A003159 -id:A003159 - 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)

A007913 Squarefree part of n: a(n) is the smallest positive number m such that n/m is a square.

Original entry on oeis.org

1, 2, 3, 1, 5, 6, 7, 2, 1, 10, 11, 3, 13, 14, 15, 1, 17, 2, 19, 5, 21, 22, 23, 6, 1, 26, 3, 7, 29, 30, 31, 2, 33, 34, 35, 1, 37, 38, 39, 10, 41, 42, 43, 11, 5, 46, 47, 3, 1, 2, 51, 13, 53, 6, 55, 14, 57, 58, 59, 15, 61, 62, 7, 1, 65, 66, 67, 17, 69, 70, 71, 2, 73, 74, 3, 19, 77

Offset: 1

R. Muller, Mar 15 1996

Also called core(n). [Not to be confused with the squarefree kernel of n, A007947.]
Sequence read mod 4 gives A065882. - Philippe Deléham, Mar 28 2004
This is an arithmetic function and is undefined if n <= 0.
A note on square roots of numbers: we can write sqrt(n) = b*sqrt(c) where c is squarefree. Then b = A000188(n) is the "inner square root" of n, c = A007913(n), lcm(A007947(b),c) = A007947(n) = "squarefree kernel" of n and bc = A019554(n) = "outer square root" of n. [Corrected by M. F. Hasler, Mar 01 2018]
If n > 1, the quantity f(n) = log(n/core(n))/log(n) satisfies 0 <= f(n) <= 1; f(n) = 0 when n is squarefree and f(n) = 1 when n is a perfect square. One can define n as being "epsilon-almost squarefree" if f(n) < epsilon. - Kurt Foster (drsardonicus(AT)earthlink.net), Jun 28 2008
a(n) is the smallest natural number m such that product of geometric mean of the divisors of n and geometric mean of the divisors of m are integers. Geometric mean of the divisors of number n is real number b(n) = Sqrt(n). a(n) = 1 for infinitely many n. a(n) = 1 for numbers from A000290: a(A000290(n)) = 1. For n = 8; b(8) = sqrt(8), a(n) = 2 because b(2) = sqrt(2); sqrt(8) * sqrt(2) = 4 (integer). - Jaroslav Krizek, Apr 26 2010
Dirichlet convolution of A010052 with the sequence of absolute values of A055615. - R. J. Mathar, Feb 11 2011
Booker, Hiary, & Keating outline a method for bounding (on the GRH) a(n) for large n using L-functions. - Charles R Greathouse IV, Feb 01 2013
According to the formula a(n) = n/A000188(n)^2, the scatterplot exhibits the straight lines y=x, y=x/4, y=x/9, ..., i.e., y=x/k^2 for all k=1,2,3,... - M. F. Hasler, May 08 2014
The Dirichlet inverse of this sequence is A008836(n) * A063659(n). - Álvar Ibeas, Mar 19 2015
a(n) = 1 if n is a square, a(n) = n if n is a product of distinct primes. - Zak Seidov, Jan 30 2016
All solutions of the Diophantine equation n*x=y^2 or, equivalently, G(n,x)=y, with G being the geometric mean, are of the form x=k^2*a(n), y=k*sqrt(n*a(n)), where k is a positive integer. - Stanislav Sykora, Feb 03 2016
If f is a multiplicative function then Sum_{d divides n} f(a(d)) is also multiplicative. For example, A010052(n) = Sum_{d divides n} mu(a(d)) and A046951(n) = Sum_{d divides n} mu(a(d)^2). - Peter Bala, Jan 24 2024

Daniel Forgues, Table of n, a(n) for n = 1..100000 (first 1000 terms from T. D. Noe)
Krassimir T. Atanassov, On Some of Smarandache's Problems.
Krassimir T. Atanassov, On the 22nd, 23rd, and the 24th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5 (1999), No. 2, 80-82.
Andrew Booker, Ghaith Hiary, and Jon Keating, Detecting squarefree numbers, CNTA XII (2012).
Henry Bottomley, Some Smarandache-type multiplicative sequences.
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Vlad Copil and Laurenţiu Panaitopol, Properties of a sequence generated by positive integers, Bulletin mathématique de la Société des Sciences Mathématiques de Roumanie, Nouvelle Série, Vol. 50 (98), No. 2 (2007), pp. 131-137; alternative link.
Florentin Smarandache, Only Problems, Not Solutions!, Xiquan Publ., Phoenix-Chicago, 1993.
Eric Weisstein's World of Mathematics, Squarefree Part.

See A000188, A007947, A008833, A019554, A117811 for related information, specific to n.
See A027746, A027748, A124010 for factorization data for n.
Analogous sequences: A050985, A053165, A055231.
Cf. A002734, A005117 (range of values), A059897, A069891 (partial sums), A090699, A350389.
Related to A006519 via A225546.

a007913 n = product $
            zipWith (^) (a027748_row n) (map (`mod` 2) $ a124010_row n)
-- Reinhard Zumkeller, Jul 06 2012

[ Squarefree(n) : n in [1..256] ]; // N. J. A. Sloane, Dec 23 2006

A007913 := proc(n) local f,a,d; f := ifactors(n)[2] ; a := 1 ; for d in f do if type(op(2,d),'odd') then a := a*op(1,d) ; end if; end do: a; end proc: # R. J. Mathar, Mar 18 2011
# second Maple program:
a:= n-> mul(i[1]^irem(i[2], 2), i=ifactors(n)[2]):
seq(a(n), n=1..100);  # Alois P. Heinz, Jul 20 2015
seq(n / expand(numtheory:-nthpow(n, 2)), n=1..77);  # Peter Luschny, Jul 12 2022

data = Table[Sqrt[n], {n, 1, 100}]; sp = data /. Sqrt[] -> 1; sfp = data/sp /. Sqrt[x] -> x (* Artur Jasinski, Nov 03 2008 *)
Table[Times@@Power@@@({#[[1]],Mod[ #[[2]],2]}&/@FactorInteger[n]),{n,100}] (* Zak Seidov, Apr 08 2009 *)
Table[{p, e} = Transpose[FactorInteger[n]]; Times @@ (p^Mod[e, 2]), {n, 100}] (* T. D. Noe, May 20 2013 *)
Sqrt[#] /. (c_:1)*a_^(b_:0) -> (c*a^b)^2& /@ Range@100 (* Bill Gosper, Jul 18 2015 *)

PARI
```
a(n)=core(n)
```

from sympy import factorint, prod
def A007913(n):
    return prod(p for p, e in factorint(n).items() if e % 2)
# Chai Wah Wu, Feb 03 2015

[squarefree_part(n) for n in (1..77)] # Peter Luschny, Feb 04 2015

Multiplicative with a(p^k) = p^(k mod 2). - David W. Wilson, Aug 01 2001
a(n) modulo 2 = A035263(n); a(A036554(n)) is even; a(A003159(n)) is odd. - Philippe Deléham, Mar 28 2004
Dirichlet g.f.: zeta(2s)*zeta(s-1)/zeta(2s-2). - R. J. Mathar, Feb 11 2011
a(n) = n/( Sum_{k=1..n} floor(k^2/n)-floor((k^2 -1)/n) )^2. - Anthony Browne, Jun 06 2016
a(n) = rad(n)/a(n/rad(n)), where rad = A007947. This recurrence relation together with a(1) = 1 generate the sequence. - Velin Yanev, Sep 19 2017
From Peter Munn, Nov 18 2019: (Start)
a(k*m) = A059897(a(k), a(m)).
a(n) = n / A008833(n).
(End)
a(A225546(n)) = A225546(A006519(n)). - Peter Munn, Jan 04 2020
From Amiram Eldar, Mar 14 2021: (Start)
Theorems proven by Copil and Panaitopol (2007):
Lim sup_{n->oo} a(n+1)-a(n) = oo.
Lim inf_{n->oo} a(n+1)-a(n) = -oo.
Sum_{k=1..n} 1/a(k) ~ c*sqrt(n) + O(log(n)), where c = zeta(3/2)/zeta(3) (A090699). (End)
a(n) = A019554(n)^2/n. - Jianing Song, May 08 2022
Sum_{k=1..n} a(k) ~ c * n^2, where c = Pi^2/30 = 0.328986... . - Amiram Eldar, Oct 25 2022
a(n) = A007947(A350389(n)). - Amiram Eldar, Jan 20 2024

More terms from Michael Somos, Nov 24 2001

Definition reformulated by Daniel Forgues, Mar 24 2009

cp's OEIS Frontend

A091855 Odious numbers (see A000069) in A003159.

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A091785 Evil numbers (see A001969) in A003159.

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A071992 a(n) = 3*n^2 + 2*n - 4 * Sum_{k=1..n} A003159(k).

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A160217 Minimal increasing sequence with a(1)=3 and the property that a(n) and n are both in or both not in A003159.

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A292118 G.f.: 1 + 2*Sum_{k >= 1} (-1)^k*q^A003159(k).

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A187417 Rank transform of A003159; complement of A187418.

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A071994 a(n) = Sum_{k=1..n} A003159(k).

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A071993 a(n) = 3*n - 2* A003159(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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A071992 a(n) = 3n^2 + 2n - 4 * Sum_{k=1..n} A003159(k).

A292118 G.f.: 1 + 2Sum_{k >= 1} (-1)^kq^A003159(k).

A071993 a(n) = 3n - 2 A003159(n).