A014550 -id:A014550 - cp's OEIS frontend

A000975 a(2n) = 2a(2n-1), a(2n+1) = 2a(2n)+1 (also a(n) is the n-th number without consecutive equal binary digits).

0, 1, 2, 5, 10, 21, 42, 85, 170, 341, 682, 1365, 2730, 5461, 10922, 21845, 43690, 87381, 174762, 349525, 699050, 1398101, 2796202, 5592405, 11184810, 22369621, 44739242, 89478485, 178956970, 357913941, 715827882, 1431655765, 2863311530, 5726623061, 11453246122

Offset: 0

Mira Bernstein, N. J. A. Sloane, Robert G. Wilson v, Sep 13 1996

Might be called the "Lichtenberg sequence" after Georg Christoph Lichtenberg, who discussed it in 1769 in connection with the Chinese Rings puzzle (baguenaudier). - Andreas M. Hinz, Feb 15 2017
Number of steps to change from a binary string of n 0's to n 1's using a Gray code. - Jon Stadler (jstadler(AT)coastal.edu)
Popular puzzles such as Spin-Out and The Brain Puzzler are based on the Gray binary system and require a(n) steps to complete for some number n.
Conjecture: {a(n)} also gives all j for which A048702(j) = A000217(j); i.e., if we take the a(n)-th triangular number (a(n)^2 + a(n))/2 and multiply it by 3, we get a(n)-th even-length binary palindrome A048701(a(n)) concatenated from a(n) and its reverse. E.g., a(4) = 10, which is 1010 in binary; the tenth triangular number is 55, and 55*3 = 165 = 10100101 in binary. - Antti Karttunen, circa 1999. (This has been now proved by Paul K. Stockmeyer in his arXiv:1608.08245 paper.) - Antti Karttunen, Aug 31 2016
Number of ways to tie a tie of n or fewer half turns, excluding mirror images. Also number of walks of length n or less on a triangular lattice with the following restrictions; given l, r and c as the lattice axes. 1. All steps are taken in the positive axis direction. 2. No two consecutive steps are taken on the same axis. 3. All walks begin with l. 4. All walks end with rlc or lrc. - Bill Blewett, Dec 21 2000
a(n) is the minimal number of vertices to be selected in a vertex-cover of the balanced tree B_n. - Sen-peng Eu, Jun 15 2002
A087117(a(n)) = A038374(a(n)) = 1 for n > 1; see also A090050. - Reinhard Zumkeller, Nov 20 2003
Intersection of A003754 and A003714; complement of A107907. - Reinhard Zumkeller, May 28 2005
Equivalently, numbers m whose binary representation is effectively, for some number k, both the lazy Fibonacci and the Zeckendorf representation of k (in which case k = A022290(m)). - Peter Munn, Oct 06 2022
a(n+1) gives row sums of Riordan array (1/(1-x), x(1+2x)). - Paul Barry, Jul 18 2005
Total number of initial 01's in all binary words of length n+1. Example: a(3) = 5 because the binary words of length 4 that start with 01 are (01)00, (01)(01), (01)10 and (01)11 and the total number of initial 01's is 5 (shown between parentheses). a(n) = Sum_{k >= 0} k*A119440(n+1, k). - Emeric Deutsch, May 19 2006
In Norway we call the 10-ring puzzle "strikketoy" or "knitwear" (see link). It takes 682 moves to free the two parts. - Hans Isdahl, Jan 07 2008
Equals A002450 and A020988 interlaced. - Zak Seidov, Feb 10 2008
For n > 1, let B_n = the complete binary tree with vertex set V where |V| = 2^n - 1. If VC is a minimum-size vertex cover of B_n, Sen-Peng Eu points out that a(n) = |VC|. It also follows that if IS = V\VC, a(n+1) = |IS|. - K.V.Iyer, Apr 13 2009
Starting with 1 and convolved with [1, 2, 2, 2, ...] = A000295. - Gary W. Adamson, Jun 02 2009
a(n) written in base 2 is sequence A056830(n). - Jaroslav Krizek, Aug 05 2009
This is the sequence A(0, 1; 1, 2; 1) 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
From Vladimir Shevelev, Jan 30 2012, Feb 13 2012: (Start)
1) Denote by {n, k} the number of permutations of 1, ..., n with the up-down index k (for definition, see comment in A203827). Then max_k{n, k} = {n, a(n)} = A000111(n).
2) a(n) is the minimal number > a(n-1) with the Hamming distance d_H(a(n-1), a(n)) = n. Thus this sequence is the Hamming analog of triangular numbers 0, 1, 3, 6, 10, ... (End)
From Hieronymus Fischer, Nov 22 2012: (Start)
Represented in binary form each term after the second one contains every previous term as a substring.
The terms a(2) = 2 and a(3) = 5 are the only primes. Proof: For even n we get a(n) = 2*(2^(2*n) - 1)/3, which shows that a(n) is even, too, and obviously a(n) > 2 for all even n > 2. For odd n we have a(n) = (2^(n+1) - 1)/3 = (2^((n+1)/2) - 1) * (2^((n+1)/2) + 1)/3. Evidently, at least one of these factors is divisible by 3, both are greater than 6, provided n > 3. Hence it follows that a(n) is composite for all odd n > 3.
Represented as a binary number, a(n+1) has exactly n prime substrings. Proof: Evidently, a(1) = 1_2 has zero and a(2) = 10_2 has 1 prime substring. Let n > 1. Written in binary, a(n+1) is 101010101...01 (if n + 1 is odd) and is 101010101...10 (if n + 1 is even) with n + 1 digits. Only the 2- and 3-digits substrings 10_2 (=2) and 101_2 (=5) are prime substrings. All the other substrings are nonprime since every substring is a previous term and all terms unequal to 2 and 5 are nonprime. For even n + 1, the number of prime substrings equal to 2 = 10_2 is (n+1)/2, and the number of prime substrings equal to 5 = 101_2 is (n-1)/2, makes a sum of n. For odd n + 1 we get n/2 for both, the number of 2's and 5's prime substrings, in any case, the sum is n. (End)
Number of different 3-colorings for the vertices of all triangulated planar polygons on a base with n+2 vertices if the colors of the two base vertices are fixed. - Patrick Labarque, Feb 09 2013
A090079(a(n)) = a(n) and A090079(m) <> a(n) for m < a(n). - Reinhard Zumkeller, Feb 16 2013
a(n) is the number of length n binary words containing at least one 1 and ending in an even number (possibly zero) of 0's. a(3) = 5 because we have: 001, 011, 100, 101, 111. - Geoffrey Critzer, Dec 15 2013
a(n) is the number of permutations of length n+1 having exactly one descent such that the first element of the permutation is an even number. - Ran Pan, Apr 18 2015
a(n) is the sequence of the last row of the Hadamard matrix H(2^n) obtained via Sylvester's construction: H(2) = [1,1;1,-1], H(2^n) = H(2^(n-1))*H(2), where * is the Kronecker product. - William P. Orrick, Jun 28 2015
Conjectured record values of A264784: a(n) = A264784(A155051(n-1)). - Reinhard Zumkeller, Dec 04 2015. (This is proved by Paul K. Stockmeyer in his arXiv:1608.08245 paper.) - Antti Karttunen, Aug 31 2016
Decimal representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 131", based on the 5-celled von Neumann neighborhood. See A279053 for references and links. - Robert Price, Dec 05 2016
For n > 4, a(n-2) is the second-largest number in row n of A127824. - Dmitry Kamenetsky, Feb 11 2017
Conjecture: a(n+1) is the number of compositions of n with two kinds of parts, n and n', where the order of the 1 and 1' does not matter. For n=2, a(3) = 5 compositions, enumerated as follows:  2; 2'; 1,1; 1',1 = 1',1; 1',1'. - Gregory L. Simay, Sep 02 2017
Conjecture proved by recognizing the appropriate g.f. is x/(1 - x)(1 - x)(1 - 2*x^2 - 2x^3 - ...) =  x/(1 - 2*x - x^2 + 2x^3). - Gregory L. Simay, Sep 10 2017
a(n) = 2^(n-1) + 2^(n-3) + 2^(n-5) + ... a(2*k -1) = A002450(k) is the sum of the powers of 4. a(2*k) = 2*A002450(k). - Gregory L. Simay, Sep 27 2017
a(2*n) = n times the string [10] in binary representation, a(2*n+1) = n times the string [10] followed with [1] in binary representation. Example: a(7) = 85 = (1010101) in binary, a(8) = 170 = (10101010) in binary. - Ctibor O. Zizka, Nov 06 2018
Except for 0, these are the positive integers whose binary expansion has cuts-resistance 1. For the operation of shortening all runs by 1, cuts-resistance is the number of applications required to reach an empty word. Cuts-resistance 2 is A329862. - Gus Wiseman, Nov 27 2019
From Markus Sigg, Sep 14 2020: (Start)
Let s(k) be the length of the Collatz orbit of k, e.g. s(1) = 1, s(2) = 2, s(3) = 5. Then s(a(n)) = n+3 for n >= 3. Proof by induction: s(a(3)) = s(5) = 6 = 3+3. For odd n >= 5 we have s(a(n)) = s(4*a(n-2)+1) = s(12*a(n-2)+4)+1 = s(3*a(n-2)+1)+3 = s(a(n-2))+2 = (n-2)+3+2 = n+3, and for even n >= 4 this gives s(a(n)) = s(2*a(n-1)) = s(a(n-1))+1 = (n-1)+3+1 = n+3.
Conjecture: For n >= 3, a(n) is the second largest natural number whose Collatz orbit has length n+3. (End)
From Gary W. Adamson, May 14 2021: (Start)
With offset 1 the sequence equals the numbers of 1's from n = 1 to 3, 3 to 7, 7 to 15, ...; of A035263; as shown below:
..1     3           7                      15...
..1  0  1  1  1  0  1  0  1  0  1  1  1  0  1...
..1.....2...........5......................10...; a(n) = Sum_(k=1..2n-1)A035263(k)
.....1...........2.......................5...; as to zeros.
..1's in the Tower of Hanoi game represent CW moves On disks in the pattern:
..0,  1,  2,  0,  1,  2, ... whereas even numbered disks move in the pattern:
..0,  2,  1,  0,  2,  1, ... (End)
Except for 0, numbers that are repunits in Gray-code representation (A014550). - Amiram Eldar, May 21 2021
From Gus Wiseman, Apr 20 2023: (Start)
Also the number of nonempty subsets of {1..n} with integer median, where the median of a multiset is the middle part in the odd-length case, and the average of the two middle parts in the even-length case. For example, the a(1) = 1 through a(4) = 10 subsets are:
  {1}  {1}  {1}      {1}
       {2}  {2}      {2}
            {3}      {3}
            {1,3}    {4}
            {1,2,3}  {1,3}
                     {2,4}
                     {1,2,3}
                     {1,2,4}
                     {1,3,4}
                     {2,3,4}
The complement is counted by A005578.
For mean instead of median we have A051293, counting empty sets A327475.
For normal multisets we have A056450, strongly normal A361202.
For partitions we have A325347, strict A359907, complement A307683.
(End)
In binary form, because the sequence 10101_2... keeps repeating, one can represent a(n) = floor(0.(10)2 * 2^n). Because 0.(10)_2 = 0.(10)_2 * 4 - 2, then 0.(10)_2 = 2/3. Substituting this value in the first expression, we obtain a(n) = floor(2^(n+1)/3). This also agrees with the formula of Paul Barry, Oct 08 2005. - _Giovanni Ciriani, Mar 31 2025

			a(4)=10 since 0001, 0011, 0010, 0110, 0111, 0101, 0100, 1100, 1101, 1111 are the 10 binary strings switching 0000 to 1111.
a(3) = 1 because "lrc" is the only way to tie a tie with 3 half turns, namely, pass the business end of the tie behind the standing part to the left, bring across the front to the right, then behind to the center. The final motion of tucking the loose end down the front behind the "lr" piece is not considered a "step".
a(4) = 2 because "lrlc" is the only way to tie a tie with 4 half turns. Note that since the number of moves is even, the first step is to go to the left in front of the tie, not behind it. This knot is the standard "four in hand", the most commonly known men's tie knot. By contrast, the second most well known tie knot, the Windsor, is represented by "lcrlcrlc".
a(n) = (2^0 - 1) XOR (2^1 - 1) XOR (2^2 - 1) XOR (2^3 - 1) XOR ... XOR (2^n - 1). - _Paul D. Hanna_, Nov 05 2011
G.f. = x + 2*x^2 + 5*x^3 + 10*x^4 + 21*x^5 + 42*x^6 + 85*x^7 + 170*x^8 + ...
a(9) = 341 = 2^8 + 2^6 + 2^4 + 2^2 + 2^0 = 4^4 + 4^3 + 4^2 + 4^1 + 4^0 = A002450(5). a(10) = 682 = 2*a(9) = 2*A002450(5). - _Gregory L. Simay_, Sep 27 2017

Thomas Fink and Yong Mao, The 85 Ways to Tie a Tie, Broadway Books, New York (1999), p. 138.
Clifford A. Pickover, The Math Book, From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics, Sterling Publ., NY, 2009.

G. C. Greubel, Table of n, a(n) for n = 0..3300 (terms 0..300 from T. D. Noe)
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
Thomas Baruchel, Properties of the cumulated deficient binary digit sum, arXiv:1908.02250 [math.NT], 2019.
Sergei L. Bezrukov et al., The congestion of n-cube layout on a rectangular grid, Discrete Mathematics 213.1-3 (2000): 13-19. See Theorem 1.
F. Chapoton and S. Giraudo, Enveloping operads and bicoloured noncrossing configurations, arXiv:1310.4521 [math.CO], 2013. Is the sequence in Table 2 this sequence? - _N. J. A. Sloane_, Jan 04 2014. (Yes, it is. See Stockmeyer's arXiv:1608.08245 2016 paper for the proof.)
Ji Young Choi, Ternary Modified Collatz Sequences And Jacobsthal Numbers, Journal of Integer Sequences, Vol. 19 (2016), #16.7.5.
Ji Young Choi, A Generalization of Collatz Functions and Jacobsthal Numbers, J. Int. Seq., Vol. 21 (2018), Article 18.5.4.
Madeleine Goertz and Aaron Williams, The Quaternary Gray Code and How It Can Be Used to Solve Ziggurat and Other Ziggu Puzzles, arXiv:2411.19291 [math.CO], 2024. See pp. 1, 17, 42.
David Hayes, Kaveh Khodjasteh, Lorenza Viola and Michael J. Biercuk, Reducing sequencing complexity in dynamical quantum error suppression by Walsh modulation, arXiv:1109.6002 [quant-ph], 2011.
Clemens Heuberger and Daniel Krenn, Esthetic Numbers and Lifting Restrictions on the Analysis of Summatory Functions of Regular Sequences, arXiv:1808.00842 [math.CO], 2018. See p. 10.
Clemens Heuberger and Daniel Krenn, Asymptotic Analysis of Regular Sequences, arXiv:1810.13178 [math.CO], 2018. See p. 29.
Andreas M. Hinz, The Lichtenberg sequence, Fib. Quart., 55 (2017), 2-12.
Andreas M. Hinz, Sandi Klavžar, Uroš Milutinović, and Ciril Petr, The Tower of Hanoi - Myths and Maths, Birkhäuser 2013. See page 56. Book's website
Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
Jia Huang, Nonassociativity of the Norton Algebras of some distance regular graphs, arXiv:2001.05547 [math.CO], 2020.
Jia Huang, Norton algebras of the Hamming Graphs via linear characters, arXiv:2101.05711 [math.CO], 2021.
Jia Huang and Erkko Lehtonen, Associative-commutative spectra for some varieties of groupoids, arXiv:2401.15786 [math.CO], 2024. See p. 17.
Jia Huang, Madison Mickey, and Jianbai Xu, The Nonassociativity of the Double Minus Operation, Journal of Integer Sequences, Vol. 20 (2017), #17.10.3.
Ryota Inagaki, Tanya Khovanova, and Austin Luo, On Chip-Firing on Undirected Binary Trees, Ann. Comb. (2025). See p. 10.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 394
Hans Isdahl, "Knitwear" puzzle
D. E. Knuth and O. P. Lossers, Partitions of a circular set, Problem 11151 in Amer. Math. Monthly 114 (3) (2007) p 265, E_3.
S. Lafortune, A. Ramani, B. Grammaticos, Y. Ohta and K.M. Tamizhmani, Blending two discrete integrability criteria: singularity confinement and algebraic entropy, arXiv:nlin/0104020 [nlin.SI], 2001.
Robert L. Lamphere, A Recurrence Relation in the Spinout Puzzle, The College Mathematics Journal, Vol. 27, Nbr. 4, Page 289, Sep. 96.
Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.
Georg Christoph Lichtenberg, Vermischte Schriften, Band 6 (1805). See chapter 6, p. 257.
Saad Mneimneh, Simple Variations on the Tower of Hanoi to Guide the Study of Recurrences and Proofs by Induction, Department of Computer Science, Hunter College, CUNY, 2019.
Richard Moot, Partial Orders, Residuation, and First-Order Linear Logic, arXiv:2008.06351 [cs.LO], 2020.
Gregg Musiker and Son Nguyen, Labeled Chip-firing on Binary Trees, arXiv:2206.02007 [math.CO], 2022.
Kival Ngaokrajang, Illustration of initial terms
Ahmet Öteleş, On the sum of Pell and Jacobsthal numbers by the determinants of Hessenberg matrices, AIP Conference Proceedings 1863, 310003 (2017).
Vladimir Shevelev, On the Basis Polynomials in the Theory of Permutations with Prescribed Up-Down Structure, arXiv:0801.0072 [math.CO], 2007-2010. See Example 3.
Chloe E. Shiff and Noah A. Rosenberg, Enumeration of rooted binary perfect phylogenies, arXiv:2410.14915 [q-bio.PE], 2024. See pp. 15, 17.
A. V. Sills and H. Wang, On the maximal Wiener index and related questions, Discrete Applied Mathematics, Volume 160, Issues 10-11, July 2012, Pages 1615-1623 - _N. J. A. Sloane_, Sep 21 2012
N. J. A. Sloane, Transforms
Elen Viviani Pereira Spreafico, Francisco Regis Vieira Alves, Paula Maria Machado Cruz Catarino, and Eudes Antonio Costa, A brief study of the one parameter Lichtenberg numbers, VI Int'l Conf. Math. Appl. Sci. Eng. (ICMASE 2025).
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Index entries for sequences related to binary expansion of n
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

Partial sums of A001045.
Row sums of triangle A013580.
Equals A026644/2.
Union of the bijections A002450 and A020988. - Robert G. Wilson v, Jun 09 2014
Column k=3 of A261139.
Cf. A000295, A005578, A015441, A043291, A053404, A059260, A077854, A119440, A127824, A130125, A135228, A155051, A179970, A264784.
Complement of A107907.
Row 3 of A300653.
Other sequences that relate to the binary representation of the terms: A003714, A003754, A007088, A022290, A056830, A104161, A107909.
Cf. A000120, A001511, A003242, A027383, A070939, A164707, A295235, A319416.
Cf. A005186, A033491, A153772.
Cf. A014550, A035263
Cf. A051293, A067659, A079309, A231147, A325347, A359893, A359907, A361801.

List([0..35],n->(2^(n+1)-2+(n mod 2))/3); # Muniru A Asiru, Nov 01 2018

a000975 n = a000975_list !! n
a000975_list = 0 : 1 : map (+ 1)
   (zipWith (+) (tail a000975_list) (map (* 2) a000975_list))
-- Reinhard Zumkeller, Mar 07 2012

[(2^(n+1) - 2 + (n mod 2))/3: n in [0..40]]; // Vincenzo Librandi, Mar 18 2015

A000975 := proc(n) option remember; if n <= 1 then n else if n mod 2 = 0 then 2*A000975(n-1) else 2*A000975(n-1)+1 fi; fi; end;
seq(iquo(2^n,3),n=1..33); # Zerinvary Lajos, Apr 20 2008
f:=n-> if n mod 2 = 0 then (2^n-1)/3 else (2^n-2)/3; fi; [seq(f(n),n=0..40)]; # N. J. A. Sloane, Mar 21 2017

Array[Ceiling[2(2^# - 1)/3] &, 41, 0]
RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == a[n - 1] + 2a[n - 2] + 1}, a, {n, 40}] (* or *)
LinearRecurrence[{2, 1, -2}, {0, 1, 2}, 40] (* Harvey P. Dale, Aug 10 2013 *)
f[n_] := Block[{exp = n - 2}, Sum[2^i, {i, exp, 0, -2}]]; Array[f, 33] (* Robert G. Wilson v, Oct 30 2015 *)
f[s_List] := Block[{a = s[[-1]]}, Append[s, If[OddQ@ Length@ s, 2a + 1, 2a]]]; Nest[f, {0}, 32] (* Robert G. Wilson v, Jul 20 2017 *)
NestList[2# + Boole[EvenQ[#]] &, 0, 39] (* Alonso del Arte, Sep 21 2018 *)

{a(n) = if( n<0, 0, 2 * 2^n \ 3)}; /* Michael Somos, Sep 04 2006 */

a(n)=if(n<=0,0,bitxor(a(n-1),2^n-1)) \\ Paul D. Hanna, Nov 05 2011

concat(0, Vec(x/(1-2*x-x^2+2*x^3) + O(x^100))) \\ Altug Alkan, Oct 30 2015

{a(n) = (4*2^n - 3 - (-1)^n) / 6}; /* Michael Somos, Jul 23 2017 */

def a(n): return (2**(n+1) - 2 + (n%2))//3
print([a(n) for n in range(35)]) # Michael S. Branicky, Dec 19 2021

a(n) = ceiling(2*(2^n-1)/3).
Alternating sum transform (PSumSIGN) of {2^n - 1} (A000225).
a(n) = a(n-1) + 2*a(n-2) + 1.
a(n) = 2*2^n/3 - 1/2 - (-1)^n/6.
a(n) = Sum_{i = 0..n} A001045(i), partial sums of A001045. - Bill Blewett
a(n) = n + 2*Sum_{k = 1..n-2} a(k).
G.f.: x/((1+x)*(1-x)*(1-2*x)) = x/(1-2*x-x^2+2*x^3). - Paul Barry, Feb 11 2003
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3). - Paul Barry, Feb 11 2003
a(n) = Sum_{k = 0..floor((n-1)/2)} 2^(n-2*k-1). - Paul Barry, Nov 11 2003
a(n+1) = Sum_{k=0..floor(n/2)} 2^(n-2*k); a(n+1) = Sum_{k = 0..n} Sum_{j = 0..k} (-1)^(j+k)*2^j. - Paul Barry, Nov 12 2003
(-1)^(n+1)*a(n) = Sum_{i = 0..n} Sum_{k = 1..i} k!*k* Stirling2(i, k)*(-1)^(k-1) = (1/3)*(-2)^(n+1)-(1/6)(3*(-1)^(n+1)-1). - Mario Catalani (mario.catalani(AT)unito.it), Dec 22 2003
a(n+1) = (n!/3)*Sum_{i - (-1)^i + j = n, i = 0..n, j = 0..n} 1/(i - (-1)^i)!/j!. - Benoit Cloitre, May 24 2004
a(n) = A001045(n+1) - A059841(n). - Paul Barry, Jul 22 2004
a(n) = Sum_{k = 0..n} 2^(n-k-1)*(1-(-1)^k), row sums of A130125. - Paul Barry, Jul 28 2004
a(n) = Sum_{k = 0..n} binomial(k, n-k+1)2^(n-k); a(n) = Sum_{k = 0..floor(n/2)} binomial(n-k, k+1)2^k. - Paul Barry, Oct 07 2004
a(n) = A107909(A104161(n)); A007088(a(n)) = A056830(n). - Reinhard Zumkeller, May 28 2005
a(n) = floor(2^(n+1)/3) = ceiling(2^(n+1)/3) - 1 = A005578(n+1) - 1. - Paul Barry, Oct 08 2005
Convolution of "Number of fixed points in all 231-avoiding involutions in S_n." (A059570) with "1-n" (A024000), treating the result as if offset was 0. - Graeme McRae, Jul 12 2006
a(n) = A081254(n) - 2^n. - Philippe Deléham, Oct 15 2006
Starting (1, 2, 5, 10, 21, 42, ...), these are the row sums of triangle A135228. - Gary W. Adamson, Nov 23 2007
Let T = the 3 X 3 matrix [1,1,0; 1,0,1; 0,1,1]. Then T^n * [1,0,0] = [A005578(n), A001045(n), a(n-1)]. - Gary W. Adamson, Dec 25 2007
2^n = 2*A005578(n-1) + 2*A001045(n) + 2*a(n-2). - Gary W. Adamson, Dec 25 2007
If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*stirling2(k,j)*x^(m-k) then a(n-3) = (-1)^(n-1)*f(n,3,-2), (n >= 3). - Milan Janjic, Apr 26 2009
a(n) + A001045(n) = A166920(n). a(n) + A001045(n+2) = A051049(n+1). - Paul Curtz, Oct 29 2009
a(n) = floor(A051049(n+1)/3). - Gary Detlefs, Dec 19 2010
a(n) = round((2^(n+2)-3)/6) = floor((2^(n+1)-1)/3) = round((2^(n+1)-2)/3); a(n) = a(n-2) + 2^(n-1), n > 1. - Mircea Merca, Dec 27 2010
a(n) = binary XOR of 2^k-1 for k=0..n. - Paul D. Hanna, Nov 05 2011
E.g.f.: 2/3*exp(2*x) - 1/2*exp(x) - 1/6*exp(-x) = 2/3*U(0); U(k) = 1 - 3/(4*(2^k) - 4*(2^k)/(1+3*(-1)^k - 24*x*(2^k)/(8*x*(2^k)*(-1)^k - (k+1)/U(k+1)))); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2011
Starting with "1" = triangle A059260 * [1, 2, 2, 2, ...] as a vector. - Gary W. Adamson, Mar 06 2012
a(n) = 2*(2^n - 1)/3, for even n; a(n) = (2^(n+1) - 1)/3 = (1/3)*(2^((n+1)/2) - 1)*(2^((n+1)/2) + 1), for odd n. - Hieronymus Fischer, Nov 22 2012
a(n) + a(n+1) = 2^(n+1) - 1. - Arie Bos, Apr 03 2013
G.f.: Q(0)/(3*(1-x)), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k + 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 21 2013
floor(a(n+2)*3/5) = A077854(n), for n >= 0. - Armands Strazds, Sep 21 2014
a(n) = (2^(n+1) - 2 + (n mod 2))/3. - Paul Toms, Mar 18 2015
a(0) = 0, a(n) = 2*(a(n-1)) + (n mod 2). - Paul Toms, Mar 18 2015
Binary: a(n) = (a(n-1) shift left 1) + (a(n-1)) NOR (...11110). - Paul Toms, Mar 18 2015
Binary: for n > 1, a(n) = 2*a(n-1) OR a(n-2). - Stanislav Sykora, Nov 12 2015
a(n) = A266613(n) - 20*2^(n-5), for n > 2. - Andres Cicuttin, Mar 31 2016
From Michael Somos, Jul 23 2017: (Start)
a(n) = -(2^n)*a(-n) for even n; a(n) = -(2^(n+1))*a(-n) + 1 for odd n.
0 = +a(n)*(+2 +4*a(n) -4*a(n+1)) + a(n+1)*(-1 +a(n+1)) for all n in Z. (End)
G.f.: (x^1+x^3+x^5+x^7+...)/(1-2*x). - Gregory L. Simay, Sep 27 2017
a(n+1) = A051049(n) + A001045(n). - Yuchun Ji, Jul 12 2018
a(n) = A153772(n+3)/4. - Markus Sigg, Sep 14 2020
a(4*k+d) = 2^(d+1)*a(4*k-1) + a(d), a(n+4) = a(n) + 2^n*10, a(0..3) = [0,1,2,5]. So the last digit is always 0,1,2,5 repeated. - Yuchun Ji, May 22 2023

Additional comments from Barry E. Williams, Jan 10 2000

A003188 Decimal equivalent of Gray code for n.

Original entry on oeis.org

0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 10, 11, 9, 8, 24, 25, 27, 26, 30, 31, 29, 28, 20, 21, 23, 22, 18, 19, 17, 16, 48, 49, 51, 50, 54, 55, 53, 52, 60, 61, 63, 62, 58, 59, 57, 56, 40, 41, 43, 42, 46, 47, 45, 44, 36, 37, 39, 38, 34, 35, 33, 32, 96, 97, 99, 98, 102, 103, 101

Offset: 0

N. J. A. Sloane

Inverse of sequence A006068 considered as a permutation of the nonnegative integers, i.e., A006068(A003188(n)) = n = A003188(A006068(n)). - Howard A. Landman, Sep 25 2001
Restricts to a permutation of each {2^(i - 1) .. 2^i - 1}. - Jason Kimberley, Apr 02 2012
a(n) mod 2 = floor(((n + 1) mod 4) / 2), see also A021913. - Reinhard Zumkeller, Apr 28 2012
Invented by Emile Baudot (1845-1903), originally called a "cyclic-permuted" code. Gray codes are named after Frank Gray, who patented their use for shaft encoders in 1953. [F. Gray, "Pulse Code Communication", U.S. Patent 2,632,058, March 17, 1953.] - Robert G. Wilson v, Jun 22 2014
For n >= 2, let G_n be the graph whose vertices are labeled as 0,1,...,2^n-1, and two vertices are adjacent if and only if their binary expansions differ in exactly one bit, then a(0),a(1),...,a(2^n-1),a(0) is a Hamilton cycle in G_n. - Jianing Song, Jun 01 2022

			For n = 13, the binary reflected Gray code representation of n is '1011' and 1011_2 = 11_10. So, a(13) = 11. - _Indranil Ghosh_, Jan 23 2017

M. Gardner, Mathematical Games, Sci. Amer. Vol. 227 (No. 2, Feb. 1972), p. 107.
M. Gardner, Knotted Doughnuts and Other Mathematical Entertainments. Freeman, NY, 1986, p. 15.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Indranil Ghosh, Table of n, a(n) for n = 0..32767 (first 1001 terms from N. J. A. Sloane)
M. W. Bunder, K. P. Tognetti, and G. E. Wheeler, On binary reflected Gray codes and functions, Discr. Math., 308 (2008), 1690-1700.
Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint 2016.
Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, p. 39.
P. Mathonet, M. Rigo, M. Stipulanti and N. Zénaïdi, On digital sequences associated with Pascal's triangle, arXiv:2201.06636 [math.NT], 2022.
J. A. Oteo and J. Ros, A Fractal Set from the Binary Reflected Gray Code, J. Phys. A: Math Gen. 38 (2005) 8935-8949.
Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003.
Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003 [Cached copy, with permission (pdf only)]
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Paul Tarau, Isomorphic Data Encodings and their Generalization to Hylomorphisms on Hereditarily Finite Data Types
Pierluigi Vellucci and Alberto Maria Bersani, Ordering of nested square roots of 2 according to Gray code, arXiv:1604.00222 [math.NT], 2016.
Index entries for sequences that are permutations of the natural numbers

a(2*A003714(n)) = 3*A003714(n) for all n. - Antti Karttunen, Apr 26 1999
Cf. A014550 (in binary), A055975 (first differences), A048724 (even bisection), A065621 (odd bisection).
Cf. A006068, A038554, A048641, A048642.

C
```
int a(int n) { return n ^ (n>>1); }
```

import Data.Bits (xor, shiftR)
a003188 n = n `xor` (shiftR n 1) :: Integer
-- Reinhard Zumkeller, May 26 2013, Apr 28 2012

// A recursive algorithm
N := 10; s := [[]];
for n in [1..N] do
for j in [#s..1 by -1] do
   Append(~s,Append(s[j],1));
   Append(~s[j],0);
end for;
end for;
[SequenceToInteger(b,2):b in s]; // Jason Kimberley, Apr 02 2012

// A direct algorithm
I2B := func< i | [b eq 1: b in IntegerToSequence(i,2)]>;
B2I := func< s | SequenceToInteger([b select 1 else 0:b in s],2)>;
[B2I(Xor(I2B(i),I2B(i div 2)cat[false])):i in [1..127]]; //Jason Kimberley, Apr 02 2012

with(combinat); graycode(6); # to produce first 64 terms
printf(cat(` %.6d`$64), op(map(convert, graycode(6), binary))); lprint(); # to produce binary strings
# alternative:
read("transforms"):
A003188 := proc(n)
    XORnos(n,floor(n/2)) ;
end proc: # R. J. Mathar, Mar 09 2015
# another Maple program:
a:= n-> Bits[Xor](n, iquo(n, 2)):
seq(a(n), n=0..70);  # Alois P. Heinz, Aug 16 2020

f[n_] := BitXor[n, Floor[n/2]]; Array[f, 70, 0] (* Robert G. Wilson v, Jun 09 2010 *)

PARI
```
a(n)=bitxor(n,n>>1);
```

a(n)=sum(k=1,n,(-1)^((k/2^valuation(k,2)-1)/2)*2^valuation(k,2))

def A003188(n):
    return int(bin(n^(n//2))[2:],2) # Indranil Ghosh, Jan 23 2017

def A003188(n): return n^ n>>1 # Chai Wah Wu, Jun 29 2022

maxn <- 63 # by choice
a <- 1
for(n in 1:maxn){ a[2*n  ] <- 2*a[n] + (n%%2 != 0)
                  a[2*n+1] <- 2*a[n] + (n%%2 == 0)}
(a <- c(0,a))
# Yosu Yurramendi, Apr 10 2020
(C#)
static uint a(this uint n) => (n >> 1) ^ n; // Frank Hollstein, Mar 12 2021

a(n) = 2*a(floor(n/2)) + A021913(n - 1). - Henry Bottomley, Apr 05 2001
a(n) = n XOR floor(n/2), where XOR is the binary exclusive OR operator. - Paul D. Hanna, Jun 04 2002
G.f.: (1/(1-x)) * Sum_{k>=0} 2^k*x^2^k/(1 + x^2^(k+1)). - Ralf Stephan, May 06 2003
a(0) = 0, a(2n) = 2a(n) + [n odd], a(2n + 1) = 2a(n) + [n even]. - Ralf Stephan, Oct 20 2003
a(0) = 0, a(n) = 2 a(floor(n/2)) + mod(floor((n + 1)/2), 2).
a(n) = Sum_{k=1..n} 2^A007814(k) * (-1)^((k/2^A007814(k) - 1)/2). - Ralf Stephan, Oct 29 2003
a(0) = 0, a(n + 1) = a(n) XOR 2^A007814(n) - Jaume Simon Gispert (jaume(AT)nuem.com), Sep 11 2004
Inverse of sequence A006068. - Philippe Deléham, Apr 29 2005
a(n) = a(n-1) XOR A006519(n). - Franklin T. Adams-Watters, Jul 18 2011
From Mikhail Kurkov, Sep 27 2023: (Start)
a(2^m + k) = a(2^m - k - 1) + 2^m for 0 <= k < 2^m, m >= 0.
a(n) = a(A053645(A054429(n))) + A053644(n) for n > 0.
a(n) = A063946(a(A053645(n)) + A053644(n)) for n > 0. (End)

A006068 a(n) is Gray-coded into n.

Original entry on oeis.org

0, 1, 3, 2, 7, 6, 4, 5, 15, 14, 12, 13, 8, 9, 11, 10, 31, 30, 28, 29, 24, 25, 27, 26, 16, 17, 19, 18, 23, 22, 20, 21, 63, 62, 60, 61, 56, 57, 59, 58, 48, 49, 51, 50, 55, 54, 52, 53, 32, 33, 35, 34, 39, 38, 36, 37, 47, 46, 44, 45, 40, 41, 43, 42, 127, 126, 124, 125, 120, 121

Offset: 0

N. J. A. Sloane

Equivalently, if binary expansion of n has m bits (say), compute derivative of n (A038554), getting sequence n' of length m-1; sort on n'.
Inverse of sequence A003188 considered as a permutation of the nonnegative integers, i.e., a(A003188(n)) = n = A003188(a(n)). - Howard A. Landman, Sep 25 2001
The sequence exhibits glide reflections that grow fractally. These show up well on the scatterplot, also audibly using the "listen" link. - Peter Munn, Aug 18 2019
Each bit at bit-index k (counted from the right hand end, with the least significant bit having bit-index 0) in the binary representation of a(n) is the parity of the number of 1's among the bits of the binary representation of n that have a bit-index of k or higher. - Frederik P.J. Vandecasteele, May 26 2025

			The first few values of n' are -,-,1,0,10,11,01,00,100,101,111,110,010,011,001,000,... (for n=0..15) and to put these in lexicographic order we must take n in the order 0,1,3,2,7,6,4,5,15,14,12,13,8,9,11,10,...

M. Gardner, Mathematical Games, Sci. Amer. Vol. 227 (No. 2, Feb. 1972), p. 107.
M. Gardner, Knotted Doughnuts and Other Mathematical Entertainments. Freeman, NY, 1986, p. 15.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1023
Joerg Arndt, Matters Computational (The Fxtbook), section 1.16 Gray code, C code inverse_gray_code() with "version 3" giving the PARI code here.
Dana G. Korssjoen, Biyao Li, Stefan Steinerberger, Raghavendra Tripathi, and Ruimin Zhang, Finding structure in sequences of real numbers via graph theory: a problem list, arXiv:2012.04625, Dec 08, 2020
Eric Rowland and Reem Yassawi, Profinite automata, arXiv:1403.7659 [math.DS], 2014. See p. 22.
Paul Tarau, Isomorphic Data Encodings and their Generalization to Hylomorphisms on Hereditarily Finite Data Types; see also alternate link.
Index entries for sequences that are permutations of the natural numbers

Cf. A038554, A005811, A003188, A014550, A003100, A265263, A377404.
Cf. A054429, A180200. - Reinhard Zumkeller, Aug 15 2010
Cf. A000079, A055975 (first differences), A209281 (binary weight).
A003987, A010060 are used to express relationship between terms of this sequence.

a006068 n = foldl xor 0 $
                  map (div n) $ takeWhile (<= n) a000079_list :: Integer
-- Reinhard Zumkeller, Apr 28 2012

a:= proc(n) option remember; `if`(n<2, n,
      Bits[Xor](n, a(iquo(n, 2))))
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Apr 17 2018

a[n_] := BitXor @@ Table[Floor[n/2^m], {m, 0, Floor[Log[2, n]]}]; a[0]=0; Table[a[n], {n, 0, 69}] (* Jean-François Alcover, Jul 19 2012, after Paul D. Hanna *)
Table[Fold[BitXor, n, Quotient[n, 2^Range[BitLength[n] - 1]]], {n, 0, 70}] (* Jan Mangaldan, Mar 20 2013 *)

{a(n)=local(B=n);for(k=1,floor(log(n+1)/log(2)),B=bitxor(B,n\2^k));B} /* Paul D. Hanna, Jan 18 2012 */

/* the following routine needs only O(log_2(n)) operations */
a(n)= {
    my( s=1, ns );
    while ( 1,
        ns = n >> s;
        if ( 0==ns, break() );
        n = bitxor(n, ns);
        s <<= 1;
    );
    return ( n );
} /* Joerg Arndt, Jul 19 2012 */

def a(n):
    s=1
    while True:
        ns=n>>s
        if ns==0: break
        n=n^ns
        s<<=1
    return n
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 07 2017, after PARI code by Joerg Arndt

nmax <- 63 # by choice
a <- vector()
for(n in 1:nmax){
  ones <- which(as.integer(intToBits(n)) == 1)
  nbit <- as.integer(intToBits(n))[1:tail(ones, n = 1)]
  level <- 0; anbit <- nbit; anbit.s <- nbit
  while(sum(anbit.s) > 0){
    s <- 2^level; if(s > length(anbit.s)) break
    anbit.s <- c(anbit[-(1:s)], rep(0,s))
    anbit <- bitwXor(anbit, anbit.s)
    level <- level + 1
  }
  a <- c(a, sum(anbit*2^(0:(length(anbit) - 1))))
}
(a <- c(0,a))
# Yosu Yurramendi, Oct 12 2021, after PARI code by Joerg Arndt

a(n) = 2*a(ceiling((n+1)/2)) + A010060(n-1). If 3*2^(k-1) < n <= 2^(k+1), a(n) = 2^(k+1) - 1 - a(n-2^k); if 2^(k+1) < n <= 3*2^k, a(n) = a(n-2^k) + 2^k. - Henry Bottomley, Jan 10 2001
a(n) = n XOR [n/2] XOR [n/4] XOR [n/8] ... XOR [n/2^m] where m = [log(n)/log(2)] (for n>0) and [x] is integer floor of x. - Paul D. Hanna, Jun 04 2002
a(n) XOR [a(n)/2] = n. - Paul D. Hanna, Jan 18 2012
A066194(n) = a(n-1) + 1, n>=1. - Philippe Deléham, Apr 29 2005
a(n) = if n<2 then n else 2*m + (n mod 2 + m mod 2) mod 2, with m=a(floor(n/2)). - Reinhard Zumkeller, Aug 10 2010
a(n XOR m) = a(n) XOR a(m), where XOR is the bitwise exclusive-or operator, A003987. - Peter Munn, Dec 14 2019
a(0) = 0. For all n >= 0 if a(n) is even a(2*n) = 2*a(n), a(2*n+1) = 2*a(n)+1, else a(2*n) = 2*a(n)+1, a(2*n+1) = 2*a(n). - Yosu Yurramendi, Oct 12 2021
Conjecture: a(n) = a(A053645(A063946(n))) + A053644(n) for n > 0 with a(0) = 0. - Mikhail Kurkov, Sep 09 2023
a(n) = 2*A265263(n) - 2*A377404(n) - A010060(n). - Alan Michael Gómez Calderón, Jun 26 2025

More terms from Henry Bottomley, Jan 10 2001

A014601 Numbers congruent to 0 or 3 mod 4.

Original entry on oeis.org

0, 3, 4, 7, 8, 11, 12, 15, 16, 19, 20, 23, 24, 27, 28, 31, 32, 35, 36, 39, 40, 43, 44, 47, 48, 51, 52, 55, 56, 59, 60, 63, 64, 67, 68, 71, 72, 75, 76, 79, 80, 83, 84, 87, 88, 91, 92, 95, 96, 99, 100, 103, 104, 107, 108, 111, 112, 115, 116, 119, 120, 123, 124

Offset: 0

Eric Rains (rains(AT)caltech.edu)

Discriminants of orders in imaginary quadratic fields (negated). [Comment corrected by Christopher E. Thompson, Dec 11 2016]
Numbers such that Langford-Skolem problem has a solution - see A014552.
Complement of A042963. - Reinhard Zumkeller, Oct 04 2004
Also called skew amenable numbers; a number k is skew amenable if there exist a set {a(i)} of integers satisfying the relations k = Sum_{i=1..k} a(i) = -Product_{i=1..k} a(i). Thus we have 8 = 1 + 1 + 1 + 1 + 1 + 1 - 2 + 4 = -(1*1*1*1*1*1*(-2)*4). - Lekraj Beedassy, Jan 07 2005
Possible nonpositive discriminants of quadratic equation a*x^2 + b*x + c or discriminants of binary quadratic forms a*x^2 + b*x*y + c*y^2. - Artur Jasinski, Apr 28 2008
Also, disregarding the 0 term, positive integers m such that, equivalently,
  (i) +-1 +-2 +-... +-m is even for all choices of signs,
  (ii) +-1 +-2 +-... +-m = 0 for some choices of signs,
  (iii) for all -m <= k <= m, k = +-1 +-2 +-... +-(k-1) +-(k+1) +-(k+2) +-... +-m for at least one choice of signs. - Rick L. Shepherd, Oct 29 2008
A145768(a(n)) is even. - Reinhard Zumkeller, Jun 05 2012
Multiples of 4 interleaved with 1 less than multiples of 4. - Wesley Ivan Hurt, Nov 08 2013
((2*k+0) + (2*k+1) + ... + (2*k+m-1) + (2*k+m)) is even if and only if m = a(n) for some n where k is any nonnegative integer. - Gionata Neri, Jul 24 2015
Numbers whose binary reflected Gray code (A014550) ends with 0. - Amiram Eldar, May 17 2021

			G.f. = 3*x + 4*x^2 + 7*x^3 + 8*x^4 + 11*x^5 + 12*x^6 + 15*x^7 + 16*x^8 + ...

H. Cohen, Course in Computational Alg. No. Theory, Springer, 1993, pp. 514-5.
A. Scholz and B. Schoeneberg, Einführung in die Zahlentheorie, 5. Aufl., de Gruyter, Berlin, New York, 1973, p. 108.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
S. F. Barger, Solution to problem 10454: Amenable Numbers, Amer. Math. Monthly, Vol. 105, No. 4 (April 1998), p. 368.
Steven R. Finch, Class number theory [Cached copy, with permission of the author]
Heiko Harborth, Solution of Steinhaus's problem with plus and minus signs, Journal of Combinatorial Theory, Series A, Volume 12, Issue 2 (March 1972), Pages 253-259.
Mickaël Launay, Les routes de Numland, L’énigme maths du "Monde" n°2. In French.
Michael Penn, The most beautiful arrangement of numbers, YouTube video, 2024.
Rick L. Shepherd, Binary quadratic forms and genus theory, Master of Arts Thesis, University of North Carolina at Greensboro, 2013.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A004676, A014550, A042948, A079896.

Cf. A274406. - Bruno Berselli, Jun 26 2016

a014601 n = a014601_list !! n
a014601_list = [x | x <- [0..], mod x 4 `elem` [0, 3]]
-- Reinhard Zumkeller, Jun 05 2012

[n: n in [0..200]|n mod 4 in {0,3}]; // Vincenzo Librandi, Dec 24 2010

A014601:=n->3*n-2*floor(n/2); seq(A014601(k), k=0..100); # Wesley Ivan Hurt, Nov 08 2013

aa = {}; Do[Do[Do[d = b^2 - 4 a c; If[d <= 0, AppendTo[aa, -d]], {a, 0, 50}], {b, 0, 50}], {c, 0, 50}]; Union[aa] (* Artur Jasinski, Apr 28 2008 *)
Select[Range[0, 124], Or[Mod[#, 4] == 0, Mod[#, 4] == 3] &] (* Ant King, Nov 18 2010 *)
CoefficientList[Series[2 x/(1 - x)^2 + (1/(1 - x) + 1/(1 + x)) x/2, {x, 0, 100}], x] (* Vincenzo Librandi, May 18 2014 *)
a[ n_] := 2 n + Mod[n, 2]; (* Michael Somos, Jul 24 2015 *)

{a(n) = 2*n + n%2}; /* Michael Somos, Dec 27 2010 */

a(n) = (n + 1)*2 + 1 - n mod 2. - Reinhard Zumkeller, Apr 21 2003
A014494(n) = A000217(a(n)). - Reinhard Zumkeller, Oct 04 2004
a(n) = Sum_{k=1..n} (2 - (-1)^k). - William A. Tedeschi, Mar 20 2008
A139131(a(n)) = A078636(a(n)). - Reinhard Zumkeller, Apr 10 2008
From R. J. Mathar, Sep 25 2009: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3) for n > 2.
G.f.: x*(3+x)/((1+x)*(x-1)^2). (End)
a(n) = 2*n + (n mod 2). - Paolo Valzasina (p.valzasina(AT)gmail.com), Nov 24 2009
a(n) = (4*n - (-1)^n + 1)/2. - Bruno Berselli, Oct 06 2010
a(n) = 4*n - a(n-1) - 1 (with a(0) = 0). - Vincenzo Librandi, Dec 24 2010
a(n) = -A042948(-n) for all n in Z. - Michael Somos, Dec 27 2010
G.f.: 2*x / (1 - x)^2 + (1 / (1 - x) + 1 / (1 + x)) * x/2. - Michael Somos, Dec 27 2010
a(n) = Sum_{k>=0} A030308(n,k)*b(k) with b(0) = 3 and b(k) = 2^(k+1) for k > 0. - Philippe Deléham, Oct 17 2011
a(n) = ceiling((4/3)*ceiling(3*n/2)). - Clark Kimberling, Jul 04 2012
a(n) = 3n - 2*floor(n/2). - Wesley Ivan Hurt, Nov 08 2013
a(n) = A042948(n+1) - 1 for all n in Z. - Michael Somos, Jul 24 2015
a(n) + a(n+1) = A004767(n) for all n in Z. - Michael Somos, Jul 24 2015
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*log(2)/4 - Pi/8. - Amiram Eldar, Dec 05 2021
E.g.f.: ((4*x + 1)*exp(x) - exp(-x))/2. - David Lovler, Aug 04 2022

A042963 Numbers congruent to 1 or 2 mod 4.

Original entry on oeis.org

1, 2, 5, 6, 9, 10, 13, 14, 17, 18, 21, 22, 25, 26, 29, 30, 33, 34, 37, 38, 41, 42, 45, 46, 49, 50, 53, 54, 57, 58, 61, 62, 65, 66, 69, 70, 73, 74, 77, 78, 81, 82, 85, 86, 89, 90, 93, 94, 97, 98, 101, 102, 105, 106, 109, 110, 113, 114, 117, 118, 121, 122, 125, 126, 129, 130, 133, 134, 137, 138

Offset: 1

N. J. A. Sloane

Complement of A014601. - Reinhard Zumkeller, Oct 04 2004
Let S(x) = (1, 2, 2, 2, ...). Then A042963 = ((S(x))^2 + S(x^2))/2 = ((1, 4, 8, 12, 16, 20, ...) + (1, 0, 2, 0, 2, 0, 2, ...))/2 = (1, 2, 5, 6, 9, 10, ...). - Gary W. Adamson, Jan 03 2011
(a(n)*(a(n) + 1 + 4*k))/2 is odd, for k >= 0. - Gionata Neri, Jul 19 2015
Equivalent to the following variation on Fermat's Diophantine m-tuple: 1 + the product of any two distinct terms is not a square; this sequence, which we'll call sequence S, is produced by the following algorithm. At the start, S is initially empty. At stage n, starting at n = 1, the algorithm checks whether there exists a number m already in the sequence, such that mn+1 is a perfect square. If such a number m is found, then n is not added to the sequence; otherwise, n is added. Then n is incremented to n + 1, and we repeat the procedure. Proof by Clark R. Lyons: We prove by strong induction that n is in the sequence S if and only if n == 1 (mod 4) or n == 2 (mod 4). Suppose now that this holds for all k < n. In case 1, either n == 1 (mod 4) or n == 2 (mod 4), and we wish to show that n does indeed enter the sequence S. That is, we wish to show that there does not exist m < n, already in the sequence at this point such that mn+1 is a square. By the inductive hypothesis m == 1 (mod 4) or m == 2 (mod 4). This means that both m and n are one of 1, 2, 5, or 6 mod 8. Using a multiplication table mod 8, we see that this implies mn+1 is congruent to one of 2, 3, 5, 6, or 7 mod 8. But we also see that mod 8, a perfect square is congruent to 0, 1, or 4. Thus mn+1 is not a perfect square, so n is added to the sequence. In case 2, n == 0 (mod 4) or n == 3 (mod 4), and we wish to show that n is not added to the sequence. That is, we wish to show that there exists m < n already in the sequence such that mn+1 is a perfect square. For this we let m = n - 2, which is positive since n >= 3. By the inductive hypothesis, since m == 1 (mod 4) or m == 2 (mod 4) and m < n, m is already in the sequence. And we have m*n + 1 = (n - 2)*n + 1 = n^2 - 2*n + 1 = (n - 1)^2, so mn+1 is indeed a perfect square, and so n is not added to the sequence. Thus n is added to the sequence if and only if n == 1 (mod 4) or n == 2 (mod 4). This completes the proof. - Robert C. Lyons, Jun 30 2016
Also the number of maximal cliques in the (n + 1) X (n + 1) black bishop graph. - Eric W. Weisstein, Dec 01 2017
Lexicographically earliest sequence of distinct positive integers such that the average of any two or more consecutive terms is never an integer. (For opposite property see A005408.) - Ivan Neretin, Dec 21 2017
Numbers whose binary reflected Gray code (A014550) ends with 1. - Amiram Eldar, May 17 2021
Also: append its negated last bit to n-1. - M. F. Hasler, Oct 17 2022

David Lovler, Table of n, a(n) for n = 1..10000 (first 1000 terms from Reinhard Zumkeller)
Eric Weisstein's World of Mathematics, Black Bishop Graph.
Eric Weisstein's World of Mathematics, Maximal Clique.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A153284 (first differences), A014848 (partial sums).
Cf. A014550, A046712 (subsequence).
Union of A016813 and A016825.

a042963 n = a042963_list !! (n-1)
a042963_list = [x | x <- [0..], mod x 4 `elem` [1,2]]
-- Reinhard Zumkeller, Feb 14 2012

[ n : n in [1..165] | n mod 4 eq 1 or n mod 4 eq 2 ]; // Vincenzo Librandi, Jan 25 2011

A046923:=n->(n mod 2) + 2n - 2; seq(A046923(n), n=1..100); # Wesley Ivan Hurt, Oct 10 2013

Select[Range[109], Or[Mod[#, 4] == 1, Mod[#, 4] == 2] &] (* Ant King, Nov 17 2010 *)
Table[(4 n - 3 - (-1)^n)/2, {n, 20}] (* Eric W. Weisstein, Dec 01 2017 *)
LinearRecurrence[{1, 1, -1}, {1, 2, 5}, 20] (* Eric W. Weisstein, Dec 01 2017 *)
CoefficientList[Series[(1 + x + 2 x^2)/((-1 + x)^2 (1 + x)), {x, 0, 20}], x] (* Eric W. Weisstein, Dec 01 2017 *)

a(n)=2*n-1-(n-1)%2 \\ Jianing Song, Oct 06 2018; adapted to offset by Michel Marcus, Sep 09 2022

apply( A042963(n)=n*2-2+n%2, [1..99]) \\ M. F. Hasler, Oct 17 2022

a(n) = 1 + A042948(n-1). [Corrected by Jianing Song, Oct 06 2018]
From Michael Somos, Jan 12 2000: (Start)
G.f.: x*(1 + x + 2*x^2)/((1 - x)^2*(1 + x)).
a(n) = a(n-1) + 2 + (-1)^n, a(0) = 1. (End) [This uses offset 0. - Jianing Song, Oct 06 2018]
A014493(n) = A000217(a(n)). - Reinhard Zumkeller, Oct 04 2004, Feb 14 2012
a(n) = Sum_{k=0..n} (A001045(k) mod 4). - Paul Barry, Mar 12 2004
A145768(a(n)) is odd. - Reinhard Zumkeller, Jun 05 2012
a(n) = A005843(n-1) + A059841(n-1). - Philippe Deléham, Mar 31 2009 [Corrected by Jianing Song, Oct 06 2018]
a(n) = 4*n - a(n-1) - 5 for n > 1. [Corrected by Jerzy R Borysowicz, Jun 09 2023]
From Ant King, Nov 17 2010: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3).
a(n) = (4*n - 3 - (-1)^n)/2. (End)
a(n) = (n mod 2) + 2*n - 2. - Wesley Ivan Hurt, Oct 10 2013
A163575(a(n)) = n - 1. - Reinhard Zumkeller, Jul 22 2014
E.g.f.: 2 + (2*x - 1)*sinh(x) + 2*(x - 1)*cosh(x). - Ilya Gutkovskiy, Jun 30 2016
E.g.f.: 2 + (2*x - 1)*exp(x) - cosh(x). - David Lovler, Jul 19 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/8 + log(2)/4. - Amiram Eldar, Dec 05 2021

Offset corrected by Reinhard Zumkeller, Feb 14 2012

More terms by David Lovler, Jul 19 2022

cp's OEIS Frontend

A344346 Numbers k which have an odd number of trailing zeros in their binary reflected Gray code A014550(k).

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A373082 Number of runs of 0's in A014550(n).

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A000975 a(2n) = 2*a(2n-1), a(2n+1) = 2*a(2n)+1 (also a(n) is the n-th number without consecutive equal binary digits).

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A003188 Decimal equivalent of Gray code for n.

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A006068 a(n) is Gray-coded into n.

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A000975 a(2n) = 2a(2n-1), a(2n+1) = 2a(2n)+1 (also a(n) is the n-th number without consecutive equal binary digits).