cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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).

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

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Comments

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)

Examples

			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
		

References

  • 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.

Crossrefs

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.
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.

Programs

  • GAP
    List([0..35],n->(2^(n+1)-2+(n mod 2))/3); # Muniru A Asiru, Nov 01 2018
    
  • Haskell
    a000975 n = a000975_list !! n
    a000975_list = 0 : 1 : map (+ 1)
       (zipWith (+) (tail a000975_list) (map (* 2) a000975_list))
    -- Reinhard Zumkeller, Mar 07 2012
    
  • Magma
    [(2^(n+1) - 2 + (n mod 2))/3: n in [0..40]]; // Vincenzo Librandi, Mar 18 2015
    
  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    {a(n) = if( n<0, 0, 2 * 2^n \ 3)}; /* Michael Somos, Sep 04 2006 */
    
  • PARI
    a(n)=if(n<=0,0,bitxor(a(n-1),2^n-1)) \\ Paul D. Hanna, Nov 05 2011
    
  • PARI
    concat(0, Vec(x/(1-2*x-x^2+2*x^3) + O(x^100))) \\ Altug Alkan, Oct 30 2015
    
  • PARI
    {a(n) = (4*2^n - 3 - (-1)^n) / 6}; /* Michael Somos, Jul 23 2017 */
    
  • Python
    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

Formula

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

Extensions

Additional comments from Barry E. Williams, Jan 10 2000

A003714 Fibbinary numbers: if n = F(i1) + F(i2) + ... + F(ik) is the Zeckendorf representation of n (i.e., write n in Fibonacci number system) then a(n) = 2^(i1 - 2) + 2^(i2 - 2) + ... + 2^(ik - 2). Also numbers whose binary representation contains no two adjacent 1's.

Original entry on oeis.org

0, 1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37, 40, 41, 42, 64, 65, 66, 68, 69, 72, 73, 74, 80, 81, 82, 84, 85, 128, 129, 130, 132, 133, 136, 137, 138, 144, 145, 146, 148, 149, 160, 161, 162, 164, 165, 168, 169, 170, 256, 257, 258, 260, 261, 264
Offset: 0

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Comments

The name "Fibbinary" is due to Marc LeBrun.
"... integers whose binary representation contains no consecutive ones and noticed that the number of such numbers with n bits was fibonacci(n)". [posting to sci.math by Bob Jenkins (bob_jenkins(AT)burtleburtle.net), Jul 17 2002]
From Benoit Cloitre, Mar 08 2003: (Start)
A number m is in the sequence if and only if C(3m, m) (or equally, C(3m, 2m)) is odd.
a(n) == A003849(n) (mod 2). (End)
Numbers m such that m XOR 2*m = 3*m. - Reinhard Zumkeller, May 03 2005. [This implies that A003188(2*a(n)) = 3*a(n) holds for all n.]
Numbers whose base-2 representation contains no two adjacent ones. For example, m = 17 = 10001_2 belongs to the sequence, but m = 19 = 10011_2 does not. - Ctibor O. Zizka, May 13 2008
m is in the sequence if and only if the central Stirling number of the second kind S(2*m, m) = A007820(m) is odd. - O-Yeat Chan (math(AT)oyeat.com), Sep 03 2009
A000120(3*a(n)) = 2*A000120(a(n)); A002450 is a subsequence.
Every nonnegative integer can be expressed as the sum of two terms of this sequence. - Franklin T. Adams-Watters, Jun 11 2011
Subsequence of A213526. - Arkadiusz Wesolowski, Jun 20 2012
This is also the union of A215024 and A215025 - see the Comment in A014417. - N. J. A. Sloane, Aug 10 2012
The binary representation of each term m contains no two adjacent 1's, so we have (m XOR 2m XOR 3m) = 0, and thus a two-player Nim game with three heaps of (m, 2m, 3m) stones is a losing configuration for the first player. - V. Raman, Sep 17 2012
Positions of zeros in A014081. - John Keith, Mar 07 2022
These numbers are similar to Fibternary numbers A003726, Tribbinary numbers A060140 and Tribternary numbers. This sequence is a subsequence of Fibternary numbers A003726. The number of Fibbinary numbers less than any power of two is a Fibonacci number. We can generate this sequence recursively: start with 0 and 1; then, if x is in the sequence add 2x and 4x+1 to the sequence. The Fibbinary numbers have the property that the n-th Fibbinary number is even if the n-th term of the Fibonacci word is a. Respectively, the n-th Fibbinary number is odd (of the form 4x+1) if the n-th term of the Fibonacci word is b. Every number has a Fibbinary multiple. - Tanya Khovanova and PRIMES STEP Senior, Aug 30 2022
This is the ordered set S of numbers defined recursively by: 0 is in S; if x is in S, then 2*x and 4*x + 1 are in S. See Kimberling (2006) Example 3, in references below. - Harry Richman, Jan 31 2024

Examples

			From _Joerg Arndt_, Jun 11 2011: (Start)
In the following, dots are used for zeros in the binary representation:
  a(n)  binary(a(n))  n
    0:    .......     0
    1:    ......1     1
    2:    .....1.     2
    4:    ....1..     3
    5:    ....1.1     4
    8:    ...1...     5
    9:    ...1..1     6
   10:    ...1.1.     7
   16:    ..1....     8
   17:    ..1...1     9
   18:    ..1..1.    10
   20:    ..1.1..    11
   21:    ..1.1.1    12
   32:    .1.....    13
   33:    .1....1    14
   34:    .1...1.    15
   36:    .1..1..    16
   37:    .1..1.1    17
   40:    .1.1...    18
   41:    .1.1..1    19
   42:    .1.1.1.    20
   64:    1......    21
   65:    1.....1    22
(End)
		

References

  • Donald E. Knuth, The Art of Computer Programming: Fundamental Algorithms, Vol. 1, 2nd ed., Addison-Wesley, 1973, pp. 85, 493.

Crossrefs

A007088(a(n)) = A014417(n) (same sequence in binary). Complement: A004780. Char. function: A085357. Even terms: A022340, odd terms: A022341. First difference: A129761.
Other sequences based on similar restrictions on binary expansion: A003726 & A278038, A003754, A048715, A048718, A107907, A107909.
3*a(n) is in A001969.
Cf. A014081 (count 11 bits).

Programs

  • Haskell
    import Data.Set (Set, singleton, insert, deleteFindMin)
    a003714 n = a003714_list !! n
    a003714_list = 0 : f (singleton 1) where
       f :: Set Integer -> [Integer]
       f s = m : (f $ insert (4*m + 1) $ insert (2*m) s')
             where (m, s') = deleteFindMin s
    -- Reinhard Zumkeller, Jun 03 2012, Feb 07 2012
    
  • Maple
    A003714 := proc(n)
        option remember;
        if n < 3 then
            n ;
        else
            2^(A072649(n)-1) + procname(n-combinat[fibonacci](1+A072649(n))) ;
        end if;
    end proc:
    seq(A003714(n),n=0..10) ;
    # To produce a table giving n, a(n) (base 10), a(n) (base 2) - from N. J. A. Sloane, Sep 30 2018
    # binary: binary representation of n, in human order
    binary:=proc(n) local t1,L;
    if n<0 then ERROR("n must be nonnegative"); fi;
    if n=0 then return([0]); fi;
    t1:=convert(n,base,2); L:=nops(t1);
    [seq(t1[L+1-i],i=1..L)];
    end;
    for n from 0 to 100 do t1:=A003714(n); lprint(n, t1, binary(t1)); od:
  • Mathematica
    fibBin[n_Integer] := Block[{k = Ceiling[Log[GoldenRatio, n Sqrt[5]]], t = n, fr = {}}, While[k > 1, If[t >= Fibonacci[k], AppendTo[fr, 1]; t = t - Fibonacci[k], AppendTo[fr, 0]]; k--]; FromDigits[fr, 2]]; Table[fibBin[n], {n, 0, 61}] (* Robert G. Wilson v, Sep 18 2004 *)
    Select[Range[0, 270], ! MemberQ[Partition[IntegerDigits[#, 2], 2, 1], {1, 1}] &] (* Harvey P. Dale, Jul 17 2011 *)
    Select[Range[256], BitAnd[#, 2 #] == 0 &] (* Alonso del Arte, Jun 18 2012 *)
    With[{r = Range[10^5]}, Pick[r, BitAnd[r, 2 r], 0]] (* Eric W. Weisstein, Aug 18 2017 *)
    Select[Range[0, 299], SequenceCount[IntegerDigits[#, 2], {1, 1}] == 0 &] (* Requires Mathematica version 10 or later. -- Harvey P. Dale, Dec 06 2018 *)
  • PARI
    msb(n)=my(k=1); while(k<=n, k<<=1); k>>1
    for(n=1,1e4,k=bitand(n,n<<1);if(k,n=bitor(n,msb(k)-1),print1(n", "))) \\ Charles R Greathouse IV, Jun 15 2011
    
  • PARI
    select( is_A003714(n)=!bitand(n,n>>1), [0..266])
    {(next_A003714(n,t)=while(t=bitand(n+=1,n<<1), n=bitor(n,1<A003714(t)) \\ M. F. Hasler, Nov 30 2021
    
  • Python
    for n in range(300):
        if 2*n & n == 0:
            print(n, end=",") # Alex Ratushnyak, Jun 21 2012
    
  • Python
    def A003714(n):
        tlist, s = [1,2], 0
        while tlist[-1]+tlist[-2] <= n:
            tlist.append(tlist[-1]+tlist[-2])
        for d in tlist[::-1]:
            s *= 2
            if d <= n:
                s += 1
                n -= d
        return s # Chai Wah Wu, Jun 14 2018
    
  • Python
    def fibbinary():
        x = 0
        while True:
            yield x
            y = ~(x >> 1)
            x = (x - y) & y # Falk Hüffner, Oct 23 2021
    (C++)
    /* start with x=0, then repeatedly call x=next_fibrep(x): */
    ulong next_fibrep(ulong x)
    {
        // 2 examples:         //  ex. 1             //  ex.2
        //                     // x == [*]0 010101   // x == [*]0 01010
        ulong y = x | (x>>1);  // y == [*]? 011111   // y == [*]? 01111
        ulong z = y + 1;       // z == [*]? 100000   // z == [*]? 10000
        z = z & -z;            // z == [0]0 100000   // z == [0]0 10000
        x ^= z;                // x == [*]0 110101   // x == [*]0 11010
        x &= ~(z-1);           // x == [*]0 100000   // x == [*]0 10000
        return x;
    }
    /* Joerg Arndt, Jun 22 2012 */
    
  • Scala
    (0 to 255).filter(n => (n & 2 * n) == 0) // Alonso del Arte, Apr 12 2020
    (C#)
    public static bool IsFibbinaryNum(this int n) => ((n & (n >> 1)) == 0) ? true : false; // Frank Hollstein, Jul 07 2021

Formula

No two adjacent 1's in binary expansion.
Let f(x) := Sum_{n >= 0} x^Fibbinary(n). (This is the generating function of the characteristic function of this sequence.) Then f satisfies the functional equation f(x) = x*f(x^4) + f(x^2).
a(0) = 0, a(1) = 1, a(2) = 2, a(n) = 2^(A072649(n) - 1) + a(n - A000045(1 + A072649(n))). - Antti Karttunen
It appears that this sequence gives m such that A082759(3*m) is odd; or, probably equivalently, m such that A037011(3*m) = 1. - Benoit Cloitre, Jun 20 2003
If m is in the sequence then so are 2*m and 4*m + 1. - Henry Bottomley, Jan 11 2005
A116361(a(n)) <= 1. - Reinhard Zumkeller, Feb 04 2006
A085357(a(n)) = 1; A179821(a(n)) = a(n). - Reinhard Zumkeller, Jul 31 2010
a(n)/n^k is bounded (but does not tend to a limit), where k = 1.44... = A104287. - Charles R Greathouse IV, Sep 19 2012
a(n) = a(A193564(n+1))*2^(A003849(n) + 1) + A003849(n) for n > 0. - Daniel Starodubtsev, Aug 05 2021
There are Fibonacci(n+1) terms with up to n bits in this sequence. - Charles R Greathouse IV, Oct 22 2021
Sum_{n>=1} 1/a(n) = 3.704711752910469457886531055976801955909489488376627037756627135425780134020... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 12 2022

Extensions

Edited by Antti Karttunen, Feb 21 2006
Cross reference to A007820 added (into O-Y.C. comment) by Jason Kimberley, Sep 14 2009
Typo corrected by Jeffrey Shallit, Sep 26 2014

A000295 Eulerian numbers (Euler's triangle: column k=2 of A008292, column k=1 of A173018).

Original entry on oeis.org

0, 0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, 2036, 4083, 8178, 16369, 32752, 65519, 131054, 262125, 524268, 1048555, 2097130, 4194281, 8388584, 16777191, 33554406, 67108837, 134217700, 268435427, 536870882, 1073741793, 2147483616, 4294967263, 8589934558
Offset: 0

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Author

Keywords

Comments

There are 2 versions of Euler's triangle:
* A008292 Classic version of Euler's triangle used by Comtet (1974).
* A173018 Version of Euler's triangle used by Graham, Knuth and Patashnik in Concrete Math. (1990).
Euler's triangle rows and columns indexing conventions:
* A008292 The rows and columns of the Eulerian triangle are both indexed starting from 1. (Classic version: used in the classic books by Riordan and Comtet.)
* A173018 The rows and columns of the Eulerian triangle are both indexed starting from 0. (Graham et al.)
Number of Dyck paths of semilength n having exactly one long ascent (i.e., ascent of length at least two). Example: a(4)=11 because among the 14 Dyck paths of semilength 4, the paths that do not have exactly one long ascent are UDUDUDUD (no long ascent), UUDDUUDD and UUDUUDDD (two long ascents). Here U=(1,1) and D=(1,-1). Also number of ordered trees with n edges having exactly one branch node (i.e., vertex of outdegree at least two). - Emeric Deutsch, Feb 22 2004
Number of permutations of {1,2,...,n} with exactly one descent (i.e., permutations (p(1),p(2),...,p(n)) such that #{i: p(i)>p(i+1)}=1). E.g., a(3)=4 because the permutations of {1,2,3} with one descent are 132, 213, 231 and 312.
a(n+1) is the convolution of nonnegative integers (A001477) and powers of two (A000079). - Graeme McRae, Jun 07 2006
Partial sum of main diagonal of A125127. - Jonathan Vos Post, Nov 22 2006
Number of partitions of an n-set having exactly one block of size > 1. Example: a(4)=11 because, if the partitioned set is {1,2,3,4}, then we have 1234, 123|4, 124|3, 134|2, 1|234, 12|3|4, 13|2|4, 14|2|3, 1|23|4, 1|24|3 and 1|2|34. - Emeric Deutsch, Oct 28 2006
k divides a(k+1) for k in A014741. - Alexander Adamchuk, Nov 03 2006
(Number of permutations avoiding patterns 321, 2413, 3412, 21534) minus one. - Jean-Luc Baril, Nov 01 2007, Mar 21 2008
The chromatic invariant of the prism graph P_n for n >= 3. - Jonathan Vos Post, Aug 29 2008
Decimal integer corresponding to the result of XORing the binary representation of 2^n - 1 and the binary representation of n with leading zeros. This sequence and a few others are syntactically similar. For n > 0, let D(n) denote the decimal integer corresponding to the binary number having n consecutive 1's. Then D(n).OP.n represents the n-th term of a sequence when .OP. stands for a binary operator such as '+', '-', '*', 'quotentof', 'mod', 'choose'. We then get the various sequences A136556, A082495, A082482, A066524, A000295, A052944. Another syntactically similar sequence results when we take the n-th term as f(D(n)).OP.f(n). For example if f='factorial' and .OP.='/', we get (A136556)(A000295) ; if f='squaring' and .OP.='-', we get (A000295)(A052944). - K.V.Iyer, Mar 30 2009
Chromatic invariant of the prism graph Y_n.
Number of labelings of a full binary tree of height n-1, such that each path from root to any leaf contains each label from {1,2,...,n-1} exactly once. - Michael Vielhaber (vielhaber(AT)gmail.com), Nov 18 2009
Also number of nontrivial equivalence classes generated by the weak associative law X((YZ)T)=(X(YZ))T on words with n open and n closed parentheses. Also the number of join (resp. meet)-irreducible elements in the pruning-grafting lattice of binary trees with n leaves. - Jean Pallo, Jan 08 2010
Nonzero terms of this sequence can be found from the row sums of the third sub-triangle extracted from Pascal's triangle as indicated below by braces:
1;
1, 1;
{1}, 2, 1;
{1, 3}, 3, 1;
{1, 4, 6}, 4, 1;
{1, 5, 10, 10}, 5, 1;
{1, 6, 15, 20, 15}, 6, 1;
... - L. Edson Jeffery, Dec 28 2011
For integers a, b, denote by a<+>b the least c >= a, such that the Hamming distance D(a,c) = b (note that, generally speaking, a<+>b differs from b<+>a). Then for n >= 3, a(n) = n<+>n. This has a simple explanation: for n >= 3 in binary we have a(n) = (2^n-1)-n = "anti n". - Vladimir Shevelev, Feb 14 2012
a(n) is the number of binary sequences of length n having at least one pair 01. - Branko Curgus, May 23 2012
Nonzero terms are those integers k for which there exists a perfect (Hamming) error-correcting code. - L. Edson Jeffery, Nov 28 2012
a(n) is the number of length n binary words constructed in the following manner: Select two positions in which to place the first two 0's of the word. Fill in all (possibly none) of the positions before the second 0 with 1's and then complete the word with an arbitrary string of 0's or 1's. So a(n) = Sum_{k=2..n} (k-1)*2^(n-k). - Geoffrey Critzer, Dec 12 2013
Without first 0: a(n)/2^n equals Sum_{k=0..n} k/2^k. For example: a(5)=57, 57/32 = 0/1 + 1/2 + 2/4 + 3/8 + 4/16 + 5/32. - Bob Selcoe, Feb 25 2014
The first barycentric coordinate of the centroid of the first n rows of Pascal's triangle, assuming the numbers are weights, is A000295(n+1)/A000337(n). See attached figure. - César Eliud Lozada, Nov 14 2014
Starting (0, 1, 4, 11, ...), this is the binomial transform of (0, 1, 2, 2, 2, ...). - Gary W. Adamson, Jul 27 2015
Also the number of (non-null) connected induced subgraphs in the n-triangular honeycomb rook graph. - Eric W. Weisstein, Aug 27 2017
a(n) is the number of swaps needed in the worst case to transform a binary tree with n full levels into a heap, using (bottom-up) heapify. - Rudy van Vliet, Sep 19 2017
The utility of large networks, particularly social networks, with n participants is given by the terms a(n) of this sequence. This assertion is known as Reed's Law, see the Wikipedia link. - Johannes W. Meijer, Jun 03 2019
a(n-1) is the number of subsets of {1..n} in which the largest element of the set exceeds by at least 2 the next largest element. For example, for n = 5, a(4) = 11 and the 11 sets are {1,3}, {1,4}, {1,5}, {2,4}, {2,5}, {3,5}, {1,2,4}, {1,2,5}, {1,3,5}, {2,3,5}, {1,2,3,5}. - Enrique Navarrete, Apr 08 2020
a(n-1) is also the number of subsets of {1..n} in which the second smallest element of the set exceeds by at least 2 the smallest element. For example, for n = 5, a(4) = 11 and the 11 sets are {1,3}, {1,4}, {1,5}, {2,4}, {2,5}, {3,5}, {1,3,4}, {1,3,5}, {1,4,5}, {2,4,5}, {1,3,4,5}. - Enrique Navarrete, Apr 09 2020
a(n+1) is the sum of the smallest elements of all subsets of {1..n}. For example, for n=3, a(4)=11; the subsets of {1,2,3} are {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}, and the sum of smallest elements is 11. - Enrique Navarrete, Aug 20 2020
Number of subsets of an n-set that have more than one element. - Eric M. Schmidt, Mar 13 2021
Number of individual bets in a "full cover" bet on n-1 horses, dogs, etc. in different races. Each horse, etc. can be bet on or not, giving 2^n bets. But, by convention, singles (a bet on only one race) are not included, reducing the total number bets by n. It is also impossible to bet on no horses at all, reducing the number of bets by another 1. A full cover on 4 horses, dogs, etc. is therefore 6 doubles, 4 trebles and 1 four-horse etc. accumulator. In British betting, such a bet on 4 horses etc. is a Yankee; on 5, a super-Yankee. - Paul Duckett, Nov 17 2021
From Enrique Navarrete, May 25 2022: (Start)
Number of binary sequences of length n with at least two 1's.
a(n-1) is the number of ways to choose an odd number of elements greater than or equal to 3 out of n elements.
a(n+1) is the number of ways to split [n] = {1,2,...,n} into two (possibly empty) complementary intervals {1,2,...,i} and {i+1,i+2,...,n} and then select a subset from the first interval (2^i choices, 0 <= i <= n), and one block/cell (i.e., subinterval) from the second interval (n-i choices, 0 <= i <= n).
(End)
Number of possible conjunctions in a system of n planets; for example, there can be 0 conjunctions with one planet, one with two planets, four with three planets (three pairs of planets plus one with all three) and so on. - Wendy Appleby, Jan 02 2023
Largest exponent m such that 2^m divides (2^n-1)!. - Franz Vrabec, Aug 18 2023
It seems that a(n-1) is the number of odd r with 0 < r < 2^n for which there exist u,v,w in the x-independent beginning of the Collatz trajectory of 2^n x + r with u+v = w+1, as detailed in the link "Collatz iteration and Euler numbers?". A better understanding of this might also give a formula for A374527. - Markus Sigg, Aug 02 2024
This sequence has a connection to consecutively halved positional voting (CHPV); see Mendenhall and Switkay. - Hal M. Switkay, Feb 25 2025
a(n) is the number of subsets of size 2 and more of an n-element set. Equivalently, a(n) is the number of (hyper)edges of size 2 and more in a complete hypergraph of n vertices. - Yigit Oktar, Apr 05 2025

Examples

			G.f. = x^2 + 4*x^3 + 11*x^4 + 26*x^5 + 57*x^6 + 120*x^7 + 247*x^8 + 502*x^9 + ...
		

References

  • O. Bottema, Problem #562, Nieuw Archief voor Wiskunde, 28 (1980) 115.
  • L. Comtet, "Permutations by Number of Rises; Eulerian Numbers." Section 6.5 in Advanced Combinatorics: The Art of Finite and Infinite Expansions, rev. enl. ed. Dordrecht, Netherlands: Reidel, pp. 51 and 240-246, 1974.
  • F. N. David and D. E. Barton, Combinatorial Chance. Hafner, NY, 1962, p. 151.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990.
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 3, p. 34.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 215.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A008292 (classic version of Euler's triangle used by Comtet (1974)).
Cf. A173018 (version of Euler's triangle used by Graham, Knuth and Patashnik in Concrete Math. (1990)).
Cf. A002662 (partial sums).
Partial sums of A000225.
Row sums of A014473 and of A143291.
Second column of triangles A112493 and A112500.
Sequences A125128 and A130103 are essentially the same.
Column k=1 of A124324.

Programs

  • Haskell
    a000295 n = 2^n - n - 1  -- Reinhard Zumkeller, Nov 25 2013
    
  • Magma
    [2^n-n-1: n in [0..40]]; // Vincenzo Librandi, Jul 29 2015
    
  • Magma
    [EulerianNumber(n, 1): n in [0..40]]; // G. C. Greubel, Oct 02 2024
    
  • Maple
    [ seq(2^n-n-1, n=1..50) ];
    A000295 := -z/(2*z-1)/(z-1)**2; # Simon Plouffe in his 1992 dissertation
    # Grammar specification:
    spec := [S, { B = Set(Z, 1 <= card), C = Sequence(B, 2 <= card), S = Prod(B, C) }, unlabeled]:
    struct := n -> combstruct[count](spec, size = n+1);
    seq(struct(n), n = 0..33); # Peter Luschny, Jul 22 2014
  • Mathematica
    a[n_] = If[n==0, 0, n*(HypergeometricPFQ[{1, 1-n}, {2}, -1] - 1)];
    Table[a[n], {n,0,40}] (* Olivier Gérard, Mar 29 2011 *)
    LinearRecurrence[{4, -5, 2}, {0, 0, 1}, 40] (* Vincenzo Librandi, Jul 29 2015 *)
    Table[2^n -n-1, {n,0,40}] (* Eric W. Weisstein, Nov 16 2017 *)
  • PARI
    a(n)=2^n-n-1 \\ Charles R Greathouse IV, Jun 10 2011
    
  • SageMath
    [2^n -(n+1) for n in range(41)] # G. C. Greubel, Oct 02 2024

Formula

a(n) = 2^n - n - 1.
G.f.: x^2/((1-2*x)*(1-x)^2).
A107907(a(n+2)) = A000079(n+2). - Reinhard Zumkeller, May 28 2005
E.g.f.: exp(x)*(exp(x)-1-x). - Emeric Deutsch, Oct 28 2006
a(0)=0, a(1)=0, a(n) = 3*a(n-1) - 2*a(n-2) + 1. - Miklos Kristof, Mar 09 2005
a(0)=0, a(n) = 2*a(n-1) + n - 1 for all n in Z.
a(n) = Sum_{k=2..n} binomial(n, k). - Paul Barry, Jun 05 2003
a(n+1) = Sum_{i=1..n} Sum_{j=1..i} C(i, j). - Benoit Cloitre, Sep 07 2003
a(n+1) = 2^n*Sum_{k=0..n} k/2^k. - Benoit Cloitre, Oct 26 2003
a(0)=0, a(1)=0, a(n) = Sum_{i=0..n-1} i+a(i) for i > 1. - Gerald McGarvey, Jun 12 2004
a(n+1) = Sum_{k=0..n} (n-k)*2^k. - Paul Barry, Jul 29 2004
a(n) = Sum_{k=0..n} binomial(n, k+2); a(n+2) = Sum_{k=0..n} binomial(n+2, k+2). - Paul Barry, Aug 23 2004
a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-k-1, k+1)*2^(n-k-2)*(-1/2)^k. - Paul Barry, Oct 25 2004
a(0) = 0; a(n) = Stirling2(n,2) + a(n-1) = A000225(n-1) + a(n-1). - Thomas Wieder, Feb 18 2007
a(n) = A000325(n) - 1. - Jonathan Vos Post, Aug 29 2008
a(0) = 0, a(n) = Sum_{k=0..n-1} 2^k - 1. - Doug Bell, Jan 19 2009
a(n) = A000217(n-1) + A002662(n) for n>0. - Geoffrey Critzer, Feb 11 2009
a(n) = A000225(n) - n. - Zerinvary Lajos, May 29 2009
a(n) = n*(2F1([1,1-n],[2],-1) - 1). - Olivier Gérard, Mar 29 2011
Column k=1 of A173018 starts a'(n) = 0, 1, 4, 11, ... and has the hypergeometric representation n*hypergeom([1, -n+1], [-n], 2). This can be seen as a formal argument to prefer Euler's A173018 over A008292. - Peter Luschny, Sep 19 2014
E.g.f.: exp(x)*(exp(x)-1-x); this is U(0) where U(k) = 1 - x/(2^k - 2^k/(x + 1 - x^2*2^(k+1)/(x*2^(k+1) - (k+1)/U(k+1)))); (continued fraction, 3rd kind, 4-step). - Sergei N. Gladkovskii, Dec 01 2012
a(n) = A079583(n) - A000225(n+1). - Miquel Cerda, Dec 25 2016
a(0) = 0; a(1) = 0; for n > 1: a(n) = Sum_{i=1..2^(n-1)-1} A001511(i). - David Siegers, Feb 26 2019
a(n) = A007814(A028366(n)). - Franz Vrabec, Aug 18 2023
a(n) = Sum_{k=1..floor((n+1)/2)} binomial(n+1, 2*k+1). - Taras Goy, Jan 02 2025

A000325 a(n) = 2^n - n.

Original entry on oeis.org

1, 1, 2, 5, 12, 27, 58, 121, 248, 503, 1014, 2037, 4084, 8179, 16370, 32753, 65520, 131055, 262126, 524269, 1048556, 2097131, 4194282, 8388585, 16777192, 33554407, 67108838, 134217701, 268435428, 536870883, 1073741794, 2147483617
Offset: 0

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Author

Rosario Salamone (Rosario.Salamone(AT)risc.uni-linz.ac.at)

Keywords

Comments

Number of permutations of degree n with at most one fall; called "Grassmannian permutations" by Lascoux and Schützenberger. - Axel Kohnert (Axel.Kohnert(AT)uni-bayreuth.de)
Number of different permutations of a deck of n cards that can be produced by a single shuffle. [DeSario]
Number of Dyck paths of semilength n having at most one long ascent (i.e., ascent of length at least two). Example: a(4)=12 because among the 14 Dyck paths of semilength 4, the only paths that have more than one long ascent are UUDDUUDD and UUDUUDDD (each with two long ascents). Here U = (1, 1) and D = (1, -1). Also number of ordered trees with n edges having at most one branch node (i.e., vertex of outdegree at least two). - Emeric Deutsch, Feb 22 2004
Number of {12,1*2*,21*}-avoiding signed permutations in the hyperoctahedral group.
Number of 1342-avoiding circular permutations on [n+1].
2^n - n is the number of ways to partition {1, 2, ..., n} into arithmetic progressions, where in each partition all the progressions have the same common difference and have lengths at least 1. - Marty Getz (ffmpg1(AT)uaf.edu) and Dixon Jones (fndjj(AT)uaf.edu), May 21 2005
if b(0) = x and b(n) = b(n-1) + b(n-1)^2*x^(n-2) for n > 0, then b(n) is a polynomial of degree a(n). - Michael Somos, Nov 04 2006
The chromatic invariant of the Mobius ladder graph M_n for n >= 2. - Jonathan Vos Post, Aug 29 2008
Dimension sequence of the dual alternative operad (i.e., associative and satisfying the identity xyz + yxz + zxy + xzy + yzx + zyx = 0) over the field of characteristic 3. - Pasha Zusmanovich, Jun 09 2009
An elephant sequence, see A175654. For the corner squares six A[5] vectors, with decimal values between 26 and 176, lead to this sequence (without the first leading 1). For the central square these vectors lead to the companion sequence A168604. - Johannes W. Meijer, Aug 15 2010
a(n+1) is also the number of order-preserving and order-decreasing partial isometries (of an n-chain). - Abdullahi Umar, Jan 13 2011
A040001(n) = p(-1) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 0, 1, ..., n. - Michael Somos, May 12 2012
A130103(n+1) = p(n+1) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 0, 1, ..., n. - Michael Somos, May 12 2012
The number of labeled graphs with n vertices whose vertex set can be partitioned into a clique and a set of isolated points. - Alex J. Best, Nov 20 2012
For n > 0, a(n) is a B_2 sequence. - Thomas Ordowski, Sep 23 2014
See coefficients of the linear terms of the polynomials of the table on p. 10 of the Getzler link. - Tom Copeland, Mar 24 2016
Consider n points lying on a circle, then for n>=2 a(n-1) is the maximum number of ways to connect two points with non-intersecting chords. - Anton Zakharov, Dec 31 2016
Also the number of cliques in the (n-1)-triangular honeycomb rook graph. - Eric W. Weisstein, Jul 14 2017
From Eric M. Schmidt, Jul 17 2017: (Start)
Number of sequences (e(1), ..., e(n)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) != e(j) < e(k). [Martinez and Savage, 2.7]
Number of sequences (e(1), ..., e(n)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i), e(j), e(k) pairwise distinct. [Martinez and Savage, 2.7]
Number of sequences (e(1), ..., e(n)), 0 <= e(i) < i, such that there is no triple i < j < k with e(j) >= e(k) and e(i) != e(k) pairwise distinct. [Martinez and Savage, 2.7]
(End)
Number of F-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are F-equivalent iff the positions of pattern F are identical in these paths. - Sergey Kirgizov, Apr 08 2018
From Gus Wiseman, Feb 10 2019: (Start)
Also the number of connected partitions of an n-cycle. For example, the a(1) = 1 through a(4) = 12 connected partitions are:
{{1}} {{12}} {{123}} {{1234}}
{{1}{2}} {{1}{23}} {{1}{234}}
{{12}{3}} {{12}{34}}
{{13}{2}} {{123}{4}}
{{1}{2}{3}} {{124}{3}}
{{134}{2}}
{{14}{23}}
{{1}{2}{34}}
{{1}{23}{4}}
{{12}{3}{4}}
{{14}{2}{3}}
{{1}{2}{3}{4}}
(End)
Number of subsets of n-set without the single-element subsets. - Yuchun Ji, Jul 16 2019
For every prime p, there are infinitely many terms of this sequence that are divisible by p (see IMO Compendium link and Doob reference). Corresponding indices n are: for p = 2, even numbers A299174; for p = 3, A047257; for p = 5, A349767. - Bernard Schott, Dec 10 2021
Primes are in A081296 and corresponding indices in A048744. - Bernard Schott, Dec 12 2021

Examples

			G.f. = 1 + x + 2*x^2 + 5*x^3 + 12*x^4 + 27*x^5 + 58*x^6 + 121*x^7 + ...
		

References

  • Michael Doob, The Canadian Mathematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993, Canadian Mathematical Society & Société Mathématique du Canada, Problem 4, 1983, page 158, 1993.

Crossrefs

Column 1 of triangle A008518.
Row sum of triangles A184049 and A184050.

Programs

  • Haskell
    a000325 n = 2 ^ n - n
    a000325_list = zipWith (-) a000079_list [0..]
    -- Reinhard Zumkeller, Jul 17 2012
    
  • Magma
    [2^n - n: n in [0..35]]; // Vincenzo Librandi, May 13 2011
    
  • Maple
    A000325 := proc(n) option remember; if n <=1 then n+1 else 2*A000325(n-1)+n-1; fi; end;
    g:=1/(1-2*z): gser:=series(g, z=0, 43): seq(coeff(gser, z, n)-n, n=0..31); # Zerinvary Lajos, Jan 09 2009
  • Mathematica
    Table[2^n - n, {n, 0, 39}] (* Alonso del Arte, Sep 15 2014 *)
    LinearRecurrence[{4, -5, 2}, {1, 2, 5}, {0, 20}] (* Eric W. Weisstein, Jul 14 2017 *)
  • PARI
    {a(n) = 2^n - n}; /* Michael Somos, Nov 04 2006 */
    
  • Python
    def A000325(n): return (1<Chai Wah Wu, Jan 11 2023

Formula

a(n+1) = 2*a(n) + n - 1, a(0) = 1. - Reinhard Zumkeller, Apr 12 2003
Binomial transform of 1, 0, 1, 1, 1, .... The sequence starting 1, 2, 5, ... has a(n) = 1 + n + 2*Sum_{k=2..n} binomial(n, k) = 2^(n+1) - n - 1. This is the binomial transform of 1, 1, 2, 2, 2, 2, .... a(n) = 1 + Sum_{k=2..n} C(n, k). - Paul Barry, Jun 06 2003
G.f.: (1-3x+3x^2)/((1-2x)*(1-x)^2). - Emeric Deutsch, Feb 22 2004
A107907(a(n+2)) = A000051(n+2) for n > 0. - Reinhard Zumkeller, May 28 2005
a(n+1) = sum of n-th row of the triangle in A109128. - Reinhard Zumkeller, Jun 20 2005
Row sums of triangle A133116. - Gary W. Adamson, Sep 14 2007
G.f.: 1 / (1 - x / (1 - x / ( 1 - x / (1 + x / (1 - 2*x))))). - Michael Somos, May 12 2012
First difference is A000225. PSUM transform is A084634. - Michael Somos, May 12 2012
a(n) = [x^n](B(x)^n-B(x)^(n-1)), n>0, a(0)=1, where B(x) = (1+2*x+sqrt(1+4*x^2))/2. - Vladimir Kruchinin, Mar 07 2014
E.g.f.: (exp(x) - x)*exp(x). - Ilya Gutkovskiy, Aug 07 2016
a(n) = A125128(n) - A000225(n) + 1. - Miquel Cerda, Aug 12 2016
a(n) = 2*A125128(n) - A095151(n) + 1. - Miquel Cerda, Aug 12 2016
a(n) = A079583(n-1) - A000225(n-1). - Miquel Cerda, Aug 15 2016
a(n)^2 - 4*a(n-1)^2 = (n-2)*(a(n)+2*a(n-1)). - Yuchun Ji, Jul 13 2018
a(n) = 2^(-n) * A186947(n) = 2^n * A002064(-n) for all n in Z. - Michael Somos, Jul 18 2018
a(2^n) = (2^a(n) - 1)*2^n. - Lorenzo Sauras Altuzarra, Feb 01 2022

A318928 Runs-resistance of binary representation of n.

Original entry on oeis.org

1, 2, 1, 3, 2, 3, 1, 3, 3, 2, 4, 2, 4, 3, 1, 3, 3, 5, 4, 4, 2, 5, 4, 3, 4, 4, 3, 3, 4, 3, 1, 3, 3, 5, 3, 3, 5, 4, 3, 4, 5, 2, 4, 3, 4, 5, 4, 3, 3, 3, 2, 4, 4, 3, 3, 2, 3, 4, 3, 3, 4, 3, 1, 3, 3, 5, 3, 3, 5, 3, 4, 3, 3, 5, 6, 4, 5, 3, 3, 4, 5, 4, 4, 4, 2, 5, 4, 5, 5, 4, 5, 5, 4, 5, 4
Offset: 1

Views

Author

N. J. A. Sloane, Sep 09 2018

Keywords

Comments

Following Lenormand (2003), we define the "runs-resistance" of a finite list L to be the number of times the RUNS transformation must be applied to L in order to reduce L to a list with a single element.
Here it is immaterial whether we read the binary representation of n from left to right or right to left.
The RUNS transformation must be applied at least once, in order to obtain a list, so a(n) >= 1.

Examples

			11 in binary is [1, 0, 1, 1],
which has runs of lengths [1, 1, 2],
which has runs of lengths [2, 1],
which has runs of lengths [1, 1],
which has a single run of length [2].
This took four steps, so a(11) = 4.
		

Crossrefs

See A319103 for an inverse, and A319417 and A319418 for records.
Ignoring the first digit gives A329870.
Cuts-resistance is A319416.
Compositions counted by runs-resistance are A329744.
Binary words counted by runs-resistance are A319411 and A329767.

Programs

  • Maple
    with(transforms);
    # compute Lenormand's "resistance" of a list
    resist:=proc(a) local ct,i,b;
    if whattype(a) <> list then ERROR("input must be a list"); fi:
    ct:=0; b:=a; for i from 1 to 100 do
    if nops(b)=1 then return(ct); fi;
    b:=RUNS(b); ct:=ct+1; od; end;
    a:=[1];
    for n from 2 to 100 do
    b:=convert(n,base,2);
    r:=resist(b);
    a:=[op(a),r];
    od:
  • Mathematica
    Table[If[n == 1, 1, Length[NestWhileList[Length/@Split[#] &, IntegerDigits[n, 2], Length[#] > 1 &]] - 1], {n, 50}] (* Gus Wiseman, Nov 25 2019 *)

Extensions

a(1) corrected by N. J. A. Sloane, Sep 20 2018

A000247 a(n) = 2^n - n - 2.

Original entry on oeis.org

0, 3, 10, 25, 56, 119, 246, 501, 1012, 2035, 4082, 8177, 16368, 32751, 65518, 131053, 262124, 524267, 1048554, 2097129, 4194280, 8388583, 16777190, 33554405, 67108836, 134217699, 268435426, 536870881, 1073741792, 2147483615
Offset: 2

Views

Author

Keywords

Comments

Ways of placing n+1 labeled balls into 2 indistinguishable boxes with at least 2 balls in each box.
2^a(n) is an integer of the form 1/(2 - Sum_{i=1..m} i/2^i). - Benoit Cloitre, Oct 25 2002
Number of permutations avoiding 13-2 that contain the pattern 23-1 exactly twice.
Cost of ternary maximum height Huffman tree with N internal nodes (non-leaves) for minimizing absolutely ordered sequences of size n = 2N + 1. - Alex Vinokur (alexvn(AT)barak-online.net), Nov 02 2004
a(n) is the number of Dyck n-paths whose third upstep initiates the last long ascent, n >= 1. A long ascent is one consisting of 2 or more upsteps. For example, a(3)=3 counts UUDuUDDD, UDUDuUDD, UUDDuUDD (third upstep in small type). - David Callan, Dec 08 2004
Subsequence of A158581; A000120(a(n)) > 1. - Reinhard Zumkeller, Apr 16 2009
Number of vertices of the tropical Grassmannian simplicial complex G(2,n), related to phylogenetic trees. - Tom Copeland, Oct 03 2011
(Conjecture) Let a(2)=0. For n > 2, let N = 2*n + 1. For each n, define the n X n tridiagonal unit-primitive matrix (see [Jeffery]) A_{N,1}=[0,1,0,...,0; 1,0,1,0,...,0; 0,1,0,1,0,...,0; ...; 0,...,0,1,0,1; 0,...,0,1,1] associated with N. Define the n-dimensional column vectors V_N = [v_1,v_2,...,v_n]^T = [A_{N,1}]^n*[1,1,...,1]^T, where [.]^T denotes matrix transpose and [1,...,1] is the n-dimensional unit vector. Let (v_k)N denote the k-th element of V_N, k in {1,...,n}. Then a(n) = (v(n-2))N. - _L. Edson Jeffery, Jan 20 2012
For n>0, (a(n)) is row 3 of the convolution array A213568. - Clark Kimberling, Jun 20 2012
For n>2, a(n-2) is the number of connected induced (non-null) subgraphs of the n-centipede graph. - Giovanni Resta, May 04 2017
a(n) is the number of maximal boundary strata of the moduli space of stable rational curves with n+1 marked points. The closures of the maximal boundary strata are called the irreducible boundary divisors of the moduli space; see Cavalieri Section 2.1. - Harry Richman, Aug 13 2024

Examples

			a(3) = 4!/(2!*2!*2!) = 3.
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 222.
  • F. N. David and D. E. Barton, Combinatorial Chance. Hafner, NY, 1962, p. 296.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 76.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000478 (3 boxes), A058844 (4 boxes).

Programs

Formula

E.g.f.: (exp(x)-1-x)*(exp(x)-1).
G.f.: x^3*(3-2*x)/((1-2*x)*(1-x)^2).
a(n) = 2*a(n-1) + n + 3 = a(n-1) + 2^(n-1) - 1 = A000295(n) - 1 = A000295(n+1) - 2^(n+1).
A107907(a(n)) = A000225(n). - Reinhard Zumkeller, May 28 2005
Starting (3, 10, 25, 56, ...) = binomial transform of [3, 7, 8, 8, 8, ...]. - Gary W. Adamson, Nov 07 2007
a(2)=0, a(3)=3, a(4)=10, a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3). - Harvey P. Dale, Aug 23 2011
a(n) = (Sum_{k=2..floor(n/2)} binomial(n+1,k)) + if(n odd, binomial(n+1,(n+1)/2)/2, 0).
a(n) = Sum_{k=0..n-3} Sum_{i=0..n-1} C(i,k). - Wesley Ivan Hurt, Sep 20 2017

Extensions

Additional comments from Michael Steyer, Dec 02 2000
More terms from Larry Reeves (larryr(AT)acm.org), Dec 04 2000
I recently changed the beginning of this sequence so the formulas etc. may need to be adjusted. - N. J. A. Sloane, Jan 24 2006
Formulas and comments adjusted for offset by Franklin T. Adams-Watters, Nov 10 2011

A297146 Numbers having an up-first zigzag pattern in base 10; see Comments.

Original entry on oeis.org

12, 13, 14, 15, 16, 17, 18, 19, 23, 24, 25, 26, 27, 28, 29, 34, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 56, 57, 58, 59, 67, 68, 69, 78, 79, 89, 120, 121, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 145
Offset: 1

Views

Author

Clark Kimberling, Jan 15 2018

Keywords

Comments

A number n having base-b digits d(m), d(m-1),..., d(0) such that d(i) != d(i+1) for 0 <= i < m shows a zigzag pattern of one or more segments, in the following sense. Writing U for up and D for down, there are two kinds of patterns: U, UD, UDU, UDUD, ... and D, DU, DUD, DUDU, ... . In the former case, we say n has an "up-first zigzag pattern in base b"; in the latter, a "down-first zigzag pattern in base b". Example: 2,4,5,3,0,1,4,2 has segments 2,4,5; 5,3,0; 0,1,4; and 4,2, so that 24530142, with pattern UDUD, has an up-first zigzag pattern in base 10, whereas 4,2,5,3,0,1,4,2 has a down-first pattern. The sequences A297146-A297148 partition the natural numbers. In the following guide, column four, "complement" means the sequence of natural numbers not in the corresponding sequences in columns 2 and 3.
***
Base up-first down-first complement
2 (none) A000975 A107907

Examples

			Base-10 digits of 59898: 5,9,8,9,8, with pattern UDUD, so that 59898 is in the sequence.
		

Crossrefs

Programs

  • Mathematica
    a[n_, b_] := Sign[Differences[IntegerDigits[n, b]]]; z = 300;
    b = 10; t = Table[a[n, b], {n, 1, 10*z}];
    u = Select[Range[z], ! MemberQ[t[[#]], 0] && First[t[[#]]] == 1 &]   (* A297146 *)
    v = Select[Range[z], ! MemberQ[t[[#]], 0] && First[t[[#]]] == -1 &]  (* A297147 *)
    Complement[Range[z], Union[u, v]]  (* A297148 *)

A329865 Numbers whose binary expansion has the same runs-resistance as cuts-resistance.

Original entry on oeis.org

0, 8, 12, 14, 17, 24, 27, 28, 35, 36, 39, 47, 49, 51, 54, 57, 61, 70, 73, 78, 80, 99, 122, 130, 156, 175, 184, 189, 190, 198, 204, 207, 208, 215, 216, 226, 228, 235, 243, 244, 245, 261, 271, 283, 295, 304, 313, 321, 322, 336, 352, 367, 375, 378, 379, 380, 386
Offset: 1

Views

Author

Gus Wiseman, Nov 23 2019

Keywords

Comments

For the operation of taking the sequence of run-lengths of a finite sequence, runs-resistance is defined to be the number of applications required to reach a singleton.
For the operation of shortening all runs by 1, cuts-resistance is defined to be the number of applications required to reach an empty word.

Examples

			The sequence of terms together with their binary expansions begins:
    0:
    8:      1000
   12:      1100
   14:      1110
   17:     10001
   24:     11000
   27:     11011
   28:     11100
   35:    100011
   36:    100100
   39:    100111
   47:    101111
   49:    110001
   51:    110011
   54:    110110
   57:    111001
   61:    111101
   70:   1000110
   73:   1001001
   78:   1001110
   80:   1010000
For example, 36 has runs-resistance 3 because we have (100100) -> (1212) -> (1111) -> (4), while the cuts-resistance is also 3 because we have (100100) -> (00) -> (0) -> ().
Similarly, 57 has runs-resistance 3 because we have (111001) -> (321) -> (111) -> (3), while the cuts-resistance is also 3 because we have (111001) -> (110) -> (1) -> ().
		

Crossrefs

Positions of 0's in A329867.
The version for runs-resistance equal to cuts-resistance minus 1 is A329866.
Compositions with runs-resistance equal to cuts-resistance are A329864.
Runs-resistance of binary expansion is A318928.
Cuts-resistance of binary expansion is A319416.
Compositions counted by runs-resistance are A329744.
Compositions counted by cuts-resistance are A329861.
Binary words counted by runs-resistance are A319411 and A329767.
Binary words counted by cuts-resistance are A319421 and A329860.

Programs

  • Mathematica
    runsres[q_]:=Length[NestWhileList[Length/@Split[#]&,q,Length[#]>1&]]-1;
    degdep[q_]:=Length[NestWhileList[Join@@Rest/@Split[#]&,q,Length[#]>0&]]-1;
    Select[Range[0,100],#==0||runsres[IntegerDigits[#,2]]==degdep[IntegerDigits[#,2]]&]

A107909 Numbers having no consecutive zeros or no consecutive ones in binary representation.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 26, 27, 29, 30, 31, 32, 33, 34, 36, 37, 40, 41, 42, 43, 45, 46, 47, 53, 54, 55, 58, 59, 61, 62, 63, 64, 65, 66, 68, 69, 72, 73, 74, 80, 81, 82, 84, 85, 86, 87, 90, 91, 93, 94, 95, 106, 107, 109, 110, 111
Offset: 0

Views

Author

Reinhard Zumkeller, May 28 2005

Keywords

Comments

Union of A003754 and A003714, complement of A107911;
a(A023548(n+2)) = A052940(n+1) for n>0;
a(A001924(n)) = A000225(n) = 2^n - 1;
a(A000126(n)) = A000079(n) = 2^n for n>0;
A107910(n) = a(n+1) - a(n).

Crossrefs

Programs

  • Perl
    foreach $n(1..100){$_=sprintf("%b",$n); print "$n\n" if !m/11/||!m/00/}
    # Ivan Neretin, May 01 2016

A329862 Positive integers whose binary expansion has cuts-resistance 2.

Original entry on oeis.org

3, 4, 6, 9, 11, 12, 13, 18, 19, 20, 22, 25, 26, 37, 38, 41, 43, 44, 45, 50, 51, 52, 53, 74, 75, 76, 77, 82, 83, 84, 86, 89, 90, 101, 102, 105, 106, 149, 150, 153, 154, 165, 166, 169, 171, 172, 173, 178, 179, 180, 181, 202, 203, 204, 205, 210, 211, 212, 213
Offset: 1

Views

Author

Gus Wiseman, Nov 23 2019

Keywords

Comments

For the operation of shortening all runs by 1, cuts-resistance is defined to be the number of applications required to reach an empty word.

Examples

			The sequence of terms together with their binary expansions begins:
   3:      11
   4:     100
   6:     110
   9:    1001
  11:    1011
  12:    1100
  13:    1101
  18:   10010
  19:   10011
  20:   10100
  22:   10110
  25:   11001
  26:   11010
  37:  100101
  38:  100110
  41:  101001
  43:  101011
  44:  101100
  45:  101101
  50:  110010
		

Crossrefs

Positions of 2's in A319416.
Numbers whose binary expansion has cuts-resistance 1 are A000975.
Binary words with cuts-resistance 2 are conjectured to be A027383.
Compositions with cuts-resistance 2 are A329863.
Cuts-resistance of binary expansion without first digit is A319420.
Binary words counted by cuts-resistance are A319421 and A329860.
Compositions counted by cuts-resistance are A329861.

Programs

  • Mathematica
    degdep[q_]:=Length[NestWhileList[Join@@Rest/@Split[#]&,q,Length[#]>0&]]-1;
    Select[Range[100],degdep[IntegerDigits[#,2]]==2&]
Showing 1-10 of 17 results. Next