^7 - 1} ], 2 ] (from Harlan J. Brothers, Feb 05 2005)

Nest[ Flatten[ # /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 7] (* Robert G. Wilson v Sep 26 2006)

(Haskell) a = 0: interleave (complement a) (tail a) where {complement = map (1 - ); interleave (x:xs) ys = x: interleave ys xs} (from Doug McIlroy (doug(AT)cs.dartmouth.edu), Jun 29 2003)

(PARI) a(n)=if(n<1, 0, sum(k=0, length(binary(n))-1, bittest(n, k))%2)

(PARI) a(n)=if(n<1, 0, subst(Pol(binary(n)), x, 1)%2)

(PARI) { default(realprecision, 6100); x=0.0; m=20080; for (n=1, m-1, x=x+x; x=x+sum(k=0, length(binary(n))-1, bittest(n, k))%2); x=2*x/2^m; for (n=0, 20000, d=floor(x); x=(x-d)*2; write("b010060.txt", n, " ", d)); } [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), Apr 28 2009]

Cf. A001285 (for 1, 2 version), A010059 (1, 0 version), A048707. A010060(n)=A000120(n) mod 2.

Cf. A080813, A080814, A036581, A108694. See also the Thue (or Roth) constant A014578.

Cf. also A001969, A035263, A005187, A115384, A132680, A141803, A104248.

Backward first differences give A029883.

A004128, A053838 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 24 2008]

Cf. A059448.

Sequence in context: A053866 A156595 A143222 this_sequence A118247 A122257 A129950

Adjacent sequences: A010057 A010058 A010059 this_sequence A010061 A010062 A010063

nonn,core,easy,nice

N. J. A. Sloane (njas(AT)research.att.com).

Additional comments from Robert G. Wilson v (rgwv(AT)rgwv.com), Dec 29 2000