A087117 -id:A087117 - cp's OEIS frontend

A237882 Numbers k such that LR0(k) > LR1(k), where LR0(k) = A087117(k) is the length of the longest run of zeros in the binary representation of k, LR1(k) = A038374(k) is the length of the longest run of ones.

0, 4, 8, 9, 16, 17, 18, 20, 24, 32, 33, 34, 35, 36, 37, 40, 41, 48, 49, 64, 65, 66, 67, 68, 69, 70, 72, 73, 74, 80, 81, 82, 84, 88, 96, 97, 98, 99, 104, 112, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 144, 145, 146, 148, 149, 152, 160

Offset: 1

Alex Ratushnyak, Feb 14 2014

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A038374, A087117.
Cf. A090050 (numbers k such that LR0(k) = LR1(k)).
Cf. A237883 (numbers k such that LR0(k) < LR1(k)).

klrQ[n_]:=With[{sidn2=Split[IntegerDigits[n,2]]},Max[Length/@Select[sidn2,#[[1]]==0&]]>Max[Length/@Select[sidn2,#[[1]]==1&]]]; Select[Range[ 0,200],klrQ] (* Harvey P. Dale, May 05 2018 *)

for n in range(1000):
    b = bin(n).lstrip("0b")
    L0 = L1 = 0
    s = '0'
    if n==0: b=s
    while b.find(s)>=0:
        s += '0'
        L0 += 1
    s = '1'
    while b.find(s)>=0:
        s += '1'
        L1 += 1
    if L0>L1: print(n, end=', ')

A237883 Numbers k such that LR0(k) < LR1(k), where LR0(k) = A087117(k) is the length of the longest run of zeros in the binary representation of k, LR1(k) = A038374(k) is the length of the longest run of ones.

Original entry on oeis.org

1, 3, 6, 7, 11, 13, 14, 15, 22, 23, 26, 27, 28, 29, 30, 31, 39, 43, 45, 46, 47, 53, 54, 55, 57, 58, 59, 60, 61, 62, 63, 78, 79, 86, 87, 90, 91, 92, 93, 94, 95, 103, 106, 107, 109, 110, 111, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 143

Offset: 1

Alex Ratushnyak, Feb 14 2014

Cf. A038374, A087117.
Cf. A090050 (numbers k such that LR0(k) = LR1(k)).
Cf. A237882 (numbers k such that LR0(k) > LR1(k)).

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).

Original entry on oeis.org

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).
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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
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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

A007582 a(n) = 2^(n-1)*(1+2^n).

Original entry on oeis.org

1, 3, 10, 36, 136, 528, 2080, 8256, 32896, 131328, 524800, 2098176, 8390656, 33558528, 134225920, 536887296, 2147516416, 8590000128, 34359869440, 137439215616, 549756338176, 2199024304128, 8796095119360, 35184376283136, 140737496743936, 562949970198528

Offset: 0

Simon Plouffe

Let G_n be the elementary Abelian group G_n = (C_2)^n for n >= 1: A006516 is the number of times the number -1 appears in the character table of G_n and A007582 is the number of times the number 1. Together the two sequences cover all the values in the table, i.e., A006516(n) + A007582(n) = 2^(2n). - Ahmed Fares (ahmedfares(AT)my-deja.com), Jun 01 2001
Number of walks of length 2n+1 between two adjacent vertices in the cycle graph C_8. Example: a(1)=3 because in the cycle ABCDEFGH we have three walks of length 3 between A and B: ABAB, ABCB and AHAB. - Emeric Deutsch, Apr 01 2004
Smallest number containing in its binary representation two equal non-overlapping subwords of length n: A097295(a(n))=n and A097295(m)Reinhard Zumkeller, Aug 04 2004
a(n)^2 + (A006516(n))^2 = a(2n). E.g., a(3) = 36, A006516(3) = 28, a(6) = 2080. 36^2 + 28^2 = 2080. - Gary W. Adamson, Jun 17 2006
Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A) for which either x equals y or x does not equal y. - Ross La Haye, Jan 02 2008
Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A). This is just a simpler statement of my previous comment for this sequence. - Ross La Haye, Jan 10 2008
For n>0: A000120(a(n))=2, A023414(a(n))=2*(n-1), A087117(a(n))=n-1. - Reinhard Zumkeller, Jun 23 2009
a(n+1) written in base 2: 11, 1010, 100100, 10001000, 1000010000, ..., i.e., number 1, n times 0, number 1, n times 0 (A163449(n)). - Jaroslav Krizek, Jul 27 2009
a(n) for n >= 1 is a bisection of A001445(n+1). - Jaroslav Krizek, Aug 14 2009
Related to A102573: letting T(q,r) be the coefficient of n^(r+1) in the polynomial 2^(q-n)/n times sum_{k=0..n} binomial(n,k)*k^q, then A007582(x)= sum_{k=0..x-1} T(x,k)*2^k. - John M. Campbell, Nov 16 2011
a(n) gives the number of pairs (r, s) such that 0 <= r <= s <= (2^n)-1 that satisfy AND(r, s, XOR(r, s)) = 0. - Ramasamy Chandramouli, Aug 30 2012
a(n) = A000217(2^n) = 2^(2n-1) + 2^(n-1) is the nearest triangular number above 2^(2n-1); cf. A006516, A233327. - Antti Karttunen, Feb 26 2014
Consider the quantum spin-1/2 chain with even number of sites L (physics, condensed matter theory). The spectrum of the Hamiltonian can be classified according to symmetries. If the only symmetry of the spin Hamiltonian is Parity, i.e., reflection with respect to the middle of the chain (see e.g. the transverse-field Ising model with open boundary conditions), then the dimension of the p=+1 parity sector is given by a(n) with n=L/2. - Marin Bukov, Mar 11 2016
a(n) is also the total number of words of length n, over an alphabet of four letters, of which one of them appears an even number of times. See the Lekraj Beedassy, Jul 22 2003, comment on A006516 (4-letter odd case), and the Balakrishnan reference there. For the 1- to 11-letter cases, see the crossrefs. - Wolfdieter Lang, Jul 17 2017
a(n) is the number of nonisomorphic spanning trees of the cyclic snake formed with n+1 copies of the cycle on 4 vertices. A cyclic snake is a connected graph whose block-cutpoint is a path and all its n blocks are isomorphic to the cycle C_m. - Christian Barrientos, Sep 05 2024
Also, with offset 1, the cogrowth sequence of the dihedral group with 16 elements, D8 = . - Sean A. Irvine, Nov 06 2024

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..200
M. Archibald, A. Blecher, A. Knopfmacher, M. E. Mays, Inversions and Parity in Compositions of Integers, J. Int. Seq., Vol. 23 (2020), Article 20.4.1.
S. Hong and J. H. Kwak, Regular fourfold coverings with respect to the identity automorphism, J. Graph Theory, 17 (1993), 621-627.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 168
Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6.
Mircea Merca, A Note on Cosine Power Sums, J. Integer Sequences, Vol. 15 (2012), Article 12.5.3.
Index entries for linear recurrences with constant coefficients, signature (6,-8)

Cf. A000217, A049773, A049775.
Cf. A006516.
Cf. A134308.
Cf. A000225, A000392, A032263, A028243, A000079.
Cf. A102573.
The number of words of length n with m letters, one of them appearing an even number of times is for m = 1..11: A000035, A011782,  A007051, A007582, A081186, A081187, A081188, A081189, A081190, A060531, A081192. - Wolfdieter Lang, Jul 17 2017

[Binomial(2^n + 1, 2) : n in [0..30]]; // Wesley Ivan Hurt, Jul 03 2020

seq(binomial(-2^n, 2), n=0..23); # Zerinvary Lajos, Feb 22 2008

Table[ Binomial[2^n + 1, 2], {n, 0, 23}] (* Robert G. Wilson v, Jul 30 2004 *)
LinearRecurrence[{6,-8},{1,3},30] (* Harvey P. Dale, Apr 08 2013 *)

A007582(n):=2^(n-1)*(1+2^n)$ makelist(A007582(n),n,0,30); /* Martin Ettl, Nov 15 2012 */

PARI
```
a(n)=if(n<0,0,2^(n-1)*(1+2^n))
```

a(n)=sum(k=-n\4,n\4,binomial(2*n+1,n+1+4*k))

G.f.: (1-3*x)/((1-2*x)*(1-4*x)). C(1+2^n, 2) where C(n, 2) is n-th triangular number A000217.
Binomial transform of A007051. Inverse binomial transform of A081186. - Paul Barry, Apr 07 2003
E.g.f.: exp(3*x)*cosh(x). - Paul Barry, Apr 07 2003
a(n) = Sum_{k=0..floor(n/2)} C(n, 2*k)*3^(n-2*k). - Paul Barry, May 08 2003
a(n+1) = 4*a(n) - 2^n; see also A049775. a(n) = 2^(n-1)*A000051(n). - Philippe Deléham, Feb 20 2004
a(n) = 6*a(n-1) - 8*a(n-2). - Emeric Deutsch, Apr 01 2004
Row sums of triangle A134308. - Gary W. Adamson, Oct 19 2007
a(n) = StirlingS2(2^n + 1,2^n) = 1 + 2*StirlingS2(n+1,2) + 3*StirlingS2(n+1,3) + 3*StirlingS2(n+1,4) = StirlingS2(n+2,2) + 3(StirlingS2(n+1,3) + StirlingS2(n+1,4)). - Ross La Haye, Mar 01 2008
a(n) = StirlingS2(2^n + 1,2^n) = 1 + 2*StirlingS2(n+1,2) + 3*StirlingS2(n+1,3) + 3*StirlingS2(n+1,4) = StirlingS2(n+2,2) + 3(StirlingS2(n+1,3) + StirlingS2(n+1,4)). - Ross La Haye, Apr 02 2008
a(n) = A000079(n) + A006516(n). - Yosu Yurramendi, Aug 06 2008
a(n) = A028403(n+1) / 4. - Jaroslav Krizek, Jul 27 2009
a(n) = Sum_{k=-floor(n/4)..floor(n/4)} binomial(2*n,n+4*k)/2. - Mircea Merca, Jan 28 2012
G.f.: Q(0)/2 where Q(k) = 1 + 2^k/(1 - 2*x/(2*x + 2^k/Q(k+1) )); (continued fraction ). - Sergei N. Gladkovskii, Apr 10 2013
a(n) = Sum_{k=1..2^n} k. - Joerg Arndt, Sep 01 2013
a(n) = (1/3) * Sum_{k=2^n..2^(n+1)} k. - J. M. Bergot, Jan 26 2015
a(n+1) = 2*a(n) + 4^n. - Yuchun Ji, Mar 10 2017

A333766 Maximum part of the n-th composition in standard order. a(0) = 0.

Original entry on oeis.org

0, 1, 2, 1, 3, 2, 2, 1, 4, 3, 2, 2, 3, 2, 2, 1, 5, 4, 3, 3, 3, 2, 2, 2, 4, 3, 2, 2, 3, 2, 2, 1, 6, 5, 4, 4, 3, 3, 3, 3, 4, 3, 2, 2, 3, 2, 2, 2, 5, 4, 3, 3, 3, 2, 2, 2, 4, 3, 2, 2, 3, 2, 2, 1, 7, 6, 5, 5, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 5, 4, 3, 3, 3, 2, 2

Offset: 0

Gus Wiseman, Apr 05 2020

nonn

One plus the longest run of 0's in the binary expansion of n.

A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The 100th composition in standard order is (1,3,3), so a(100) = 3.

Positions of ones are A000225.
Positions of terms <= 2 are A003754.
The version for prime indices is A061395.
Positions of terms > 1 are A062289.
Positions of first appearances are A131577.
The minimum part is given by A333768.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Compositions without 1's are A022340.
- Sum is A070939.
- Product is A124758.
- Runs are counted by A124767.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Runs-resistance is A333628.
- Weakly decreasing compositions are A114994.
- Weakly increasing compositions are A225620.
- Strictly decreasing compositions are A333255.
- Strictly increasing compositions are A333256.
Cf. A029931, A048793, A087117, A228351, A328594, A333217, A333218, A333219, A333632, A333767.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[If[n==0,0,Max@@stc[n]],{n,0,100}]

For n > 0, a(n) = A087117(n) + 1.

A038374 Length of longest contiguous block of 1's in binary expansion of n.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 2, 3, 4, 1, 1, 1, 2, 1, 1, 2, 3, 2, 2, 2, 2, 3, 3, 4, 5, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 2, 3, 4, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 5, 6, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 2, 3, 4, 1, 1, 1, 2, 1, 1, 2, 3, 2, 2, 2

Offset: 1

Jeffrey Shallit

			a(157) = 3 because 157 in base 2 is 10011101 and longest contiguous block of 1's is of length 3.
May be arranged into blocks of lengths 1, 2, 4, 8, 16, ...:
1,
1, 2,
1, 1, 2, 3,
1, 1, 1, 2, 2, 2, 3, 4,
1, 1, 1, 2, 1, 1, 2, 3, 2, 2, 2, 2, 3, 3, 4, 5,
1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 2, 3, 4, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 5, 6,
... - _N. J. A. Sloane_, Jul 25 2014

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

Cf. A087117, A090000, A090001, A090002, A090003, A090050, A245196.

import Data.List (unfoldr, group)
a038374 = maximum . map length . filter ((== 1) . head) . group .
   unfoldr (\x -> if x == 0 then Nothing else Just $ swap $ divMod x 2)
-- Reinhard Zumkeller, May 01 2012

A038374 := proc(n) local nshft,thisr,resul; nshft := n ; resul :=0 ; thisr :=0 ; while nshft > 0 do if nshft mod 2 <> 0 then thisr := thisr+1 ; else resul := max(resul,thisr) ; thisr := 0 ; fi ; nshft := floor(nshft/2) ; od ; resul := max(resul,thisr) ; RETURN(resul) ; end : for n from 1 to 80 do printf("%d,",A038374(n)) ; od : # R. J. Mathar, Jun 15 2006

Table[Max[Length/@DeleteCases[Split[IntegerDigits[n,2]],?(MemberQ[ #,0] &)]],{n,120}] (* _Harvey P. Dale, Jun 10 2013 *)

a(n)=if (n==0, return (0)); n>>=valuation(n, 2); if(n<2, return(n)); my(e=valuation(n+1, 2)); max(e, a(n>>e)) \\ Charles R Greathouse IV, Jan 12 2014; edited by Michel Marcus, Apr 14 2019

from itertools import groupby
def a(n): return max(len(list(g)) for k, g in groupby(bin(n)[1:]) if k=='1')
print([a(n) for n in range(1, 91)]) # Michael S. Branicky, Jul 04 2022

a(n) >= A089309(n). a(n) >= A089310(n). a(2^i)=1. a(2^i-1)=i. - R. J. Mathar, Jun 15 2006

May be defined by the recurrence given in A245196, taking G(n)=n+1 (n>=0) and m=1. - N. J. A. Sloane, Jul 25 2014

A333768 Minimum part of the n-th composition in standard order. a(0) = 0.

Original entry on oeis.org

0, 1, 2, 1, 3, 1, 1, 1, 4, 1, 2, 1, 1, 1, 1, 1, 5, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 2, 1, 3, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 7, 1, 2, 1, 3, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1

Offset: 0

Gus Wiseman, Apr 06 2020

nonn

One plus the shortest run of 0's after a 1 in the binary expansion of n > 0.

			The 148th composition in standard order is (3,2,3), so a(148) = 2.

Positions of first appearances (ignoring index 0) are A000079.
Positions of terms > 1 are A022340.
The version for prime indices is A055396.
The  maximum part is given by A333766.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Compositions without 1's are A022340.
- Sum is A070939.
- Product is A124758.
- Runs are counted by A124767.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Runs-resistance is A333628.
- Weakly decreasing compositions are A114994.
- Weakly increasing compositions are A225620.
- Strictly decreasing compositions are A333255.
- Strictly increasing compositions are A333256.
Cf. A029931, A048793, A087117, A228351, A328594, A333217, A333218, A333219, A333632, A333767.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[If[n==0,0,Min@@stc[n]],{n,0,100}]

For n > 0, a(n) = A333767(n) + 1.

A087116 Number of maximal groups of consecutive zeros in binary representation of n.

Original entry on oeis.org

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

Offset: 0

Reinhard Zumkeller, Aug 14 2003

nonn

The following four statements are equivalent: a(n) = 0; n = 2^k - 1 for some k; A087117(n) = 0; A023416(n) = 0.

			G.f. = 1 + x^2 + x^4 + x^5 + x^6 + x^8 + x^9 + 2*x^10 + x^11 + x^12 + x^13 + x^14 + ...

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for sequences related to binary expansion of n

Cf. A087118, A087119, A087120, A023416, A007088.
Cf. A023416, A087117.
Cf. A033264, A003817, A069010.
Essentially the same as A033264.

a087116 0 = 1
a087116 n = f 0 n where
   f y 0 = y
   f y x = if r == 0 then g x' else f y x'
           where (x', r) = divMod x 2
                 g z = if r == 0 then g z' else f (y + 1) z'
                       where (z', r) = divMod z 2
-- Reinhard Zumkeller, Mar 31 2015

a[n_] := SequenceCount[IntegerDigits[n, 2], {Longest[0..]}];
Table[a[n], {n, 0, 101}] (* Jean-François Alcover, Oct 18 2021 *)

a(n) = if (n == 0, 1, hammingweight(bitxor(n, n>>1)) >> 1);
vector(102, i, a(i-1))  \\ Gheorghe Coserea, Sep 17 2015

def A087116(n):
    return sum(1 for d in bin(n)[2:].split('1') if len(d)) # Chai Wah Wu, Nov 04 2016

a(n) = A033264(n) for n > 0 since strings of 0's alternate with strings of 1's. - Jonathan Sondow, Jan 17 2016
a(n) = a(2*n + 1) = a(4*n + 2) - 1, if n > 0. - Michael Somos, Nov 04 2016
a(n) = A069010(A003817(n)-n) for n > 0. - Chai Wah Wu, Nov 04 2016

A090050 Numbers having equal length of longest contiguous block of zeros and ones in binary expansion.

Original entry on oeis.org

2, 5, 10, 12, 19, 21, 25, 38, 42, 44, 50, 51, 52, 56, 71, 75, 76, 77, 83, 85, 89, 100, 101, 102, 105, 108, 113, 142, 147, 150, 153, 154, 155, 166, 170, 172, 178, 179, 180, 184, 199, 201, 202, 203, 204, 205, 210, 211, 212, 217, 226, 227, 232, 240, 271, 279, 284

Offset: 1

Reinhard Zumkeller, Nov 20 2003

A087117(a(n)) = A038374(a(n)), see also A000975.

			180 -> '10110100' with A087117(180)=2 and A038374(180)=2, therefore 180 is a term.

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

Cf. A031443 (binary digitally balanced).

Cf. A000975, A038374, A087117.

a090050 n = a090050_list !! (n+1)
a090050_list = [x | x <- [1..], a087117 x == a038374 x]
-- Reinhard Zumkeller, May 01 2012

zobQ[n_]:=Module[{s=Split[IntegerDigits[n,2]]},Max[Length/@Select[ s, MemberQ[ #,0]&]] == Max[Length/@Select[s,MemberQ[#,1]&]]]; Select[ Range[ 300],zobQ] (* Harvey P. Dale, Aug 25 2019 *)
Select[Range@1000, (s=Split@IntegerDigits[#,2]; Length@s>1 && Last@Differences@(Length@# & /@ Union@s) == 0) &] (* Hans Rudolf Widmer, Oct 10 2023 *)

Definition corrected, thanks to Leroy Quet. - Sep 17 2008

A334591 Side length of largest triangle of zeros in the XOR-triangle with first row generated from the binary expansion of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 2, 3, 2, 2, 1, 2, 1, 2, 3, 4, 3, 2, 2, 2, 3, 1, 2, 3, 2, 2, 1, 2, 2, 3, 4, 5, 4, 3, 3, 2, 2, 3, 2, 3, 2, 4, 3, 2, 1, 2, 3, 4, 3, 2, 3, 2, 3, 1, 2, 3, 2, 2, 2, 3, 3, 4, 5, 6, 5, 4, 4, 3, 3, 3, 3, 3, 2, 3, 2, 4, 3, 2, 3, 4, 3, 2, 2, 4, 5, 3, 3

Offset: 1

Peter Kagey, May 07 2020

An XOR-triangle is an inverted 0-1 triangle formed by choosing a top row and having each entry in the subsequent rows be the XOR of the two values above it.
Records occur at a(2^n) = n.
Ones occur at 2, 3, 5, 6, 11, 13, 22, 27, 45, 54, 91, 109, 182, 219, 365, 438, 731, 877, 1462,...
a(n) <= A087117(n).

			For n = 53, a(53) = 3 because 53 = 110101_2 in binary, and the largest triangle of 0s in the corresponding XOR-triangle has size 3 (see third, fourth, and fifth rows):
  1 1 0 1 0 1
   0 1 1 1 1
    1 0 0 0
     1 0 0
      1 0
       1

Peter Kagey, Table of n, a(n) for n = 1..8191
MathOverflow user DSM, Number triangle
Index entries for sequences related to binary expansion of n

Cf. A087117, A334556, A334592, A334593, A334594, A334595, A334596.

Array[Function[w, Max@ Flatten@ Array[If[# == 1, If[First@ # == 1, Nothing, Length@ #] & /@ Split@ w[[#]], If[First@ # == -1, Length@ #, Nothing] & /@ Split[w[[#]] - Most@ w[[# - 1]] ] ] &, Length@ w] /. -Infinity -> 0]@ NestWhileList[Map[BitXor @@ # &, Partition[#, 2, 1]] &, IntegerDigits[#, 2], Length@ # > 1 &] &, 105] (* Michael De Vlieger, May 08 2020 *)

cp's OEIS Frontend

A237882 Numbers k such that LR0(k) > LR1(k), where LR0(k) = A087117(k) is the length of the longest run of zeros in the binary representation of k, LR1(k) = A038374(k) is the length of the longest run of ones.

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A237883 Numbers k such that LR0(k) < LR1(k), where LR0(k) = A087117(k) is the length of the longest run of zeros in the binary representation of k, LR1(k) = A038374(k) is the length of the longest run of ones.

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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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A007582 a(n) = 2^(n-1)*(1+2^n).

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A333766 Maximum part of the n-th composition in standard order. a(0) = 0.

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A038374 Length of longest contiguous block of 1's in binary expansion of n.

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A333768 Minimum part of the n-th composition in standard order. a(0) = 0.

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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).