A043291 -id:A043291 - 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).
Paul K. Stockmeyer, An Exploration of Sequence A000975, arXiv:1608.08245 [math.CO], 2016; Fib. Quart. 55 (2017) 174.
Eric Weisstein's World of Mathematics, Baguenaudier
A. K. Whitford, Binet's Formula Generalized, Fibonacci Quarterly, Vol. 15, No. 1, 1979, pp. 21, 24, 29.
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

A033001 Every run of digits of n in base 3 has length 2.

Original entry on oeis.org

4, 8, 36, 44, 72, 76, 328, 332, 396, 400, 652, 656, 684, 692, 2952, 2960, 2988, 2992, 3568, 3572, 3600, 3608, 5868, 5876, 5904, 5908, 6160, 6164, 6228, 6232, 26572, 26576, 26640, 26644, 26896, 26900, 26928, 26936, 32112, 32120

Offset: 1

Clark Kimberling

See A043291 for the base 2 version (which has a very simple formula), A033002 - A033014 for bases 4 through 16, A033015 - A033029 for the variants with runs of length >= 2. - M. F. Hasler, Feb 01 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..250

Select[Range[10000], Union[Length/@Split[IntegerDigits[#, 3]]]=={2}&] (* Vincenzo Librandi, Feb 05 2014 *)

is_A033001(n)=!until(!n\=9,bittest(4588304,n%27)||return)

for(n=1,9999,is_A033001(n)&&print1(n",")) \\ (End)

a(n) = my(v=binary(n+1)); v[1]=0; for(i=2,#v, v[i]+=(v[i]>=v[i-1])); 4*fromdigits(v,9); \\ Kevin Ryde, Mar 13 2021

a(n)=4*A043307(n). - M. F. Hasler, Feb 01 2014

A033015 Numbers whose base-2 expansion has no run of digits with length < 2.

Original entry on oeis.org

3, 7, 12, 15, 24, 28, 31, 48, 51, 56, 60, 63, 96, 99, 103, 112, 115, 120, 124, 127, 192, 195, 199, 204, 207, 224, 227, 231, 240, 243, 248, 252, 255, 384, 387, 391, 396, 399, 408, 412, 415, 448, 451, 455, 460, 463, 480, 483, 487, 496, 499, 504, 508, 511, 768

Offset: 1

Clark Kimberling

See A033016 and following for the variants in other bases, A043291 for run lengths equal to 2 (which has a very simple formula) and A033001 and following for the analog of the latter in other bases. - M. F. Hasler, Feb 01 2014
The number zero also satisfies the definition if we consider that its base-2 expansion is empty. - M. F. Hasler, Oct 06 2022
If we define row n as subset of terms with n bits, i.e., 2^(n-1) < a(k) < 2^n, then we get row n by duplicating the last bit (LSB) of the terms in row n-1 and appending twice the negated LSB to the terms in row n-2. This gives the FORMULA for the number of terms in row n. - M. F. Hasler, Oct 17 2022

			The first terms, written in binary, are: 11, 111, 1100, 1111, 11000, 11100, 11111, 110000, 110011, ...; cf. sequence A355280. - _M. F. Hasler_, Oct 06 2022

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A355280 (in binary).
Cf. A222813 (palindromes subsequence).
Cf. A033016, A043291.
See A033001 for further cross-references.

Select[Range[2000], Min[Length/@Split[IntegerDigits[#, 2]]]>1&] (* Vincenzo Librandi, Feb 05 2014 *)

is(n)=my(t); if(n%2, t=valuation(n+1,2); if(t==1,return(0)); n>>=t); while(n, t=valuation(n,2); if(t==1,return(0)); n>>=t; t=valuation(n+1,2); if(t==1,return(0)); n>>=t); 1 \\ Charles R Greathouse IV, Mar 29 2013

select( is_A033015(n)=!bitand(n=bitxor(n,n<<1),n<<1)&&bitand(n,3)!=2, [1..770]) \\ M. F. Hasler, Oct 06 2022 (replacing less efficient code from 2014)

{A033015_row(n)=if(n>3, setunion([x*2+x%2|x<-A033015_row(n-1)], [x*4+3-x%2*3|x<-A033015_row(n-2)]), n>1, [2^n-1], [])} \\ "Row" of n-digit terms. For (very) large n one could use memoization rather than this naive recursive definition.
concat(apply(A033015_row, [1..9])) \\ To get the "flattened" sequence. - M. F. Hasler, Oct 17 2022

from itertools import groupby
def ok(n): return all(len(list(g)) >= 2 for k, g in groupby(bin(n)[2:]))
print([i for i in range(1, 769) if ok(i)]) # Michael S. Branicky, Jan 04 2021

def A033015_row(n): # terms with n bits <=> in [2^(n-1) .. 2^n]
    return [[], [], [3], [7]][n] if n < 4 else sorted(
    [x*2+x%2 for x in A033015_row(n-1)] +
    [x*4+3-x%2*3 for x in A033015_row(n-2)]) # M. F. Hasler, Oct 17 2022
print(sum((A033015_row(n)for n in range(11)),[]))

The number of n-bit terms is Fibonacci(n-1) = A000045(n-1). - M. F. Hasler, Oct 17 2022

Extended by Ray Chandler, Dec 18 2009

A043307 a(n) = A033001(n)/4.

Original entry on oeis.org

1, 2, 9, 11, 18, 19, 82, 83, 99, 100, 163, 164, 171, 173, 738, 740, 747, 748, 892, 893, 900, 902, 1467, 1469, 1476, 1477, 1540, 1541, 1557, 1558, 6643, 6644, 6660, 6661, 6724, 6725, 6732, 6734, 8028, 8030, 8037, 8038, 8101, 8102, 8118, 8119

Offset: 1

Clark Kimberling

Also: Numbers which, written in base 9, have only digits 0, 1 or 2, and no two adjacent digits equal. - M. F. Hasler, Feb 03 2014

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A043308 - A043320, A043291, A033001 - A033014, A033016 - A033029.

A[1]:= [1,2]:
for d from 2 to 6 do
  A[d]:= map(t -> seq(9*t+j,j=subs(t mod 9 = NULL, [0,1,2])), A[d-1])
od:
seq(op(A[d]),d=1..6); # Robert Israel, Jan 29 2017

Table[FromDigits[#,9]&/@Select[Tuples[{0,1,2},n],Min[Abs[Differences[#]]]>0&],{n,2,5}]// Flatten// Union (* Harvey P. Dale, May 27 2023 *)

is_A043307(n)=(n=[n])&&!until(!n[1],((n=divrem(n[1],9))[2]<3 && n[1]%3!=n[2])||return) \\ M. F. Hasler, Feb 03 2014

a(n) = my(v=binary(n+1)); v[1]=0; for(i=2,#v, v[i]+=(v[i]>=v[i-1])); fromdigits(v,9); \\ Kevin Ryde, Mar 13 2021

From Robert Israel, Jan 29 2017: (Start)
If a(n) == 0 (mod 3) then a(2*n+1) = 9*a(n) + 1 else a(2*n+1) = 9*a(n).
If a(n) == 2 (mod 3) then a(2*n+2) = 9*a(n) + 1 else a(2*n+1) = 9*a(n)+2.
a(4k+5) = 9*a(2k+2).
(End)

A043320 Numbers which, written in base 256, have all digits less than 16 and no two adjacent digits equal.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 256, 258, 259, 260, 261, 262, 263, 264, 265, 266, 267, 268, 269, 270, 271, 512, 513, 515, 516, 517, 518, 519, 520, 521, 522, 523, 524, 525, 526, 527, 768, 769, 770, 772, 773, 774

Offset: 1

Clark Kimberling

Sequence A033014 consists of the numbers that have all base 16 digits repeated *exactly* twice. (This is equivalent to say that the base-256 digits are 0x00, 0x11, 0x22,... or 0xFF, in hex notation, and no two adjacent base-256 digits are equal.) Thus, these numbers are divisible by 0x11 = 17, and the result of the division is a number which has no other base-256 digits than 0x00, 0x01,... or 0x0F, and no two adjacent digits equal. Conversely, it is clear that exactly these numbers are terms of A033014 when multiplied by 17 = 0x11. - M. F. Hasler, Feb 05 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..1800

Cf. A043307 - A043319, A043291, A033001 - A033014, A033015 - A033029.

Select[Range[20000], Union[Length/@Split[IntegerDigits[#, 16]]]=={2}&]/17 (* Vincenzo Librandi, Feb 06 2014 *)

is_A043320(n)={(n=[n])&&!until(!n[1], ((n=divrem(n[1], 256))[2]<16 && n[1]%16!=n[2])||return)} \\ M. F. Hasler, Feb 03 2014

from itertools import count, islice, groupby
def A043320_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:set(len(list(g)) for k, g in groupby(hex(17*n)[2:]))=={2},count(max(startvalue,1)))
A043320_list = list(islice(A043320_gen(),20)) # Chai Wah Wu, Mar 10 2023

a(n) = A033014(n)/17. [This was initially the definition of the sequence. - M. F. Hasler, Feb 03 2014]

New definition by M. F. Hasler, Feb 03 2014

A140690 A positive integer n is included if n written in binary can be subdivided into a number of runs all of equal-length, the first run from the left consisting of all 1's, the next run consisting of all 0's, the next run consisting of all 1's, the next run consisting of all 0's, etc.

Original entry on oeis.org

1, 2, 3, 5, 7, 10, 12, 15, 21, 31, 42, 51, 56, 63, 85, 127, 170, 204, 240, 255, 341, 455, 511, 682, 819, 992, 1023, 1365, 2047, 2730, 3276, 3640, 3855, 4032, 4095, 5461, 8191, 10922, 13107, 16256, 16383, 21845, 29127, 31775, 32767, 43690, 52428, 61680

Offset: 1

Leroy Quet, Jul 11 2008

Also: numbers of the form (2^s-1)*[4^{s*(k+1)}-1]/(4^s-1) or 2^s(2^s-1)*[4^{s*(k+1)}-1]/(4^s-1), s>=1, k>=0. Subsequences are, with the possible exception of terms at n=0, A002450(n), A043291(n), A015565(2n), A093134(2n+1), A000225(n), A020522(n). [R. J. Mathar, Aug 04 2008]
From Emeric Deutsch, Jan 25 2018: (Start)
Also the indices of the compositions having equal parts.
We define the index of a composition to be the positive integer whose binary form has run-lengths (i.e. runs of 1's, runs of 0's, etc., from left to right) equal to the parts of the composition. Example: the composition [1,1,3,1] has index 46 since the binary form of 46 is 101110. The integer 992 is in the sequence since its binary form is 1111100000 and the composition [5,5] has equal parts. The integer 100 is not in the sequence since its binary form is 1100100 and the composition [2,2,1,2] does not have equal parts.
The command c(n) from the Maple program yields the composition having index n. (End)

			819 in binary is 1100110011. The runs of 0's and 1's are (11)(00)(11)(00)(11). Each run (alternating 1's and 0's) is the same length. So 819 is in the sequence.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for sequences related to binary expansion of n

Cf. A101211, A298644.

import Data.Set (singleton, deleteFindMin, insert)
a140690 n = a140690_list !! (n-1)
a140690_list = f $ singleton (1, 1, 2) where
   f s | k == 1 = m : f (insert (2*b-1, 1, 2*b) $ insert (b*m, k+1, b) s')
       | even k    = m : f (insert (b*m+b-1, k+1, b) s')
       | otherwise = m : f (insert (b*m, k+1, b) s')
       where ((m, k, b), s') = deleteFindMin s
-- Reinhard Zumkeller, Feb 21 2014

Runs := proc (L) local j, r, i, k: j := 1: r[j] := L[1]: for i from 2 to nops(L) do if L[i] = L[i-1] then r[j] := r[j], L[i] else j := j+1: r[j] := L[i] end if end do: [seq([r[k]], k = 1 .. j)] end proc: RunLengths := proc (L) map(nops, Runs(L)) end proc: c := proc (n) ListTools:-Reverse(convert(n, base, 2)): RunLengths(%) end proc: A := {}: for n to 62000 do if nops(convert(c(n), set)) = 1 then A := `union`(A, {n}) else  end if end do: A; # most of the Maple program is due to W. Edwin Clark. - Emeric Deutsch, Jan 25 2018

Select[Range[62000],Length[Union[Length/@Split[IntegerDigits[#,2]]]]==1&] (* Harvey P. Dale, Mar 22 2012 *)

Terms beyond 42 from R. J. Mathar, Aug 04 2008

A077854 Expansion of 1/((1-x)(1-2x)*(1+x^2)).

Original entry on oeis.org

1, 3, 6, 12, 25, 51, 102, 204, 409, 819, 1638, 3276, 6553, 13107, 26214, 52428, 104857, 209715, 419430, 838860, 1677721, 3355443, 6710886, 13421772, 26843545, 53687091, 107374182, 214748364, 429496729, 858993459, 1717986918, 3435973836, 6871947673

Offset: 0

N. J. A. Sloane, Nov 17 2002

Partial sums of A007910. - Mircea Merca, Dec 27 2010
This is the decimal representation of the middle column of "Rule 54" elementary cellular automaton. - Karl V. Keller, Jr., Sep 26 2021
This same sequence (except that the offset is changed to 4) is 2^n with the final digit chopped off. - J. Lowell, May 11 2022

			The sequence in hexadecimal shows the pattern
1, 3, 6, c,
19, 33, 66, cc,
199, 333, 666, ccc,
1999, 3333, 6666, cccc,
19999, 33333, 66666, ccccc,
199999, 333333, 666666, cccccc,
1999999, 3333333, 6666666, ccccccc,
19999999, 33333333, 66666666, cccccccc,
... - _Armands Strazds_, Oct 09 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,3,-2).

Equals A007909(n+3) - [n congruent 2, 3 mod 4].

Cf. A130306, A043291 (subsequence); A000975, A007910, A133872, A259661 (binary).

import Data.Bits (xor)
a077854 n = a077854_list !! n
a077854_list = scanl1 xor $ tail a000975_list :: [Integer]
-- Reinhard Zumkeller, Jan 04 2013

[Round((2^(n+4)-5)/10): n in [0..40]]; // Vincenzo Librandi, Jun 25 2011

a := proc(n) option remember; if n=0 then RETURN(1); fi; if n=1 then RETURN(3); fi; if n=2 then RETURN(6); fi; if n=3 then RETURN(12); fi; 3*a(n-1)-3*a(n-2)+3*a(n-3)-2*a(n-4); end;
seq(iquo(2^n,5),n=3..35); # Zerinvary Lajos, Apr 20 2008

CoefficientList[Series[1/((1 - x) (1 - 2 x) (1 + x^2)), {x, 0, 32}], x] (* Michael De Vlieger, Mar 29 2016 *)
LinearRecurrence[{3,-3,3,-2},{1,3,6,12},40] (* Harvey P. Dale, Feb 06 2019 *)

a(n)=(16<Charles R Greathouse IV, Sep 23 2012

Vec(1/(1-3*x+3*x^2-3*x^3+2*x^4)+O(x^99)) \\ Derek Orr, Oct 26 2014

print([2**(n+3)//5 for n in range(50)]) # Karl V. Keller, Jr., Sep 26 2021

a(n) = 3*a(n-1) - 3*a(n-2) + 3*a(n-3) - 2*a(n-4), with initial values a(0) = 1, a(1) = 3, a(2) = 6, a(3) = 12.
a(n) = (1/10)*(2^(n+4) + (-1)^floor(n/2) - 2*(-1)^floor((n+1)/2) - 5).
Row sums of A130306. - Gary W. Adamson, May 20 2007
a(n) = floor(2^(n+3)/5). - Gary Detlefs, Sep 06 2010
a(n) = round((2^(n+4)-5)/10) = floor((2^(n+3)-1)/5) = ceiling((2^(n+3)-4)/5) = round((2^(n+3)-2)/5); a(n) = a(n-4) + 3*2^(n-1), n > 3. - Mircea Merca, Dec 27 2010
a(n) = 2^(n+1) - 1 - a(n-2); a(n) = a(n-1)/2 for n == 2, 3 (mod 4); a(n) = (a(n-1)-1)/2 for n == 0, 1 (mod 4). - Arie Bos, Apr 06 2013
a(n) = floor(A000975(n+2)*3/5). - Armands Strazds, Oct 18 2014
a(n) = Sum_{k=1..n+3} floor(1 + sin(k*Pi/2 + 3*Pi/4))*2^(n-k+3). - Andres Cicuttin, Mar 28 2016
a(n) = (-15 + 3*2^(3+n) + 2^(1 + n - 4*floor((1+n)/4)) + 2^(2 + n - 4*floor((2+n)/4)))/15. - Andres Cicuttin, Mar 28 2016
a(n) = (16*2^n+(-1)^((2*n-1+(-1)^n)/4)-2*(-1)^((2*n+1-(-1)^n)/4)-5)/10. - Wesley Ivan Hurt, Apr 01 2016

A033114 Base-4 digits are, in order, the first n terms of the periodic sequence with initial period 1,0.

Original entry on oeis.org

1, 4, 17, 68, 273, 1092, 4369, 17476, 69905, 279620, 1118481, 4473924, 17895697, 71582788, 286331153, 1145324612, 4581298449, 18325193796, 73300775185, 293203100740, 1172812402961, 4691249611844, 18764998447377, 75059993789508

Offset: 1

Clark Kimberling

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Thomas Baruchel, Properties of the cumulated deficient binary digit sum, arXiv:1908.02250 [math.NT], 2019.
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 p. 17.
Index entries for linear recurrences with constant coefficients, signature (4,1,-4).

Cf. A015521, A043291, A117616.

[Round((8*4^n-5)/30): n in [1..30]]; // Vincenzo Librandi, Jun 25 2011

seq(floor((4^(n+1)-1)/15),n=1..25) # Mircea Merca, Dec 28 2010

Join[{a=1,b=4},Table[c=3*b+4*a+1;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 17 2011 *)

a(n) = floor(4^(n+1)/15) = 4^(n+1)/15 - 1/6 - (-1)^n/10. - Benoit Cloitre, Apr 18 2003
G.f.: 1/((1-x)*(1+x)*(1-4*x)); a(n) = 3*a(n-1) + 4*a(n-2)+1. Partial sum of A015521. - Paul Barry, Nov 12 2003
a(n) = Sum_{k=0..floor(n/2)} 4^(n-2*k); a(n) = Sum_{k=0..n} Sum_{j=0..k} (-1)^(j+k)*4^j. - Paul Barry, Nov 12 2003
Convolution of A000302 and A059841 (4^n and periodic{1, 0}). a(n) = Sum_{k=0..n} (1 + (-1)^(n-k))*4^k/2. - Paul Barry, Jul 19 2004
a(n) = Sum_{k=0..n} (-1)^(n-k)*(J(2*k+1)-1)/2, J(n)=A001045(n). - Paul Barry, Mar 06 2008
a(n) = round((8*4^n-5)/30) = ceiling((4*4^n-4)/15) = round((4*4^n-4)/15); a(n) = a(n-2) + 4^(n-1), n > 1. - Mircea Merca, Dec 28 2010
a(n) = A117616(n)/2. - J. M. Bergot, Apr 22 2015
a(n) = A043291(n)/3; a(n+1) = 4*a(n) + A000035(n). - Robert Israel, Apr 22 2015
a(n)+a(n+1) = A002450(n+1). - R. J. Mathar, Feb 27 2019

A153435 Numbers with 2n binary digits where every run length is 2, written in binary.

Original entry on oeis.org

11, 1100, 110011, 11001100, 1100110011, 110011001100, 11001100110011, 1100110011001100, 110011001100110011, 11001100110011001100, 1100110011001100110011, 110011001100110011001100

Offset: 1

Omar E. Pol, Dec 26 2008

A043291 written in base 2.

			n ... a(n) ....... A043291(n)
1 ... 11 ............. 3
2 ... 1100 .......... 12
3 ... 110011 ........ 51
4 ... 11001100 ..... 204
5 ... 1100110011 ... 819

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (100,1,-100).

Cf. A043291.

A153435:=n->(-101-99*(-1)^n+2^(3+2*n)*25^(1+n))/1818; seq(A153435(n), n=1..20); # Wesley Ivan Hurt, Apr 19 2014

Table[(-101 - 99*(-1)^n + 2^(3 + 2*n)*25^(1 + n))/1818, {n, 20}] (* Wesley Ivan Hurt, Apr 19 2014 *)
CoefficientList[Series[11/((x - 1) (x + 1) (100 x - 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Apr 20 2014 *)

Vec(11*x / ((x-1)*(x+1)*(100*x-1)) + O(x^100)) \\ Colin Barker, Apr 19 2014

From Colin Barker, Apr 19 2014: (Start)
a(n) = (-101-99*(-1)^n+2^(3+2*n)*25^(1+n))/1818.
a(n) = 100*a(n-1)+a(n-2)-100*a(n-3).
G.f.: 11*x / ((x-1)*(x+1)*(100*x-1)).(End).

A033002 Every run of digits of n in base 4 has length 2.

Original entry on oeis.org

5, 10, 15, 80, 90, 95, 160, 165, 175, 240, 245, 250, 1285, 1290, 1295, 1440, 1445, 1455, 1520, 1525, 1530, 2565, 2570, 2575, 2640, 2650, 2655, 2800, 2805, 2810, 3845, 3850, 3855, 3920, 3930, 3935, 4000, 4005, 4015, 20560

Offset: 1

Clark Kimberling

See A043291 and A033001 through A033014 for the analog in other bases, A033015 - A033029 for the variants with run lengths >= 2. - M. F. Hasler, Feb 04 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..600

Select[Range[10000], Union[Length/@Split[IntegerDigits[#, 4]]]=={2}&] (* Vincenzo Librandi, Feb 05 2014 *)

a(n) = 5*A043308(n) (= 5*n for n<4). - M. F. Hasler, Feb 04 2014

cp's OEIS Frontend

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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A033001 Every run of digits of n in base 3 has length 2.

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A033015 Numbers whose base-2 expansion has no run of digits with length < 2.

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A043307 a(n) = A033001(n)/4.

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A043320 Numbers which, written in base 256, have all digits less than 16 and no two adjacent digits equal.

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A140690 A positive integer n is included if n written in binary can be subdivided into a number of runs all of equal-length, the first run from the left consisting of all 1's, the next run consisting of all 0's, the next run consisting of all 1's, the next run consisting of all 0's, etc.

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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).

A077854 Expansion of 1/((1-x)(1-2x)*(1+x^2)).