A010693 -id:A010693 - cp's OEIS frontend

A262392 a(n) = A007504(n) + A010693(n).

2, 5, 7, 13, 19, 31, 43, 61, 79, 103, 131, 163, 199, 241, 283, 331, 383, 443, 503, 571, 641, 715, 793, 877, 965, 1063, 1163, 1267, 1373, 1483, 1595, 1723, 1853, 1991, 2129, 2279, 2429, 2587, 2749, 2917, 3089, 3269, 3449, 3641, 3833, 4031, 4229

Offset: 0

Altug Alkan, Sep 21 2015

Sequence is interesting because of the fact that a(n) is a prime number for n = 0..20.

Main inspiration of sequence was indices of prime numbers in A036439 and A227547.

			a(0) = A007504(0) + A010693(0) = 0 + 2 = 2.
a(1) = A007504(1) + A010693(1) = 2 + 3 = 5.
a(2) = A007504(2) + A010693(2) = 5 + 2 = 7.
a(3) = A007504(3) + A010693(3) = 10 + 3 = 13.
a(4) = A007504(4) + A010693(4) = 17 + 2 = 19.
a(5) = A007504(5) + A010693(5) = 28 + 3 = 31.

Cf. A036439, A227547, A007504, A010693, A071148.

s = Accumulate@ Prime@ Range@ 1200; {2}~Join~Table[s[[n]] + If[OddQ@ n, 3, 2], {n, 46}] (* Michael De Vlieger, Sep 21 2015 *)

a(n) = sum(k=1, n, prime(k)) + (n%2) + 2;
vector(50, n, a(n-1))

a(n) = Sum_{k=1..n} prime(k) + n mod 2 + 2 for n>0, a(0)=2 (from Pari code).

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.

Original entry on oeis.org

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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C. D. Offner, Repetitions of Words and the Thue-Morse sequence. Preprint, no date.
Matt Parker, The Fairest Sharing Sequence Ever, YouTube video, Nov 27 2015
A. Parreau, M. Rigo, E. Rowland and E. Vandomme, A new approach to the 2-regularity of the l-abelian complexity of 2-automatic sequences, arXiv preprint arXiv:1405.3532 [cs.FL], 2014.
C. A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Zentralblatt review
Saúl Pilatowsky-Cameo, Soonwon Choi, and Wen Wei Ho, Critically slow Hilbert-space ergodicity in quantum morphic drives, arXiv:2502.06936 [quant-ph], 2025. See pp. 6, 15.
E. Prouhet, Mémoire sur quelques relations entre les puissances des nombres, Comptes Rendus Acad. des Sciences, 33 (No. 8, 1851), p. 225. [Said to implicitly mention this sequence]
Michel Rigo, Relations on words, arXiv preprint arXiv:1602.03364 [cs.FL], 2016.
Luke Schaeffer and Jeffrey Shallit, Closed, Palindromic, Rich, Privileged, Trapezoidal, and Balanced Words in Automatic Sequences, Electronic Journal of Combinatorics 23(1) (2016), #P1.25.
M. R. Schroeder & N. J. A. Sloane, Correspondence, 1991
R. Schroeppel and R. W. Gosper, HACKMEM #122 (1972).
Vladimir Shevelev, Two analogs of Thue-Morse sequence, arXiv preprint arXiv:1603.04434 [math.NT], 2016-2017.
N. J. A. Sloane, The first 1000 terms as a string
L. Spiegelhofer, Normality of the Thue-Morse Sequence along Piatetski-Shapiro sequences, Quart. J. Math. 66 (3) (2015).
Hassan Tarfaoui, Concours Général 1990 - Exercice 1 (in French).
Eric Weisstein's World of Mathematics, Thue-Morse Sequence
Eric Weisstein's World of Mathematics, Thue-Morse Constant
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)

A000034 Period 2: repeat [1, 2]; a(n) = 1 + (n mod 2).

Original entry on oeis.org

1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2

Offset: 0

N. J. A. Sloane

Also continued fraction for (sqrt(3)+1)/2 (cf. A040001) and base-3 digital root of n+1 (cf. A007089, A010888). - Henry Bottomley, Jul 05 2001
The sequence 1,-2,-1,2,1,-2,-1,2,... with g.f. (1-2x)/(1+x^2) has a(n) = cos(Pi*n/2)-2*sin(Pi*n/2). - Paul Barry, Oct 18 2004
Hankel transform is [1,-3,0,0,0,0,0,0,0,...]. - Philippe Deléham, Mar 29 2007
4/33 = 0.121212... - Eric Desbiaux, Nov 03 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=A[i,i]:=1, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1) = charpoly(A,2). - Milan Janjic, Jan 24 2010
First differences of A032766. - Tom Edgar, Jul 17 2014
Denominator of the harmonic mean of the first n triangular numbers. - Colin Barker, Nov 13 2014
This is the lexicographically earliest sequence of positive integers such that no polynomial of degree d can be fitted to d+2 consecutive terms (equivalently, such that no iterated difference is zero). - Pontus von Brömssen, Dec 26 2021 [See A300002 for the case where not only consecutive terms are considered. - Pontus von Brömssen, Jan 03 2023]
Number of maximum antichains in the power set of {1,2,...,n} partially ordered by set inclusion. For even n, there is a unique maximum antichain formed by all subsets of size n/2; for odd n, there are two maximum antichains, one formed by all subsets of size (n-1)/2 and the other formed by all subsets of size (n+1)/2. See the David Guichard link below for a proof. - Jianing Song, Jun 19 2022

Jozsef Beck, Combinatorial Games, Cambridge University Press, 2008.
J.-M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 545 pages 73 and 260, Ellipses, Paris 2004.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Sean A. Irvine, Table of n, a(n) for n = 0..9999
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Glen Joyce C. Dulatre, Jamilah V. Alarcon, Vhenedict M. Florida and Daisy Ann A. Disu, On Fractal Sequences, DMMMSU-CAS Science Monitor (2016-2017) Vol. 15 No. 2, 109-113.
Daniele A. Gewurz and Francesca Merola, Sequences realized as Parker vectors of oligomorphic permutation groups, J. Integer Seqs., Vol. 6, 2003.
David Guichard, Sperner's Theorem
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 383
Agustín Moreno Cañadas, Hernán Giraldo and Robinson Julian Serna Vanegas, Some integer partitions induced by orbits of Dynkin type, Far East Journal of Mathematical Sciences (FJMS), Vol. 101, No. 12 (2017), pp. 2745-2766.
Wikipedia, Collatz conjecture
Index entries for linear recurrences with constant coefficients, signature (0,1).

Cf. A000035, A003945 (binomial transf), A007089, A010693, A010704, A010888, A032766, A040001, A123344, A134451, A300002.

Cf. sequences listed in Comments section of A283393.

List([0..120],n->1+(n mod 2)); # Muniru A Asiru, Feb 01 2019

a000034 = (+ 1) . (`mod` 2)
a000034_list = cycle [1,2]
-- Reinhard Zumkeller, Jul 03 2012, Dec 02 2011 and corrected by James Spahlinger, Oct 08 2012

[1+(n mod 2) : n in [0..100]]; // Wesley Ivan Hurt, Sep 11 2014

(1+2*x)/(1-x^2);
A000034 := proc(n) op((n mod 2)+1,[1,2]) ; end proc: # R. J. Mathar, Feb 05 2011

a[n_] := If[OddQ[n], 2, 1]; Table[a[n], {n, 0, 90}] (* Stefan Steinerberger, Apr 17 2006 *)
Nest[ Flatten[# /. {    0 -> {1}, 1 -> {2}, 2 -> {1, 2, 1}  }] &, {1}, 8] (* Robert G. Wilson v, May 20 2014 *)

PARI
```
a(n)=1+n%2
```

a(n)=1+bittest(n,0) \\ M. F. Hasler, Jan 13 2012

def A000034(n): return 1 + (n & 1) # Chai Wah Wu, May 25 2022

G.f.: (1+2*x)/(1-x^2).
a(n) = 2^((1-(-1)^n)/2) = 2^(ceiling(n/2) - floor(n/2)). - Paul Barry, Jun 03 2003
a(n) = (3-(-1)^n)/2; a(n) = 1 + (n mod 2) = 3-a(n-1) = a(n-2) = a(-n).
a(n) = gcd(n-1, n+1). - Paul Barry, Sep 16 2004
Binomial transform of A123344, inverse binomial transform of A003945. - Philippe Deléham, Jun 04 2007
a(n) = A134451(n+1). - Reinhard Zumkeller, Oct 27 2007
a(n) = if(n=0,1,if(mod(a(n-1),2)=0,a(n-1)/2,(3*a(n-1)+1)/2)). See Collatz conjecture. - Paul Barry, Mar 31 2008
a(n) = 2^n (mod 3). - Vincenzo Librandi, Feb 05 2011
a(n) = A000035(n) + 1. - M. F. Hasler, Jan 13 2012
a(n) = abs(sin(n*Pi/2) - 2*cos(n*Pi/2)). - Mohammad K. Azarian, Mar 12 2012
a(n) = A010704(n) / 3. - Reinhard Zumkeller, Jul 03 2012
a(n) = floor((4/33)*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 03 2013
a(n) = floor((5/8)*3^(n+1)) mod 3. - Hieronymus Fischer, Jan 03 2013
a(n) = floor((n+1)*3/2) - floor((n)*3/2). - Hailey R. Olafson, Jul 23 2014
a(n) = denominator(n/2). - Wesley Ivan Hurt, Sep 11 2014
Dirichlet g.f.: zeta(s)*(1 + 1/2^s). - Mats Granvik, Jul 18 2016
E.g.f.: 2*sinh(x) + cosh(x). - Ilya Gutkovskiy, Jul 18 2016
a(n) = A010693(n) - 1. - Filip Zaludek, Oct 29 2016
a(n) = n + 1 - 2*floor(n/2). - Lorenzo Sauras Altuzarra, Jun 28 2019
Limit_{n->oo} (1/n)*Sum_{k=1..n} a(k) = 3/2 (De Koninck reference). - Bernard Schott, Nov 09 2021

Better definition from M. F. Hasler, Jan 13 2012

A047215 Numbers that are congruent to {0, 2} mod 5.

Original entry on oeis.org

0, 2, 5, 7, 10, 12, 15, 17, 20, 22, 25, 27, 30, 32, 35, 37, 40, 42, 45, 47, 50, 52, 55, 57, 60, 62, 65, 67, 70, 72, 75, 77, 80, 82, 85, 87, 90, 92, 95, 97, 100, 102, 105, 107, 110, 112, 115, 117, 120, 122, 125, 127, 130, 132, 135, 137, 140, 142, 145, 147, 150, 152, 155, 157

Offset: 0

N. J. A. Sloane

Number of partitions of 5n into exactly 2 parts. - Colin Barker, Mar 23 2015

Numbers k such that k^2/5 + k*(k + 1)/5 = k*(2*k + 1)/5 is a nonnegative integer. - Bruno Berselli, Feb 14 2017

David Lovler, Table of n, a(n) for n = 0..1000
Robbert Fokkink and Gandhar Joshi, Anti-recurrence sequences, arXiv:2506.13337 [math.NT], 2025. See p. 4.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Different from A038126.
Cf. A001622, A010693, A030308, A084215
Cf. A249547 (partial sums), A010693 (1st differences).

[(5*n - (n mod 2))/2: n in [1..100]]; // G. C. Greubel, Jun 23 2024

Table[Floor[5 n/2], {n, 0, 100}] (* or *) LinearRecurrence[{1, 1, -1}, {0, 2, 5},101] (* Vladimir Joseph Stephan Orlovsky, Jan 28 2012 *)

a(n)=5*n\2 \\ Charles R Greathouse IV, Oct 17 2011

[int(5*n//2) for n in range(1,101)] # G. C. Greubel, Jun 23 2024

a(n) = floor(5*n/2).
From R. J. Mathar, Sep 23 2008: (Start)
G.f.: x*(2 + 3*x)/((1 + x)*(1 - x)^2).
a(n) = 5*n/2 + ((-1)^n-1)/4.
a(n+1) - a(n) = A010693(n+1). (End)
a(n) = 5*n - a(n-1) - 8 with a(1)=0. - Vincenzo Librandi, Aug 05 2010
a(n+1) = Sum_{k>=0} A030308(n,k)*A084215(k+1). - Philippe Deléham, Oct 17 2011
a(n) = 2*n + floor(n/2). - Arkadiusz Wesolowski, Sep 19 2012
Sum_{n>=1} (-1)^(n+1)/a(n) = log(5)/4 - sqrt(5)*log(phi)/10 + sqrt(1-2/sqrt(5))*Pi/10, where phi is the golden ratio (A001622). - Amiram Eldar, Dec 07 2021
E.g.f.: (5*x*exp(x) - sinh(x))/2. - David Lovler, Aug 22 2022

A176059 Periodic sequence: Repeat 3, 2.

Original entry on oeis.org

3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3

Offset: 0

Klaus Brockhaus, Apr 07 2010

Interleaving of A010701 and A007395.
Also continued fraction expansion of (3+sqrt(15))/2.
Also decimal expansion of 32/99.
a(n) = A010693(n+1).
Essentially first differences of A047218.
Binomial transform of 3 followed by -A122803.
Inverse binomial transform of 3 followed by A020714.
Second inverse binomial transform of A057198 without initial term 1.

Index entries for linear recurrences with constant coefficients, signature (0, 1).

Cf. A010701 (all 3's sequence), A007395 (all 2's sequence), A176058 (decimal expansion of (3+sqrt(15))/2), A010693 (repeat 2, 3), A047218 (congruent to {0, 3} mod 5), A122803 (powers of -2), A020714 (5*2^n), A057198 ((5*3^(n-1)+1)/2, n > 0).

Cf. A026532 (partial products).

a176059 = (3 -) . (`mod` 2) -- Reinhard Zumkeller, Nov 27 2012

a176059_list = cycle [3,2]  -- Reinhard Zumkeller, Apr 04 2012

&cat[ [3, 2]: n in [0..52] ];
[ (5+(-1)^n)/2: n in [0..104] ];

A176059:=n->(5+(-1)^n)/2; seq(A176059(n), n=0..100); # Wesley Ivan Hurt, Feb 26 2014

a[n_] := {3, 2}[[Mod[n, 2] + 1]]; Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Jul 19 2013 *)
PadRight[{},120,{3,2}] (* Harvey P. Dale, Oct 06 2019 *)

a(n)=3-n%2 \\ Charles R Greathouse IV, Jul 13 2016

a(n) = (5+(-1)^n)/2.
a(n) = a(n-2) for n > 1; a(0) = 3, a(1) = 2.
a(n) = -a(n-1)+5 for n > 0; a(0) = 3.
a(n) = 3*((n+1) mod 2)+2*(n mod 2).
G.f.: (3+2*x)/((1-x)*(1+x)).
E.g.f.: 3*cosh(x) + 2*sinh(x). - Stefano Spezia, Aug 04 2025

A026549 Ratios of successive terms are 2, 3, 2, 3, 2, 3, 2, 3, ...

Original entry on oeis.org

1, 2, 6, 12, 36, 72, 216, 432, 1296, 2592, 7776, 15552, 46656, 93312, 279936, 559872, 1679616, 3359232, 10077696, 20155392, 60466176, 120932352, 362797056, 725594112, 2176782336, 4353564672, 13060694016, 26121388032, 78364164096, 156728328192, 470184984576, 940369969152

Offset: 0

Clark Kimberling

Appears to be the number of permutations p of {1,2,...,n} such that p(i)+p(i+1)>=n for every i=1,2,...,n-1 (if offset is 1). - Vladeta Jovovic, Dec 15 2003
Equals eigensequence of a triangle with 1's in even columns and (1,3,3,3,...) in odd columns. a(5) = 72 = (1, 3, 1, 3, 1, 1) dot (1, 1, 2, 6, 12, 36) = (1 + 3 + 2 + 18 + 12 + 36), where (1, 3, 1, 3, 1, 1) = row 5 of the generating triangle. - Gary W. Adamson, Aug 02 2010
Partial products of A010693. - Reinhard Zumkeller, Mar 29 2012
Satisfies Benford's law [Theodore P. Hill, Personal communication, Feb 06, 2017]. - N. J. A. Sloane, Feb 08 2017
For n >= 2, a(n) is the least k > a(n-1) such that both k and a(n-2) + a(n-1) + k have exactly n prime factors, counted with multiplicity. - Robert Israel, Aug 06 2024

			G.f. = 1 + 2*x + 6*x^2 + 12*x^3 + 36*x^4 + 72*x^5 + 216*x^6 + ... - _Michael Somos_, Apr 09 2022

Arno Berger and Theodore P. Hill, An Introduction to Benford's Law, Princeton University Press, 2015.

Vincenzo Librandi, Table of n, a(n) for n = 0..700
Paul Barry, Embedding structures associated with Riordan arrays and moment matrices, International Journal of Combinatorics, Vol. 2014 (2014), Article ID 301394, 7 pages; arXiv preprint, arXiv:1312.0583 [math.CO], 2013.
Sean A. Irvine, Walks on Graphs.
Index entries for linear recurrences with constant coefficients, signature (0,6).
Index entries for sequences related to Benford's law.

Cf. A026532, A026536, A026551, A026567.

Cf. A010693, A208131, A109827.

a026549 n = a026549_list !! n
a026549_list = scanl (*) 1 $ a010693_list
-- Reinhard Zumkeller, Mar 29 2012

[(1/2)*(3-(-1)^n)*6^Floor(n/2): n in [0..30]]; // Vincenzo Librandi, Jun 08 2011

seq(seq(2^i*3^j, i=j..j+1),j=0..30); # Robert Israel, Aug 06 2024

LinearRecurrence[{0,6},{1,2},30] (* Harvey P. Dale, May 29 2016 *)

{a(n) = 6^(n\2) * (n%2+1)}; /* Michael Somos, Apr 09 2022 */

[(1+(n%2))*6^(n//2) for n in (0..30)] # G. C. Greubel, Apr 09 2022

Equals T(n, 0) + T(n, 1) + ... + T(n, 2n), T given by A026536.
a(n) = 2*A026532(n), for n > 0.
G.f.: (1+2*x)/(1-6*x^2) - Paul Barry, Aug 25 2003
a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011
a(n) = (1/2)*(3 - (-1)^n)*6^floor(n/2), or a(n) = 6*a(n-2). - Vincenzo Librandi, Jun 08 2011
a(n) = 1/a(-n) if n is even and (2/3)/a(-n) if n is odd for all n in Z. - Michael Somos, Apr 09 2022
Sum_{n>=0} 1/a(n) = 9/5. - Amiram Eldar, Feb 13 2023

New definition from Ralf Stephan, Dec 01 2004

A195013 Multiples of 2 and of 3 interleaved: a(2n-1) = 2n, a(2n) = 3n.

Original entry on oeis.org

2, 3, 4, 6, 6, 9, 8, 12, 10, 15, 12, 18, 14, 21, 16, 24, 18, 27, 20, 30, 22, 33, 24, 36, 26, 39, 28, 42, 30, 45, 32, 48, 34, 51, 36, 54, 38, 57, 40, 60, 42, 63, 44, 66, 46, 69, 48, 72, 50, 75, 52, 78, 54, 81, 56, 84, 58, 87, 60, 90, 62, 93, 64, 96, 66, 99, 68, 102

Offset: 1

Omar E. Pol, Sep 09 2011

First differences of A195014.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
D. H. Bailey, J. M. Borwein, and J. S. Kimberley, Discovery of large Poisson polynomials using a new arbitrary precision software package, Slides, 2015.
D. H. Bailey, J. M. Borwein, and J. S. Kimberley, Computer discovery and analysis of large Poisson polynomials, 2016.
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A005843, A008585, A195014, A195019.

Cf. A111712 (partial sums of this sequence prepended with 1).

import Data.List (transpose)
a195013 n = a195013_list !! (n-1)
a195013_list = concat $ transpose [[2, 4 ..], [3, 6 ..]]
-- Reinhard Zumkeller, Apr 06 2015

&cat[[2*n,3*n]: n in [1..34]]; // Bruno Berselli, Sep 25 2011

With[{r = Range[50]}, Riffle[2*r, 3*r]] (* or *)
LinearRecurrence[{0, 2, 0, -1}, {2, 3, 4, 6}, 100] (* Paolo Xausa, Feb 09 2024 *)

a(n)=(5*n+(n-2)*(-1)^n+2)/4 \\ Charles R Greathouse IV, Sep 24 2015

Pair(2*n, 3*n).
From Bruno Berselli, Sep 26 2011: (Start)
G.f.: x*(2+3*x)/(1-x^2)^2.
a(n) = (5*n+(n-2)*(-1)^n+2)/4.
a(n) = 2*a(n-2) - a(n-4) = a(n-2) + A010693(n-1).
a(n)+a(-n) = A010673(n).
a(n)-a(-n) = A106832(n). (End)

A158478 Number of colors needed to paint n sectors of a circle.

Original entry on oeis.org

0, 1, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2

Offset: 0

Jaume Oliver Lafont, Mar 20 2009

nonn

No pair of adjacent sectors can have the same color.
For n > 0: smallest prime factor of A098548(n).
Decimal expansion of 61/4950. - Elmo R. Oliveira, May 05 2024

Stefano Spezia, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (0,1).

Cf. A010693, A020639, A098548, A158411.

a158478 n = if n < 4 then n else 2 + mod n 2
a158478_list = [0..3] ++ cycle [2,3]
-- Reinhard Zumkeller, Nov 30 2014

Join[Range[0, 1], ConstantArray[{2, 3}, 20]] // Flatten (* Robert Price, Jun 15 2020 *)

PARI
```
a(n)=if(n<4,n,2+n%2)
```

G.f.: x*(1+2*x+2*x^2)/(1-x^2).
E.g.f.: 2*cosh(x) + 3*sinh(x) - 2*(1 + x). - Stefano Spezia, Mar 24 2020
a(n) = a(n-2) for n > 3. - Elmo R. Oliveira, May 05 2024
a(n) = (n mod 2) + (2 mod (n+1)). - Aaron J Grech, Sep 02 2024

A016767 a(n) = (3*n)^3.

Original entry on oeis.org

0, 27, 216, 729, 1728, 3375, 5832, 9261, 13824, 19683, 27000, 35937, 46656, 59319, 74088, 91125, 110592, 132651, 157464, 185193, 216000, 250047, 287496, 328509, 373248, 421875, 474552, 531441, 592704

Offset: 0

N. J. A. Sloane

			G.f. = 27*x + 216*x^2 + 729*x^3 * 1728*x^4 + 3375*x^5 + 5832*x^6 + ... - _Michael Somos_, Sep 16 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..250
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

[(3*n)^3: n in [0..40]]; // Vincenzo Librandi, Jul 29 2011

Table[(3*n)^3, {n,0,30}] (* G. C. Greubel, Sep 15 2018 *)

a(n)=27*n^3 \\ Charles R Greathouse IV, Jul 29 2011

From Reinhard Zumkeller, Jan 29 2009: (Start)
a(n) = 27*A000578(n).
A020639(a(n)) = A010693(n), for n>0. (End)
a(n) = A155955(n,3) for n>2. - Reinhard Zumkeller, Jan 31 2009
G.f.: 27*x*(1+4*x+x^2)/(x-1)^4 . - R. J. Mathar, Jul 07 2017
E.g.f.: 27*x*(1 + 3*x + x^2)*exp(x). - G. C. Greubel, Sep 15 2018
a(n) = -a(-n) for all n in Z. - Michael Somos, Sep 16 2018

A101622 A Horadam-Jacobsthal sequence.

Original entry on oeis.org

0, 1, 6, 13, 30, 61, 126, 253, 510, 1021, 2046, 4093, 8190, 16381, 32766, 65533, 131070, 262141, 524286, 1048573, 2097150, 4194301, 8388606, 16777213, 33554430, 67108861, 134217726, 268435453, 536870910, 1073741821, 2147483646, 4294967293, 8589934590

Offset: 0

Paul Barry, Dec 10 2004

Companion sequence to A084639.
This is the sequence A(0,1;1,2;5) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
Except for the initial three terms, the decimal representation of the x-axis, from the left edge to the origin, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 961", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. - Robert Price, Mar 27 2017
Named after the Australian mathematician Alwyn Francis Horadam (1923-2016) and the German mathematician Ernst Jacobsthal (1882-1965). - Amiram Eldar, Jun 10 2021

Stephen Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. F. Horadam, Jacobsthal Representation Numbers, Fib Quart., Vol. 34, No. 1 (1996), pp. 40-54.
Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences. [_Wolfdieter Lang_, Oct 18 2010]
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton.
Stephen Wolfram, A New Kind of Science.
Wolfram Research, Wolfram Atlas of Simple Programs.
Index entries for sequences related to cellular automata.
Index to 2D 5-Neighbor Cellular Automata.
Index to Elementary Cellular Automata.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

Cf. A131953.

[(2^(n+2)+(-1)^n-5)/2: n in [0..35]]; // Vincenzo Librandi, Aug 12 2011

LinearRecurrence[{2,1,-2},{0,1,6},40] (* Harvey P. Dale, Jul 08 2014 *)

concat(0, Vec(x*(1+4*x)/((1-x)*(1+x)*(1-2*x)) + O(x^30))) \\ Colin Barker, Mar 28 2017

a(n) = (2^(n+2) + (-1)^n - 5)/2.
G.f.: x*(1+4*x)/((1-x)*(1+x)*(1-2*x)).
a(n) = (A014551(n+2)-5)/2.
(1, 6, 13, 30, 61, ...) are the row sums of A131953. - Gary W. Adamson, Jul 31 2007
From Paul Curtz, Jan 01 2009: (Start)
a(n) = a(n-1) + 2*a(n-2) + 5.
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3).
a(n) = A000079(n+1) - A010693(n).
a(n+1) = A141722(n) + 5 = A141722(n) + A010716(n).
a(2n+1) - a(2n) = 1, 7, 31, ... = A083420.
a(2n+1) - 2*a(2n) = 1.
a(2n) = A002446 = 6*A002450, a(2n+1) = A141725. (End)
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) for n>2. - Colin Barker, Mar 28 2017
a(n) = (1/2) * Sum_{k=1..n} binomial(n+1,k) * (2+(-1)^k). - Wesley Ivan Hurt, Sep 23 2017

cp's OEIS Frontend

A262392 a(n) = A007504(n) + A010693(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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A000034 Period 2: repeat [1, 2]; a(n) = 1 + (n mod 2).

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A047215 Numbers that are congruent to {0, 2} mod 5.

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A176059 Periodic sequence: Repeat 3, 2.

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