A193231 -id:A193231 - cp's OEIS frontend

A010060 Thue-Morse sequence: let A_k denote the first 2^k terms; then A_0 = 0 and for k >= 0, A_{k+1} = A_k B_k, where B_k is obtained from A_k by interchanging 0's and 1's.

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, 1, 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, 1, 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

Offset: 0

N. J. A. Sloane

Named after Axel Thue, whose name is pronounced as if it were spelled "Tü" where the ü sound is roughly as in the German word üben. (It is incorrect to say "Too-ee" or "Too-eh".) - N. J. A. Sloane, Jun 12 2018
Also called the Thue-Morse infinite word, or the Morse-Hedlund sequence, or the parity sequence.
Fixed point of the morphism 0 --> 01, 1 --> 10, see example. - Joerg Arndt, Mar 12 2013
The sequence is cubefree (does not contain three consecutive identical blocks) [see Offner for a direct proof] 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, Jan 10 2003
Characteristic function of A000069 (odious numbers). - Ralf Stephan, 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 generalized Thue-Morse sequence. The classical Thue-Morse sequence is the case k=2, m=2, F(t)= 1*t. - Ctibor O. Zizka, Feb 12 2008 (with correction from Daniel Hug, May 19 2017)
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, 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, Jun 15 2008
Generalized Thue-Morse sequences mod n (n>1) = the array shown in A141803. As n -> infinity the sequences -> (1, 2, 3, ...). - Gary W. Adamson, 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. - Gary W. Adamson, 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 0's and 1's. 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 0's and 1's. Interchanging the 0's and 1's 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 0's and 1's. 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 0's and 1's, as well. If two subsequences are formed from the same subsequence by interchanging its 0's and 1's 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.
According to the German chess rules of 1929 a game of chess was drawn if the same sequence of moves was repeated three times consecutively. Euwe, see the references, proved that this rule could lead to infinite games. For his proof he reinvented the Thue-Morse sequence. - Johannes W. Meijer, Feb 04 2010
"Thue-Morse 0->01 & 1->10, at each stage append the previous with its complement. Start with 0, 1, 2, 3 and write them in binary. Next calculate the sum of the digits (mod 2) - that is divide the sum by 2 and use the remainder." Pickover, The Math Book.
Let s_2(n) be the sum of the base-2 digits of n and epsilon(n) = (-1)^s_2(n), the Thue-Morse sequence, then prod(n >= 0, ((2*n+1)/(2*n+2))^epsilon(n) ) = 1/sqrt(2). - Jonathan Vos Post, Jun 06 2012
Dekking shows that the constant obtained by interpreting this sequence as a binary expansion is transcendental; see also "The Ubiquitous Prouhet-Thue-Morse Sequence". - Charles R Greathouse IV, Jul 23 2013
Drmota, Mauduit, and Rivat proved that the subsequence a(n^2) is normal--see A228039. - Jonathan Sondow, Sep 03 2013
Although the probability of a 0 or 1 is equal, guesses predicated on the latest bit seen produce a correct match 2 out of 3 times. - Bill McEachen, Mar 13 2015
From a(0) to a(2n+1), there are n+1 terms equal to 0 and n+1 terms equal to 1 (see Hassan Tarfaoui link, Concours Général 1990). - Bernard Schott, Jan 21 2022

			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.
From _Joerg Arndt_, Mar 12 2013: (Start)
The first steps of the iterated substitution are
Start: 0
Rules:
  0 --> 01
  1 --> 10
-------------
0:   (#=1)
  0
1:   (#=2)
  01
2:   (#=4)
  0110
3:   (#=8)
  01101001
4:   (#=16)
  0110100110010110
5:   (#=32)
  01101001100101101001011001101001
6:   (#=64)
  0110100110010110100101100110100110010110011010010110100110010110
(End)
From _Omar E. Pol_, Oct 28 2013: (Start)
Written as an irregular triangle in which row lengths is A011782, the sequence begins:
  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;
  1,0,0,1,0,1,1,0,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,0;
It appears that: row j lists the first A011782(j) terms of A010059, with j >= 0; row sums give A166444 which is also 0 together with A011782; right border gives A000035.
(End)

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Eric Weisstein's World of Mathematics, Parity
Joost Winter, Marcello M. Bonsangue, and Jan J. M. M. Rutten, Context-free coalgebras, Journal of Computer and System Sciences, 81.5 (2015): 911-939.
Hans Zantema, Complexity of Automatic Sequences, International Conference on Language and Automata Theory and Applications (LATA 2020): Language and Automata Theory and Applications, 260-271.
Index entries for sequences that are fixed points of mappings
Index entries for "core" sequences
Index entries for sequences related to binary expansion of n
Index entries for characteristic functions
Index to sequences related to Olympiads and other Mathematical competitions.

Cf. A001285 (for 1, 2 version), A010059 (for 1, 0 version), A106400 (for +1, -1 version), A048707. A010060(n)=A000120(n) mod 2.
Cf. A007413, A080813, A080814, A036581, A108694. See also the Thue (or Roth) constant A014578, also A014571.
Cf. also A001969, A035263, A005187, A115384, A132680, A141803, A104248, A193231.
Run lengths give A026465. Backward first differences give A029883.
Cf. A004128, A053838, A059448, A171900, A161916, A214212, A005942 (subword complexity), A010693 (Abelian complexity), A225186 (squares), A228039 (a(n^2)), A282317.
Sequences mentioned in the Allouche et al. "Taxonomy" paper, listed by example number: 1: A003849, 2: A010060, 3: A010056, 4: A020985 and A020987, 5: A191818, 6: A316340 and A273129, 18: A316341, 19: A030302, 20: A063438, 21: A316342, 22: A316343, 23: A003849 minus its first term, 24: A316344, 25: A316345 and A316824, 26: A020985 and A020987, 27: A316825, 28: A159689, 29: A049320, 30: A003849, 31: A316826, 32: A316827, 33: A316828, 34: A316344, 35: A043529, 36: A316829, 37: A010060.

a010060 n = a010060_list !! n
a010060_list =
   0 : interleave (complement a010060_list) (tail a010060_list)
   where complement = map (1 - )
         interleave (x:xs) ys = x : interleave ys xs
-- Doug McIlroy (doug(AT)cs.dartmouth.edu), Jun 29 2003
-- Edited by Reinhard Zumkeller, Oct 03 2012

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]]]]
A010060:=proc(n)
    add(i,i=convert(n, base, 2)) mod 2 ;
end proc:
seq(A010060(n),n=0..104); # Emeric Deutsch, Mar 19 2005
map(`-`,convert(StringTools[ThueMorse](1000),bytes),48); # Robert Israel, Sep 22 2014

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 ] (* Harlan J. Brothers, Feb 05 2005 *)
Nest[ Flatten[ # /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 7] (* Robert G. Wilson v Sep 26 2006 *)
a[n_] := If[n == 0, 0, If[Mod[n, 2] == 0, a[n/2], 1 - a[(n - 1)/2]]] (* Ben Branman, Oct 22 2010 *)
a[n_] := Mod[Length[FixedPointList[BitAnd[#, # - 1] &, n]], 2] (* Jan Mangaldan, Jul 23 2015 *)
Table[2/3 (1 - Cos[Pi/3 (n - Sum[(-1)^Binomial[n, k], {k, 1, n}])]), {n, 0, 100}] (* or, for version 10.2 or higher *) Table[ThueMorse[n], {n, 0, 100}] (* Vladimir Reshetnikov, May 06 2016 *)
ThueMorse[Range[0, 100]] (* The program uses the ThueMorse function from Mathematica version 11 *) (* Harvey P. Dale, Aug 11 2016 *)
Nest[Join[#, 1 - #] &, {0}, 7] (* Paolo Xausa, Oct 25 2024 *)

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

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

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)); \\ Harry J. Smith, Apr 28 2009

a(n)=hammingweight(n)%2 \\ Charles R Greathouse IV, Mar 22 2013

A010060_list = [0]
for _ in range(14):
    A010060_list += [1-d for d in A010060_list] # Chai Wah Wu, Mar 04 2016

def A010060(n): return n.bit_count()&1 # Chai Wah Wu, Mar 01 2023

maxrow <- 8 # by choice
b01 <- 1
for(m in 0:maxrow) for(k in 0:(2^m-1)){
b01[2^(m+1)+    k] <-   b01[2^m+k]
b01[2^(m+1)+2^m+k] <- 1-b01[2^m+k]
}
(b01 <- c(0,b01))
# Yosu Yurramendi, Apr 10 2017

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.
If n = Sum b_i*2^i is the binary expansion of n then a(n) = Sum b_i (mod 2).
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, 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, Dec 10 2003
a(n) = -1 + (Sum_{k=0..n} binomial(n,k) mod 2) mod 3 = -1 + A001316(n) mod 3. - Benoit Cloitre, May 09 2004
Let b(1) = 1 and b(n) = b(ceiling(n/2)) - b(floor(n/2)) then a(n-1) = (1/2)*(1 - b(2n-1)). - Benoit Cloitre, Apr 26 2005
a(n) = 1 - A010059(n) = A001285(n) - 1. - Ralf Stephan, Jun 20 2003
a(n) = A001969(n) - 2n. - Franklin T. Adams-Watters, Aug 28 2006
a(n) = A115384(n) - A115384(n-1) for n > 0. - Reinhard Zumkeller, Aug 26 2007
For n >= 0, a(A004760(n+1)) = 1 - a(n). - Vladimir Shevelev, Apr 25 2009
a(A160217(n)) = 1 - a(n). - Vladimir Shevelev, May 05 2009
a(n) == A000069(n) (mod 2). - Robert G. Wilson v, Jan 18 2012
a(n) = A000035(A000120(n)). - Omar E. Pol, Oct 26 2013
a(n) = A000035(A193231(n)). - Antti Karttunen, Dec 27 2013
a(n) + A181155(n-1) = 2n for n >= 1. - Clark Kimberling, Oct 06 2014
G.f. A(x) satisfies: A(x) = x / (1 - x^2) + (1 - x) * A(x^2). - Ilya Gutkovskiy, Jul 29 2021
From Bernard Schott, Jan 21 2022: (Start)
a(n) = a(n*2^k) for k >= 0.
a((2^m-1)^2) = (1-(-1)^m)/2 (see Hassan Tarfaoui link, Concours Général 1990). (End)

A001317 Sierpiński's triangle (Pascal's triangle mod 2) converted to decimal.

Original entry on oeis.org

1, 3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535, 65537, 196611, 327685, 983055, 1114129, 3342387, 5570645, 16711935, 16843009, 50529027, 84215045, 252645135, 286331153, 858993459, 1431655765, 4294967295, 4294967297, 12884901891, 21474836485, 64424509455, 73014444049, 219043332147, 365072220245, 1095216660735, 1103806595329, 3311419785987

Offset: 0

N. J. A. Sloane

The members are all palindromic in binary, i.e., a subset of A006995. - Ralf Stephan, Sep 28 2004
J. H. Conway writes (in Math Forum): at least the first 31 numbers give odd-sided constructible polygons. See also A047999. - M. Dauchez (mdzzdm(AT)yahoo.fr), Sep 19 2005 [This observation was also made in 1982 by N. L. White (see letter). - N. J. A. Sloane, Jun 15 2015]
Decimal number generated by the binary bits of the n-th generation of the Rule 60 elementary cellular automaton. Thus: 1; 0, 1, 1; 0, 0, 1, 0, 1; 0, 0, 0, 1, 1, 1, 1; 0, 0, 0, 0, 1, 0, 0, 0, 1; ... . - Eric W. Weisstein, Apr 08 2006
Limit_{n->oo} log(a(n))/n = log(2). - Bret Mulvey, May 17 2008
Equals row sums of triangle A166548; e.g., 17 = (2 + 4 + 6 + 4 + 1). - Gary W. Adamson, Oct 16 2009
Equals row sums of triangle A166555. - Gary W. Adamson, Oct 17 2009
For n >= 1, all terms are in A001969. - Vladimir Shevelev, Oct 25 2010
Let n,m >= 0 be such that no carries occur when adding them. Then a(n+m) = a(n)*a(m). - Vladimir Shevelev, Nov 28 2010
Let phi_a(n) be the number of a(k) <= a(n) and respectively prime to a(n) (i.e., totient function over {a(n)}). Then, for n >= 1, phi_a(n) = 2^v(n), where v(n) is the number of 0's in the binary representation of n. - Vladimir Shevelev, Nov 29 2010
Trisection of this sequence gives rows of A008287 mod 2 converted to decimal. See also A177897, A177960. - Vladimir Shevelev, Jan 02 2011
Converting the rows of the powers of the k-nomial (k = 2^e where e >= 1) term-wise to binary and reading the concatenation as binary number gives every (k-1)st term of this sequence. Similarly with powers p^k of any prime. It might be interesting to study how this fails for powers of composites. - Joerg Arndt, Jan 07 2011
This sequence appears in Pascal's triangle mod 2 in another way, too. If we write it as
  1111111...
  10101010...
  11001100...
  10001000...
we get (taking the period part in each row):
  .(1) (base 2) = 1
  .(10) = 2/3
  .(1100) = 12/15 = 4/5
  .(1000) = 8/15
The k-th row, treated as a binary fraction, seems to be equal to 2^k / a(k). - Katarzyna Matylla, Mar 12 2011
From Daniel Forgues, Jun 16-18 2011: (Start)
Since there are 5 known Fermat primes, there are 32 products of distinct Fermat primes (thus there are 31 constructible odd-sided polygons, since a polygon has at least 3 sides). a(0)=1 (empty product) and a(1) to a(31) are those 31 non-products of distinct Fermat primes.
It can be proved by induction that all terms of this sequence are products of distinct Fermat numbers (A000215):
a(0)=1 (empty product) are products of distinct Fermat numbers in { };
a(2^n+k) = a(k) * (2^(2^n)+1) = a(k) * F_n, n >= 0, 0 <= k <= 2^n - 1.
Thus for n >= 1, 0 <= k <= 2^n - 1, and
  a(k) = Product_{i=0..n-1} F_i^(alpha_i), alpha_i in {0, 1},
this implies
  a(2^n+k) = Product_{i=0..n-1} F_i^(alpha_i) * F_n, alpha_i in {0, 1}.
(Cf. OEIS Wiki links below.)  (End)
The bits in the binary expansion of a(n) give the coefficients of the n-th power of polynomial (X+1) in ring GF(2)[X]. E.g., 3 ("11" in binary) stands for (X+1)^1, 5 ("101" in binary) stands for (X+1)^2 = (X^2 + 1), and so on. - Antti Karttunen, Feb 10 2016

			Given a(5)=51, a(6)=85 since a(5) XOR 2*a(5) = 51 XOR 102 = 85.
From _Daniel Forgues_, Jun 18 2011: (Start)
  a(0) = 1 (empty product);
  a(1) = 3 = 1 * F_0 = a(2^0+0) = a(0) * F_0;
  a(2) = 5 = 1 * F_1 = a(2^1+0) = a(0) * F_1;
  a(3) = 15 = 3 * 5 = F_0 * F_1 = a(2^1+1) = a(1) * F_1;
  a(4) = 17 = 1 * F_2 = a(2^2+0) = a(0) * F_2;
  a(5) = 51 = 3 * 17 = F_0 * F_2 = a(2^2+1) = a(1) * F_2;
  a(6) = 85 = 5 * 17 = F_1 * F_2 = a(2^2+2) = a(2) * F_2;
  a(7) = 255 = 3 * 5 * 17 = F_0 * F_1 * F_2 = a(2^2+3) = a(3) * F_2;
  ... (End)

Jean-Paul Allouche and Jeffrey Shallit, Automatic sequences, Cambridge University Press, 2003, p. 113.
Henry Wadsworth Gould, Exponential Binomial Coefficient Series, Tech. Rep. 4, Math. Dept., West Virginia Univ., Morgantown, WV, Sept. 1961.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 136-137.

Amiram Eldar, Table of n, a(n) for n = 0..3321 (terms 0..300 from T. D. Noe, terms 301..1023 from Tilman Piesk)
Gary W. Adamson, Letter to N. J. A. Sloane, Apr 29 1994, and MS "Algorithm for the Chinese Remainder Theorem".
Gary W. Adamson and N. J. A. Sloane, Correspondence, May 1994, including Adamson's MSS "Algorithm for Generating nth Row of Pascal's Triangle, mod 2, from n", and "The Tower of Hanoi Wheel".
Cristian Cobeli and Alexandru Zaharescu, Promenade around Pascal Triangle-Number Motives, Bull. Math. Soc. Sci. Math. Roumanie, Tome 56(104), No. 1 (2013), pp. 73-98; alternative link.
Richard K. Guy, The Second Strong Law of Small Numbers, Math. Mag, Vol. 63, No. 1 (1990), pp. 3-20. [Annotated scanned copy]
Richard K. Guy and N. J. A. Sloane, Correspondence, 1988.
Denton Hewgill, A relationship between Pascal's triangle and Fermat numbers, Fib. Quart., Vol. 15, No. 2 (1977), pp. 183-184.
A. J. Macfarlane, Generating functions for integer sequences defined by the evolution of cellular automata with even rule numbers, Fig. 4.
Dr. Math, Regular polygon formulas.
P. Mathonet, M. Rigo, M. Stipulanti and N. Zénaïdi, On digital sequences associated with Pascal's triangle, arXiv:2201.06636 [math.NT], 2022.
OEIS Wiki, Constructible odd-sided polygons and Sierpinski's triangle.
Vladimir Shevelev, On Stephan's conjectures concerning Pascal triangle modulo 2, J. Alg. Number Theory, Vol. 7, No. 1 (2012) pp. 11-29, preprint, arXiv:1011.6083 [math.NT], 2010-2012.
John Frederick Sweeney, Clifford Clock and the Moolakaprithi Cube, 2014. See p. 26. - _N. J. A. Sloane_, Mar 20 2014
Eric Weisstein's World of Mathematics, Rule 60 and Rule 102.
Neil L. White, Letter to N. J. A. Sloane, Apr 15 1982.
R. G. Wilson, V, Letter to N. J. A. Sloane, Aug. 1993.
Index entries for sequences related to cellular automata
Index entries for sequences operating on polynomials in ring GF(2)[X]

Cf. A000079, A000215 (Fermat numbers), A047999, A048720, A054432, A173019, A177882, A177897, A177960, A193231, A230116, A247282, A249184, A268391.
Cf. A038183 (odd bisection, 1D Cellular Automata Rule 90).
Iterates of A048724 (starting from 1).
Row 3 of A048723.
Positions of records in A268389.
Positions of ones in A268669 and A268384 (characteristic function).
Not the same as A045544 nor as A053576.
Cf. A045544.

a001317 = foldr (\u v-> 2*v + u) 0 . map toInteger . a047999_row
-- Reinhard Zumkeller, Nov 24 2012
(Scheme, with memoization-macro definec, two variants)
(definec (A001317 n) (if (zero? n) 1 (A048724 (A001317 (- n 1)))))
(definec (A001317 n) (if (zero? n) 1 (A048720bi 3 (A001317 (- n 1))))) ;; Where A048720bi implements the dyadic function given in A048720.
;; Antti Karttunen, Feb 10 2016

[&+[(Binomial(n, i) mod 2)*2^i: i in [0..n]]: n in [0..41]]; // Vincenzo Librandi, Feb 12 2016

A001317 := proc(n) local k; add((binomial(n,k) mod 2)*2^k, k=0..n); end;

a[n_] := Nest[ BitXor[#, BitShiftLeft[#, 1]] &, 1, n]; Array[a, 42, 0] (* Joel Madigan (dochoncho(AT)gmail.com), Dec 03 2007 *)
NestList[BitXor[#,2#]&,1,50] (* Harvey P. Dale, Aug 02 2021 *)

PARI
```
a(n)=sum(i=0,n,(binomial(n,i)%2)*2^i)
```

a=1; for(n=0, 66, print1(a,", "); a=bitxor(a,a<<1) ); \\ Joerg Arndt, Mar 27 2013

A001317(n,a=1)={for(k=1,n,a=bitxor(a,a<<1));a} \\ M. F. Hasler, Jun 06 2016

a(n) = subst(lift(Mod(1+'x,2)^n), 'x, 2); \\ Gheorghe Coserea, Nov 09 2017

from sympy import binomial
def a(n): return sum([(binomial(n, i)%2)*2**i for i in range(n + 1)]) # Indranil Ghosh, Apr 11 2017

def A001317(n): return int(''.join(str(int(not(~n&k))) for k in range(n+1)),2) # Chai Wah Wu, Feb 04 2022

a(n+1) = a(n) XOR 2*a(n), where XOR is binary exclusive OR operator. - Paul D. Hanna, Apr 27 2003
a(n) = Product_{e(j, n) = 1} (2^(2^j) + 1), where e(j, n) is the j-th least significant digit in the binary representation of n (Roberts: see Allouche & Shallit). - Benoit Cloitre, Jun 08 2004
a(2*n+1) = 3*a(2*n). Proof: Since a(n) = Product_{k in K} (1 + 2^(2^k)), where K is the set of integers such that n = Sum_{k in K} 2^k, clearly K(2*n+1) = K(2*n) union {0}, hence a(2*n+1) = (1+2^(2^0))*a(2*n) = 3*a(2*n). - Emmanuel Ferrand and Ralf Stephan, Sep 28 2004
a(32*n) = 3 ^ (32 * n * log(2) / log(3)) + 1. - Bret Mulvey, May 17 2008
For n >= 1, A000120(a(n)) = 2^A000120(n). - Vladimir Shevelev, Oct 25 2010
a(2^n) = A000215(n); a(2^n-1) = a(2^n)-2; for n >= 1, m >= 0,
a(2^(n-1)-1)*a(2^n*m + 2^(n-1)) = 3*a(2^(n-1))*a(2^n*m + 2^(n-1)-2). - Vladimir Shevelev, Nov 28 2010
Sum_{k>=0} 1/a(k) = Product_{n>=0} (1 + 1/F_n), where F_n=A000215(n);
Sum_{k>=0} (-1)^(m(k))/a(k) = 1/2, where {m(n)} is Thue-Morse sequence (A010060).
If F_n is defined by F_n(z) = z^(2^n) + 1 and a(n) by (1/2)*Sum_{i>=0}(1-(-1)^{binomial(n,i)})*z^i, then, for z > 1, the latter two identities hold as well with the replacement 1/2 in the right hand side of the 2nd one by 1-1/z. - Vladimir Shevelev, Nov 29 2010
G.f.: Product_{k>=0} ( 1 + z^(2^k) + (2*z)^(2^k) ). - conjectured by Shamil Shakirov, proved by Vladimir Shevelev
a(n) = A000225(n+1) - A219843(n). - Reinhard Zumkeller, Nov 30 2012
From Antti Karttunen, Feb 10 2016: (Start)
a(0) = 1, and for n > 1, a(n) = A048720(3, a(n-1)) = A048724(a(n-1)).
a(n) = A048723(3,n).
a(n) = A193231(A000079(n)).
For all n >= 0: A268389(a(n)) = n.
(End)

A057889 Bijective bit-reverse of n: keep the trailing zeros in the binary expansion of n fixed, but reverse all the digits up to that point.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 12, 11, 14, 15, 16, 17, 18, 25, 20, 21, 26, 29, 24, 19, 22, 27, 28, 23, 30, 31, 32, 33, 34, 49, 36, 41, 50, 57, 40, 37, 42, 53, 52, 45, 58, 61, 48, 35, 38, 51, 44, 43, 54, 59, 56, 39, 46, 55, 60, 47, 62, 63, 64, 65, 66, 97, 68, 81, 98, 113

Offset: 0

Marc LeBrun, Sep 25 2000

The original name was "Bit-reverse of n, including as many leading as trailing zeros." - Antti Karttunen, Dec 25 2024
A permutation of integers consisting only of fixed points and pairs. a(n)=n when n is a binary palindrome (including as many leading as trailing zeros), otherwise a(n)=A003010(n) (i.e. n has no axis of symmetry). A057890 gives the palindromes (fixed points, akin to A006995) while A057891 gives the "antidromes" (pairs). See also A280505.
This is multiplicative in domain GF(2)[X], i.e. with carryless binary arithmetic. A193231 is another such permutation of natural numbers. - Antti Karttunen, Dec 25 2024

			a(6)=6 because 0110 is a palindrome, but a(11)=13 because 1011 reverses into 1101.

N. J. A. Sloane, Table of n, a(n) for n = 0..16384, May 30 2016 [First 8192 terms from _Ivan Neretin_, Jul 09 2015]
Index entries for sequences related to binary expansion of n
Index entries for sequences related to polynomials in ring GF(2)[X]
Index entries for sequences that are permutations of the natural numbers

Cf. A030101, A000265, A006519, A006995, A057890, A057891, A280505, A280508, A331166 [= min(n,a(n))], A366378 [k for which a(k) = k (mod 3)], A369044 [= A014963(a(n))].
Similar permutations for other bases: A263273 (base-3), A264994 (base-4), A264995 (base-5), A264979  (base-9).
Other related (binary) permutations: A056539, A193231.
Compositions of this permutation with other binary (or other base-related) permutations: A264965, A264966, A265329, A265369, A379471, A379472.
Compositions with permutations involving prime factorization: A245450, A245453, A266402, A266404, A293448, A366275, A366276.
Other derived permutations: A246200 [= a(3*n)/3], A266351, A302027, A302028, A345201, A356331, A356332, A356759, A366389.
See also A235027 (which is not a permutation).

Table[FromDigits[Reverse[IntegerDigits[n, 2]], 2]*2^IntegerExponent[n, 2], {n, 71}] (* Ivan Neretin, Jul 09 2015 *)

A030101(n) = if(n<1,0,subst(Polrev(binary(n)),x,2));
A057889(n) = if(!n,n,A030101(n/(2^valuation(n,2))) * (2^valuation(n, 2))); \\ Antti Karttunen, Dec 25 2024

def a(n):
    x = bin(n)[2:]
    y = x[::-1]
    return int(str(int(y))+(len(x) - len(str(int(y))))*'0', 2)
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 11 2017

def A057889(n): return int(bin(n>>(m:=(~n&n-1).bit_length()))[-1:1:-1],2)<Chai Wah Wu, Dec 25 2024

a(n) = A030101(A000265(n)) * A006519(n), with a(0)=0.

Clarified the name with May 30 2016 comment from N. J. A. Sloane, and moved the old name to the comments - Antti Karttunen, Dec 25 2024

A091242 Reducible polynomials over GF(2), coded in binary.

Original entry on oeis.org

4, 5, 6, 8, 9, 10, 12, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 26, 27, 28, 29, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 43, 44, 45, 46, 48, 49, 50, 51, 52, 53, 54, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 71, 72, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 88

Offset: 1

Antti Karttunen, Jan 03 2004

nonn

"Coded in binary" means that a polynomial a(n)*X^n+...+a(0)*X^0 over GF(2) is represented by the binary number a(n)*2^n+...+a(0)*2^0 in Z (where a(k)=0 or 1). - M. F. Hasler, Aug 18 2014
The reducible polynomials in GF(2)[X] are the analog to the composite numbers A002808 in the integers.
It follows that the sequence is closed under application of A048720(.,.), which effects multiplication of the coded polynomials. It is also closed under application of blue code, A193231. The majority of the terms are coded multiples of X^1 (represented by 2) and/or X^1+1 (represented by 3): see A005843 and A001969 respectively. A246157 lists the other terms. - Peter Munn, Apr 20 2021

			For example, 5 = 101 in binary encodes the polynomial x^2+1 which is factored as (x+1)^2 in the polynomial ring GF(2)[X].

Robert Israel, Table of n, a(n) for n = 1..10000
Antti Karttunen, Scheme-program for computing this sequence.
Index entries for sequences operating on GF(2)[X]-polynomials

Inverse: A091246. Almost complement of A014580. Union of A091209 & A091212. First differences: A091243. Characteristic function: A091247. In binary format: A091254.
Number of degree-n reducible polynomials: A058766.
Subsequences: A001969\{0,3}, A005843\{0,2}, A246156, A246157, A246158, A316970.
Cf. A002808, A048720, A193231.

filter:= proc(n) local L;
  L:= convert(n,base,2);
  not Irreduc(add(L[i]*x^(i-1),i=1..nops(L))) mod 2
end proc:
select(filter, [$2..200]); # Robert Israel, Aug 30 2018

okQ[n_] := Module[{x, id = IntegerDigits[n, 2] // Reverse}, !IrreduciblePolynomialQ[id.x^Range[0, Length[id]-1], Modulus -> 2]];
Select[Range[2, 200], okQ] (* Jean-François Alcover, Jan 04 2022 *)

Edited by M. F. Hasler, Aug 18 2014

A099884 XOR difference triangle of the powers of 2, read by rows; Square array A(row,col): A(0,col) = 2^col, A(row,col) = A048724(A(row-1, col)) for row > 0, read by descending antidiagonals.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 8, 12, 10, 15, 16, 24, 20, 30, 17, 32, 48, 40, 60, 34, 51, 64, 96, 80, 120, 68, 102, 85, 128, 192, 160, 240, 136, 204, 170, 255, 256, 384, 320, 480, 272, 408, 340, 510, 257, 512, 768, 640, 960, 544, 816, 680, 1020, 514, 771, 1024, 1536, 1280, 1920

Offset: 0

Paul D. Hanna, Oct 28 2004

Define an "XOR difference triangle" for a sequence A by the following process. Start with A in the leftmost column. Generate the next column by performing the XOR operation between adjacent terms of the prior column. Repeat this process to generate the XOR difference triangle for A. Further, we define the "XOR BINOMIAL transform" of A as the main diagonal in the XOR difference triangle for A. The XOR BINOMIAL transform is its self-inverse. Let a sequence B be the XOR BINOMIAL transform of A, then we may express B by: B(n) = SumXOR_{k=0..n} A047999(n,k)*A(k), which is equivalent to: B(n) = (C(n,0)mod 2)*A(0) XOR (C(n,1)mod 2)*A(1) XOR (C(n,2)mod 2)*A(2) XOR ... XOR (X(n,n)mod 2)*A(n), where the coefficients are C(n,k)(mod 2) = A047999(n,k).

This sequence is a rearrangement of the numbers which are 2^k times distinct Fermat numbers (numbers of the form 2^(2^m) + 1). This matches the sizes of polygons constructible with compass and straightedge (A003401) up to 2^32+1, which is the first nonprime Fermat number. - Franklin T. Adams-Watters, Jun 16 2006

			The main diagonal equals A001317 (Pascal's triangle mod 2 in decimal):
{1,3,5,15,17,51,85,255,257,771,1285,3855,...}, and defines the XOR BINOMIAL transform of the powers of 2.
Rows begin:
  1;
  2, 3;
  4, 6, 5;
  8, 12, 10, 15;
  16, 24, 20, 30, 17;
  32, 48, 40, 60, 34, 51;
  64, 96, 80, 120, 68, 102, 85;
  128, 192, 160, 240, 136, 204, 170, 255;
  256, 384, 320, 480, 272, 408, 340, 510, 257;
  512, 768, 640, 960, 544, 816, 680, 1020, 514, 771;
  1024, 1536, 1280, 1920, 1088, 1632, 1360, 2040, 1028, 1542, 1285;
  2048, 3072, 2560, 3840, 2176, 3264, 2720, 4080, 2056, 3084, 2570, 3855;
  ...
From _Antti Karttunen_, Sep 19 2016: (Start)
Viewed as a square array, the top left corner looks like this:
     1,    2,     4,     8,    16,     32,     64,    128
     3,    6,    12,    24,    48,     96,    192,    384
     5,   10,    20,    40,    80,    160,    320,    640
    15,   30,    60,   120,   240,    480,    960,   1920
    17,   34,    68,   136,   272,    544,   1088,   2176
    51,  102,   204,   408,   816,   1632,   3264,   6528
    85,  170,   340,   680,  1360,   2720,   5440,  10880
   255,  510,  1020,  2040,  4080,   8160,  16320,  32640
   257,  514,  1028,  2056,  4112,   8224,  16448,  32896
   771, 1542,  3084,  6168, 12336,  24672,  49344,  98688
  1285, 2570,  5140, 10280, 20560,  41120,  82240, 164480
  3855, 7710, 15420, 30840, 61680, 123360, 246720, 493440
  4369, 8738, 17476, 34952, 69904, 139808, 279616, 559232
  ...
(End)
The square array shown above can be viewed as a subtable of a multiplication table with particular relevance to the carryless multiplication defined by A048720, as the first column gives the A048720 powers of 3 (and the first row gives powers of 2, which are the same as in standard arithmetic). - _Peter Munn_, Jan 13 2020

Paul D. Hanna, First 45 Rows of Triangle, in flattened form.
SeqFan-mailing list, Discussion of about related array A255483
Index entries for sequences related to polynomials in ring GF(2)[X]

Essentially GF(2)[X] analog of table A036561. - Antti Karttunen, Jan 18 2020
Cf. A047999, A158875 (row sums).
Cf. A000215, A003401, A048675, A048720, A048724, A066117, A248663, A255483, A276586.
Cf. A000079 (first column of triangular table, the topmost row of square array).
Cf. A001317 (the rightmost diagonal of triangular table, the leftmost column of square array).
Cf. A099885, A117998 (central diagonals).
Cf. A276618 (transpose), A091202, A193231.

a[n_]:= Sum[Mod[Binomial[n, i], 2]*2^i, {i, 0, n}]; T[n_, k_]:=2^(n - k)a[k]; Table[T[n, k], {n, 0, 20}, {k, 0, n}] // Flatten (* Indranil Ghosh, Apr 11 2017 *)

{T(n,k)=local(B);B=0;for(i=0,k,B=bitxor(B,binomial(k,i)%2*2^(n-i)));B}
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))

from sympy import binomial
def a(n):
    return sum((binomial(n, i)%2)*2**i for i in range(n + 1))
def T(n, k): return 2**(n - k)*a(k)
for n in range(21): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 11 2017

(define (A099884 n) (A099884bi (A002262 n) (A025581 n)))
;; Then use either this recurrence:
(define (A099884bi row col) (if (zero? row) (A000079 col) (A048724 (A099884bi (- row 1) col))))
;; or this one:
(define (A099884bi row col) (if (zero? col) (A001317 row) (* 2 (A099884bi row (- col 1)))))
;; Antti Karttunen, Sep 19 2016

T(n, k) = 2^(n-k)*A001317(k). T(n, n) = A001317(n) = SumXOR_{k=0..n} A047999(n, k)*2^k, where SumXOR is the analog of summation under the binary XOR operation.
From Antti Karttunen, Sep 19 2016: (Start)
When viewed as a square array A(row,col), with row >= 0, col >= 0, the following recurrences and formulas are valid:
A(0,col) = A000079(col), for row > 0, A(row,col) = A048724(A(row-1, col)).
A(row,0) = A001317(row), for col > 0, A(row,col) = 2*A(row,col-1).
A(row,col) = A248663(A066117(row+1,col+1)) = A048675(A255483(row,col+1)).
(End)
With the definitions from Antti Karttunen above, A(row+1, col) = A048720(3, A(row, col)). - Peter Munn, Jan 13 2020
A(n,k) = A193231(A(k,n)) = A091202(A036561(n,k)). - Antti Karttunen, Jan 18 2020

Square array interpretation added as a second, alternative description by Antti Karttunen, Sep 19 2016

cp's OEIS Frontend

A379472 Composition of Blue code and bijective bit-reverse, in this order: a(n) = A057889(A193231(n)).

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A268675 Self-inverse permutation of natural numbers: a(1) = 1; a(n) = A000079(A193231(A007814(n))) * A250469(a(A268674(n))).

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A365810 Squareferee numbers ordered factorization-wise by Blue code: a(n) = A019565(A193231(n)).

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A010060 Thue-Morse sequence: let A_k denote the first 2^k terms; then A_0 = 0 and for k >= 0, A_{k+1} = A_k B_k, where B_k is obtained from A_k by interchanging 0's and 1's.

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A001317 Sierpiński's triangle (Pascal's triangle mod 2) converted to decimal.

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A057889 Bijective bit-reverse of n: keep the trailing zeros in the binary expansion of n fixed, but reverse all the digits up to that point.

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