A164707 -id:A164707 - cp's OEIS frontend

A164708 Those positive integers missing from sequence A164707. Those positive integers that, when written in binary, contain at least two runs of 1's that are of differing lengths.

11, 13, 19, 22, 23, 25, 26, 29, 35, 38, 39, 43, 44, 45, 46, 47, 49, 50, 52, 53, 55, 57, 58, 59, 61, 67, 70, 71, 75, 76, 77, 78, 79, 83, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 97, 98, 100, 101, 103, 104, 105, 106, 107, 109, 110, 111, 113, 114, 115, 116, 117, 118, 121

Offset: 1

Leroy Quet, Aug 23 2009

Clarification: A binary number consists of "runs" completely of 1's alternating with runs completely of 0's. No two or more runs all of the same digit are adjacent.

A164707

Select[Range[200],Length[Union[Length/@Select[Split[IntegerDigits[ #,2]], MemberQ[ #,1]&]]]>1&] (* Harvey P. Dale, May 28 2016 *)

More terms from Max Alekseyev, Feb 03 2010

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.
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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.
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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.
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Kival Ngaokrajang, Illustration of initial terms
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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

A329738 Number of compositions of n whose run-lengths are all equal.

Original entry on oeis.org

1, 1, 2, 4, 6, 8, 19, 24, 45, 75, 133, 215, 401, 662, 1177, 2035, 3587, 6190, 10933, 18979, 33339, 58157, 101958, 178046, 312088, 545478, 955321, 1670994, 2925717, 5118560, 8960946, 15680074, 27447350, 48033502, 84076143, 147142496, 257546243, 450748484, 788937192

Offset: 0

Gus Wiseman, Nov 20 2019

nonn

A composition of n is a finite sequence of positive integers with sum n.

			The a(1) = 1 through a(6) = 19 compositions:
  (1)  (2)   (3)    (4)     (5)      (6)
       (11)  (12)   (13)    (14)     (15)
             (21)   (22)    (23)     (24)
             (111)  (31)    (32)     (33)
                    (121)   (41)     (42)
                    (1111)  (131)    (51)
                            (212)    (123)
                            (11111)  (132)
                                     (141)
                                     (213)
                                     (222)
                                     (231)
                                     (312)
                                     (321)
                                     (1122)
                                     (1212)
                                     (2121)
                                     (2211)
                                     (111111)

Andrew Howroyd, Table of n, a(n) for n = 0..1000

Compositions with relatively prime run-lengths are A000740.
Compositions with equal multiplicities are A098504.
Compositions with equal differences are A175342.
Compositions with distinct run-lengths are A329739.
Cf. A003242, A008965, A107429, A164707, A238130, A242882, A274174, A329745, A329766.

Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],SameQ@@Length/@Split[#]&]],{n,0,10}]

seq(n)={my(b=Vec(1/(1 - sum(k=1, n, x^k/(1+x^k) + O(x*x^n)))-1)); concat([1], vector(n, k, sumdiv(k, d, b[d])))} \\ Andrew Howroyd, Dec 30 2020

a(n) = Sum_{d|n} A003242(d).

a(n) = A329745(n) + A000005(n).

A328592 Numbers whose binary expansion has all different lengths of runs of 1's.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 8, 11, 12, 13, 14, 15, 16, 19, 22, 23, 24, 25, 26, 28, 29, 30, 31, 32, 35, 38, 39, 44, 46, 47, 48, 49, 50, 52, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 67, 70, 71, 76, 78, 79, 88, 92, 94, 95, 96, 97, 98, 100, 103, 104, 110, 111, 112, 113, 114

Offset: 1

Gus Wiseman, Oct 20 2019

nonn

Also numbers whose binary indices have different lengths of runs of successive parts. A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

The complement is {5, 9, 10, 17, 18, 20, 21, 27, ...}.

			The sequence of terms together with their binary expansions and binary indices begins:
   0:     0 ~ {}
   1:     1 ~ {1}
   2:    10 ~ {2}
   3:    11 ~ {1,2}
   4:   100 ~ {3}
   6:   110 ~ {2,3}
   7:   111 ~ {1,2,3}
   8:  1000 ~ {4}
  11:  1011 ~ {1,2,4}
  12:  1100 ~ {3,4}
  13:  1101 ~ {1,3,4}
  14:  1110 ~ {2,3,4}
  15:  1111 ~ {1,2,3,4}
  16: 10000 ~ {5}
  19: 10011 ~ {1,2,5}
  22: 10110 ~ {2,3,5}
  23: 10111 ~ {1,2,3,5}
  24: 11000 ~ {4,5}
  25: 11001 ~ {1,4,5}
  26: 11010 ~ {2,4,5}

The version for prime indices is A130091.
The binary expansion of n has A069010(n) runs of 1's.
The lengths of runs of 1's in the binary expansion of n are row n of A245563.
Numbers whose binary expansion has equal lengths of runs of 1's are A164707.
Cf. A000120, A003714, A014081, A098859, A121016, A328593, A328595, A328596.

Select[Range[0,100],UnsameQ@@Length/@Split[Join@@Position[Reverse[IntegerDigits[#,2]],1],#2==#1+1&]&]

A245563 Table read by rows: row n gives list of lengths of runs of 1's in binary expansion of n, starting with low-order bits.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Aug 10 2014

A formula for A071053(n) depends on this table.

			Here are the run lengths for the numbers 0 through 21:
0, []
1, [1]
2, [1]
3, [2]
4, [1]
5, [1, 1]
6, [2]
7, [3]
8, [1]
9, [1, 1]
10, [1, 1]
11, [2, 1]
12, [2]
13, [1, 2]
14, [3]
15, [4]
16, [1]
17, [1, 1]
18, [1, 1]
19, [2, 1]
20, [1, 1]
21, [1, 1, 1]

Reinhard Zumkeller, Rows n = 0..1000 of triangle, flattened
Index entries for sequences related to binary expansion of n

Row sums = A000120 (the binary weight).
Cf. A245562, A071053.
Cf. A030308, A227736.
Row lengths are A069010.
The version for prime indices (instead of binary indices) is A124010.
Numbers with distinct run-lengths are A328592.
Numbers with equal run-lengths are A164707.
Cf. A003714, A014081, A328166, A328594, A328595, A328596.

import Data.List (group)
a245563 n k = a245563_tabf !! n !! k
a245563_row n = a245563_tabf !! n
a245563_tabf = [0] : map
   (map length . (filter ((== 1) . head)) . group) (tail a030308_tabf)
-- Reinhard Zumkeller, Aug 10 2014

for n from 0 to 128 do
lis:=[]; t1:=convert(n,base,2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
elif out1 = 0 and t1[i] = 1 then c:=c+1;
elif out1 = 1 and t1[i] = 0 then c:=c;
elif out1 = 0 and t1[i] = 0 then lis:=[op(lis),c]; out1:=1; c:=0;
fi;
if i = L1 and c>0 then lis:=[op(lis),c]; fi;
od:
lprint(n,lis);
od:

Join@@Table[Length/@Split[Join@@Position[Reverse[IntegerDigits[n,2]],1],#2==#1+1&],{n,0,100}] (* Gus Wiseman, Nov 03 2019 *)

from re import split
A245563_list = [0]
for n in range(1,100):
    A245563_list.extend(len(d) for d in split('0+',bin(n)[:1:-1]) if d != '')
# Chai Wah Wu, Sep 07 2014

A382857 Number of ways to permute the prime indices of n so that the run-lengths are all equal.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 2, 1, 0, 1, 2, 1, 1, 1, 6, 1, 1, 2, 2, 2, 4, 1, 2, 2, 0, 1, 6, 1, 1, 1, 2, 1, 0, 1, 1, 2, 1, 1, 0, 2, 0, 2, 2, 1, 6, 1, 2, 1, 1, 2, 6, 1, 1, 2, 6, 1, 1, 1, 2, 1, 1, 2, 6, 1, 0, 1, 2, 1, 6, 2, 2

Offset: 0

Gus Wiseman, Apr 09 2025

nonn

The first x with a(x) > 1 but A382771(x) > 0 is a(216) = 4, A382771(216) = 4.

A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798, sum A056239.

			The prime indices of 216 are {1,1,1,2,2,2} and we have permutations:
  (1,1,1,2,2,2)
  (1,2,1,2,1,2)
  (2,1,2,1,2,1)
  (2,2,2,1,1,1)
so a(216) = 4.
The prime indices of 25920 are {1,1,1,1,1,1,2,2,2,2,3} and we have permutations:
  (1,2,1,2,1,2,1,2,1,3,1)
  (1,2,1,2,1,2,1,3,1,2,1)
  (1,2,1,2,1,3,1,2,1,2,1)
  (1,2,1,3,1,2,1,2,1,2,1)
  (1,3,1,2,1,2,1,2,1,2,1)
so a(25920) = 5.

The restriction to signature representatives (A181821) is A382858, distinct A382773.
The restriction to factorials is A335407, distinct A382774.
For distinct instead of equal run-lengths we have A382771.
For run-sums instead of run-lengths we have A382877, distinct A382876.
Positions of first appearances are A382878.
Positions of 0 are A382879.
Positions of terms > 1 are A383089.
Positions of 1 are A383112.
A003963 gives product of prime indices.
A005811 counts runs in binary expansion.
A044813 lists numbers whose binary expansion has distinct run-lengths.
A056239 adds up prime indices, row sums of A112798.
A239455 counts Look-and-Say partitions, ranks A351294.
A304442 counts partitions with equal run-sums, ranks A353833.
A164707 lists numbers whose binary expansion has all equal run-lengths, distinct A328592.
A353744 ranks compositions with equal run-lengths, counted by A329738.
Cf. A000720, A000961, A001221, A001222, A003242, A008480, A047966, A238130, A238279, A351201, A351293, A351295.

Table[Length[Select[Permutations[Join@@ConstantArray@@@FactorInteger[n]], SameQ@@Length/@Split[#]&]],{n,0,100}]

A121016 Numbers whose binary expansion is properly periodic.

Original entry on oeis.org

3, 7, 10, 15, 31, 36, 42, 45, 54, 63, 127, 136, 153, 170, 187, 204, 221, 238, 255, 292, 365, 438, 511, 528, 561, 594, 627, 660, 682, 693, 726, 759, 792, 825, 858, 891, 924, 957, 990, 1023, 2047, 2080, 2145, 2184, 2210, 2275, 2340, 2405, 2457, 2470, 2535

Offset: 1

Jacob A. Siehler, Sep 08 2006

A finite sequence is aperiodic if its cyclic rotations are all different. - Gus Wiseman, Oct 31 2019

			For example, 204=(1100 1100)_2 and 292=(100 100 100)_2 belong to the sequence, but 30=(11110)_2 cannot be split into repeating periods.
From _Gus Wiseman_, Oct 31 2019: (Start)
The sequence of terms together with their binary expansions and binary indices begins:
   3:         11 ~ {1,2}
   7:        111 ~ {1,2,3}
   10:      1010 ~ {2,4}
   15:      1111 ~ {1,2,3,4}
   31:     11111 ~ {1,2,3,4,5}
   36:    100100 ~ {3,6}
   42:    101010 ~ {2,4,6}
   45:    101101 ~ {1,3,4,6}
   54:    110110 ~ {2,3,5,6}
   63:    111111 ~ {1,2,3,4,5,6}
  127:   1111111 ~ {1,2,3,4,5,6,7}
  136:  10001000 ~ {4,8}
  153:  10011001 ~ {1,4,5,8}
  170:  10101010 ~ {2,4,6,8}
  187:  10111011 ~ {1,2,4,5,6,8}
  204:  11001100 ~ {3,4,7,8}
  221:  11011101 ~ {1,3,4,5,7,8}
  238:  11101110 ~ {2,3,4,6,7,8}
  255:  11111111 ~ {1,2,3,4,5,6,7,8}
  292: 100100100 ~ {3,6,9}
(End)

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

A020330 is a subsequence.
Numbers whose binary expansion is aperiodic are A328594.
Numbers whose reversed binary expansion is Lyndon are A328596.
Numbers whose binary indices have equal run-lengths are A164707.
Cf. A000120, A003714, A014081, A065609, A069010, A275692, A328595.

PeriodicQ[n_, base_] := Block[{l = IntegerDigits[n, base]}, MemberQ[ RotateLeft[l, # ] & /@ Most@ Divisors@ Length@l, l]]; Select[ Range@2599, PeriodicQ[ #, 2] &]

is(n)=n=binary(n);fordiv(#n,d,for(i=1,#n/d-1, for(j=1,d, if(n[j]!=n[j+i*d], next(3)))); return(d<#n)) \\ Charles R Greathouse IV, Dec 10 2013

A382879 Positions of 0 in A382857 (permutations of prime indices with equal run-lengths).

Original entry on oeis.org

24, 40, 48, 54, 56, 80, 88, 96, 104, 112, 135, 136, 152, 160, 162, 176, 184, 189, 192, 208, 224, 232, 240, 248, 250, 272, 288, 296, 297, 304, 320, 328, 336, 344, 351, 352, 368, 375, 376, 384, 405, 416, 424, 448, 459, 464, 472, 480, 486, 488, 496, 513, 528, 536

Offset: 1

Gus Wiseman, Apr 09 2025

nonn

A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798, sum A056239.

			The terms together with their prime indices begin:
   24: {1,1,1,2}
   40: {1,1,1,3}
   48: {1,1,1,1,2}
   54: {1,2,2,2}
   56: {1,1,1,4}
   80: {1,1,1,1,3}
   88: {1,1,1,5}
   96: {1,1,1,1,1,2}
  104: {1,1,1,6}
  112: {1,1,1,1,4}
  135: {2,2,2,3}
  136: {1,1,1,7}
  152: {1,1,1,8}
  160: {1,1,1,1,1,3}

For distinct instead of equal the complement is A351294, counted by A239455.
For distinct instead of equal we have A351295, counted by A351293.
For run-sums instead of run-lengths we have A383100, zeros of A382877, distinct A382876.
Positions of 0 in A382857 (firsts A382878), by signature A382858 (distinct A382773).
For prime signature instead of prime indices we have A382914.
Partitions of this type are counted by A382915.
The complement is counted by A383013.
A005811 counts runs in binary expansion.
A056239 adds up prime indices, row sums of A112798.
A297770 counts distinct runs in binary expansion.
A164707 lists numbers whose binary form has equal runs of ones, distinct A328592.
A304442 counts partitions with equal run-sums, ranks A353833.
A329739 counts compositions with distinct run-lengths, ranks A351290.
A353744 ranks compositions with equal run-lengths, distinct A351596 (complement A351291).
Cf. A000720, A000961, A001221, A001222, A003242, A008480, A047966, A130091, A238279, A351201.

Select[Range[100], Select[Permutations[Join@@ConstantArray@@@FactorInteger[#]], SameQ@@Length/@Split[#]&]=={}&]

A384877 Irregular triangle read by rows where row k lists the lengths of maximal anti-runs (increasing by more than 1) in the binary indices of n.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Jun 17 2025

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

			The binary indices of 182 are {2,3,5,6,8}, with maximal anti-runs ((2),(3,5),(6,8)) so row 182 is (1,2,2).
Triangle begins:
   0: ()
   1: (1)
   2: (1)
   3: (1,1)
   4: (1)
   5: (2)
   6: (1,1)
   7: (1,1,1)
   8: (1)
   9: (2)
  10: (2)
  11: (1,2)
  12: (1,1)
  13: (2,1)
  14: (1,1,1)
  15: (1,1,1,1)

Row-sums are A000120.
Positions of rows of the form (1,1,...) are A023758.
Positions of first appearances of each distinct row appear to be A052499.
For runs instead of anti-runs we have A245563, reverse A245562.
Row-lengths are A384890.
A355394 counts partitions without a neighborless part, singleton case A355393.
A356606 counts strict partitions without a neighborless part, complement A356607.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A044813, A048793, A069010, A164707, A243815, A246029, A328592, A384177, A384877, A384879, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Table[Length/@Split[bpe[n],#2!=#1+1&],{n,0,100}]

A384890 Number of maximal anti-runs (increasing by more than 1) in the binary indices of n.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Jun 17 2025

nonn

First differs from A272604 at a(51) = 3, A272604(51) = 2.
A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.
Do all constant runs in this sequence have lengths 1, 2, or 3?

			The binary indices of 51 are {1,2,5,6}, with maximal anti-runs ((1),(2,5),(6)), so a(51) = 3.

For runs instead of anti-runs we have A069010 = run-lengths of A245563 (reverse A245562).
Row-lengths of A384877, firsts A384878.
For prime indices instead of binary indices we have A384906.
A000120 counts binary indices.
A356606 counts strict partitions without a neighborless part, complement A356607.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A044813, A048793, A052499, A164707, A243815, A300820, A328592, A384177, A384879, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Table[Length[Split[bpe[n],#2!=#1+1&]],{n,0,100}]

cp's OEIS Frontend

A164708 Those positive integers missing from sequence A164707. Those positive integers that, when written in binary, contain at least two runs of 1's that are of differing lengths.

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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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A329738 Number of compositions of n whose run-lengths are all equal.

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A328592 Numbers whose binary expansion has all different lengths of runs of 1's.

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A245563 Table read by rows: row n gives list of lengths of runs of 1's in binary expansion of n, starting with low-order bits.

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A382857 Number of ways to permute the prime indices of n so that the run-lengths are all equal.

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A121016 Numbers whose binary expansion is properly periodic.

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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).