A011782 -id:A011782 - cp's OEIS frontend

A006257 Josephus problem: a(2n) = 2a(n)-1, a(2n+1) = 2a(n)+1.

0, 1, 1, 3, 1, 3, 5, 7, 1, 3, 5, 7, 9, 11, 13, 15, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29

Offset: 0

N. J. A. Sloane

Write the numbers 1 through n in a circle, start at 1 and cross off every other number until only one number is left.
A version of the children's game "One potato, two potato, ...".
a(n)/A062383(n) = (0, 0.1, 0.01, 0.11, 0.001, ...) enumerates all binary fractions in the unit interval [0, 1). - Fredrik Johansson, Aug 14 2006
Iterating a(n), a(a(n)), ... eventually leads to 2^A000120(n) - 1. - Franklin T. Adams-Watters, Apr 09 2010
By inspection, the solution to the Josephus Problem is a sequence of odd numbers (from 1) starting at each power of 2.  This yields a direct closed form expression (see formula below). - Gregory Pat Scandalis, Oct 15 2013
Also zero together with a triangle read by rows in which row n lists the first 2^(n-1) odd numbers (see A005408), n >= 1. Row lengths give A011782. Right border gives A000225. Row sums give A000302, n >= 1. See example. - Omar E. Pol, Oct 16 2013
For n > 0: a(n) = n + 1 - A080079(n). - Reinhard Zumkeller, Apr 14 2014
In binary, a(n) = ROL(n), where ROL = rotate left = remove the leftmost digit and append it to the right. For example, n = 41 = 101001_2 => a(n) = (0)10011_2 = 19. This also explains FTAW's comment above. - M. F. Hasler, Nov 02 2016
In the under-down Australian card deck separation: top card on bottom of a deck of n cards, next card separated on the table, etc., until one card is left. The position a(n), for n >= 1, from top will be the left over card. See, e.g., the Behrends reference, pp. 156-164. For the down-under case see 2*A053645(n), for n >= 3, n not a power of 2. If n >= 2 is a power of 2 the botton card survives. - Wolfdieter Lang, Jul 28 2020

			From _Omar E. Pol_, Jun 09 2009: (Start)
Written as an irregular triangle the sequence begins:
  0;
  1;
  1,3;
  1,3,5,7;
  1,3,5,7,9,11,13,15;
  1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31;
  1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,
   43,45,47,49,51,53,55,57,59,61,63;
...
(End)
From _Omar E. Pol_, Nov 03 2018: (Start)
An illustration of initial terms, where a(n) is the area (or number of cells) in the n-th region of the structure:
   n   a(n)       Diagram
   0    0     _
   1    1    |_|_ _
   2    1      |_| |
   3    3      |_ _|_ _ _ _
   4    1          |_| | | |
   5    3          |_ _| | |
   6    5          |_ _ _| |
   7    7          |_ _ _ _|
(End)

Erhard Behrends, Der mathematische Zauberstab, Rowolth Taschenbuch Verlag, rororo 62902, 4. Auflage, 2019, pp. 156-164. [English version: The Math Behind the Magic, AMS, 2019.]
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 10.
M. S. Petković, "Josephus problem", Famous Puzzles of Great Mathematicians, page 179, Amer. Math. Soc. (AMS), 2009.
Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Paul Weisenhorn, Josephus und seine Folgen, MNU, 59(2006), pp. 18-19.

Iain Fox, Table of n, a(n) for n = 0..100000 (terms 0..1000 from T. D. Noe, terms 1001..10000 from Indranil Ghosh).
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, Theoretical Computer Sci., 98 (1992), 163-197, ex. 34.
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.
Paul Barry, Conjectures and results on some generalized Rueppel sequences, arXiv:2107.00442 [math.CO], 2021.
Daniel Erman and Brady Haran, The Josephus Problem, Numberphile video (2016)
Chris Groër, The Mathematics of Survival: From Antiquity to the Playground, Amer. Math. Monthly, 110 (No. 9, 2003), 812-825.
Alasdair MacFhraing, Aireamh Muinntir Fhinn Is Dhubhain, Agus Sgeul Josephuis Is An Da Fhichead Iudhaich, [Gaelic with English summary], Proc. Royal Irish Acad., Vol. LII, Sect. A., No. 7, 1948, 87-93.
Yuri Nikolayevsky and Ioannis Tsartsaflis, Cohomology of N-graded Lie algebras of maximal class over Z_2, arXiv:1512.87676 [math.RA], (2016), pages 2, 6.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Eric Weisstein's World of Mathematics, Josephus Problem
Wikipedia, Josephus problem
Index entries for sequences related to the Josephus Problem

Cf. A005428, A035327, A038572, A049940, A049964, A053644, A053645, A088147, A088442.
Second column, and main diagonal, of triangle A032434.
Cf. A181281 (with s=5), A054995 (with s=3).
Column k=2 of A360099.

Require Import ZArith.
Fixpoint a (n : positive) : Z :=
match n with
| xH => 1
| xI n' => (2*(a n') + 1)%Z
| xO n' => (2*(a n') - 1)%Z
end.
(* Stefan Haan, Aug 27 2023 *)

a006257 n = a006257_list !! n
a006257_list =
   0 : 1 : (map (+ 1) $ zipWith mod (map (+ 1) $ tail a006257_list) [2..])
-- Reinhard Zumkeller, Oct 06 2011

[0] cat [2*(n-2^Floor(Log(2,n)))+1: n in [1..100]]; // Vincenzo Librandi, Jan 14 2016

a(0):=0: for n from 1 to 100 do a(n):=(a(n-1)+1) mod n +1: end do:
seq(a(i),i=0..100); # Paul Weisenhorn, Oct 10 2010; corrected by Robert Israel, Jan 13 2016
A006257 := proc(n)
    convert(n,base,2) ;
    ListTools[Rotate](%,-1) ;
    add( op(i,%)*2^(i-1),i=1..nops(%)) ;
end proc: # R. J. Mathar, May 20 2016
A006257 := n -> 2*n  - Bits:-Iff(n, n):
seq(A006257(n), n=0..78); # Peter Luschny, Sep 24 2019

Table[ FromDigits[ RotateLeft[ IntegerDigits[n, 2]], 2], {n, 0, 80}] (* Robert G. Wilson v, Sep 21 2003 *)
Flatten@Table[Range[1, 2^n - 1, 2], {n, 0, 5}] (* Birkas Gyorgy, Feb 07 2011 *)
m = 5; Range[2^m - 1] + 1 - Flatten@Table[Reverse@Range[2^n], {n, 0, m - 1}] (* Birkas Gyorgy, Feb 07 2011 *)

a(n)=sum(k=1,n,if(bitxor(n,k)Paul D. Hanna

a(n)=if(n, 2*n-2^logint(2*n,2)+1, 0) \\ Charles R Greathouse IV, Oct 29 2016

import math
def A006257(n):
     return 0 if n==0 else 2*(n-2**int(math.log(n,2)))+1 # Indranil Ghosh, Jan 11 2017

def A006257(n): return bool(n&(m:=1<Chai Wah Wu, Jan 22 2023
(C#)
static long cs_A006257(this long n) => n == 0 ? 0 : 1 + (1 + (n - 1).cs_A006257()) % n; // Frank Hollstein, Feb 24 2021

To get a(n), write n in binary, rotate left 1 place.
a(n) = 2*A053645(n) + 1 = 2(n-msb(n))+1. - Marc LeBrun, Jul 11 2001. [Here "msb" = "most significant bit", A053644.]
G.f.: 1 + 2/(1-x) * ((3*x-1)/(2-2*x) - Sum_{k>=1} 2^(k-1)*x^2^k). - Ralf Stephan, Apr 18 2003
a(n) = number of positive integers k < n such that n XOR k < n. a(n) = n - A035327(n). - Paul D. Hanna, Jan 21 2006
a(n) = n for n = 2^k - 1. - Zak Seidov, Dec 14 2006
a(n) = n - A035327(n). - K. Spage, Oct 22 2009
a(2^m+k) = 1+2*k; with 0 <= m and 0 <= k < 2^m; n = 2^m+k; m = floor(log_2(n)); k = n-2^m; a(n) = ((a(n-1)+1) mod n) + 1; a(1) = 1. E.g., n=27; m=4; k=11; a(27) = 1 + 2*11 = 23. - Paul Weisenhorn, Oct 10 2010
a(n) = 2*(n - 2^floor(log_2(n))) + 1 (see comment above). - Gregory Pat Scandalis, Oct 15 2013
a(n) = 0 if n = 0 and a(n) = 2*a(floor(n/2)) - (-1)^(n mod 2) if n > 0. - Marek A. Suchenek, Mar 31 2016
G.f. A(x) satisfies: A(x) = 2*A(x^2)*(1 + x) + x/(1 + x). - Ilya Gutkovskiy, Aug 31 2019
For n > 0: a(n) = 2 * A062050(n) - 1. - Frank Hollstein, Oct 25 2021

More terms from Robert G. Wilson v, Sep 21 2003

A039599 Triangle formed from even-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 9, 5, 1, 14, 28, 20, 7, 1, 42, 90, 75, 35, 9, 1, 132, 297, 275, 154, 54, 11, 1, 429, 1001, 1001, 637, 273, 77, 13, 1, 1430, 3432, 3640, 2548, 1260, 440, 104, 15, 1, 4862, 11934, 13260, 9996, 5508, 2244, 663, 135, 17, 1

Offset: 0

N. J. A. Sloane

T(n,k) is the number of lattice paths from (0,0) to (n,n) with steps E = (1,0) and N = (0,1) which touch but do not cross the line x - y = k and only situated above this line; example: T(3,2) = 5 because we have EENNNE, EENNEN, EENENN, ENEENN, NEEENN. - Philippe Deléham, May 23 2005
The matrix inverse of this triangle is the triangular matrix T(n,k) = (-1)^(n+k)* A085478(n,k). - Philippe Deléham, May 26 2005
Essentially the same as A050155 except with a leading diagonal A000108 (Catalan numbers) 1, 1, 2, 5, 14, 42, 132, 429, .... - Philippe Deléham, May 31 2005
Number of Grand Dyck paths of semilength n and having k downward returns to the x-axis. (A Grand Dyck path of semilength n is a path in the half-plane x>=0, starting at (0,0), ending at (2n,0) and consisting of steps u=(1,1) and d=(1,-1)). Example: T(3,2)=5 because we have u(d)uud(d),uud(d)u(d),u(d)u(d)du,u(d)duu(d) and duu(d)u(d) (the downward returns to the x-axis are shown between parentheses). - Emeric Deutsch, May 06 2006
Riordan array (c(x),x*c(x)^2) where c(x) is the g.f. of A000108; inverse array is (1/(1+x),x/(1+x)^2). - Philippe Deléham, Feb 12 2007
The triangle may also be generated from M^n*[1,0,0,0,0,0,0,0,...], where M is the infinite tridiagonal matrix with all 1's in the super and subdiagonals and [1,2,2,2,2,2,2,...] in the main diagonal. - Philippe Deléham, Feb 26 2007
Inverse binomial matrix applied to A124733. Binomial matrix applied to A089942. - Philippe Deléham, Feb 26 2007
Number of standard tableaux of shape (n+k,n-k). - Philippe Deléham, Mar 22 2007
From Philippe Deléham, Mar 30 2007: (Start)
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y):
(0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970
(1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877;
(1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598;
(2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954;
(3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791;
(4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. (End)
The table U(n,k) = Sum_{j=0..n} T(n,j)*k^j is given in A098474. - Philippe Deléham, Mar 29 2007
Sequence read mod 2 gives A127872. - Philippe Deléham, Apr 12 2007
Number of 2n step walks from (0,0) to (2n,2k) and consisting of step u=(1,1) and d=(1,-1) and the path stays in the nonnegative quadrant. Example: T(3,0)=5 because we have uuuddd, uududd, ududud, uduudd, uuddud; T(3,1)=9 because we have uuuudd, uuuddu, uuudud, ududuu, uuduud, uduudu, uudduu, uduuud, uududu; T(3,2)=5 because we have uuuuud, uuuudu, uuuduu, uuduuu, uduuuu; T(3,3)=1 because we have uuuuuu. - Philippe Deléham, Apr 16 2007, Apr 17 2007, Apr 18 2007
Triangular matrix, read by rows, equal to the matrix inverse of triangle A129818. - Philippe Deléham, Jun 19 2007
Let Sum_{n>=0} a(n)*x^n = (1+x)/(1-mx+x^2) = o.g.f. of A_m, then Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n. Related expansions of A_m are: A099493, A033999, A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, A097783, A077416, A126866, A028230, A161591, for m=-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, respectively. - Philippe Deléham, Nov 16 2009
The Kn11, Kn12, Fi1 and Fi2 triangle sums link the triangle given above with three sequences; see the crossrefs. For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 20 2011
4^n = (n-th row terms) dot (first n+1 odd integer terms). Example: 4^4 = 256 = (14, 28, 20, 7, 1) dot (1, 3, 5, 7, 9) = (14 + 84 + 100 + 49 + 9) = 256. - Gary W. Adamson, Jun 13 2011
The linear system of n equations with coefficients defined by the first n rows solve for diagonal lengths of regular polygons with N= 2n+1 edges; the constants c^0, c^1, c^2, ... are on the right hand side, where c = 2 + 2*cos(2*Pi/N). Example: take the first 4 rows relating to the 9-gon (nonagon), N = 2*4 + 1; with c = 2 + 2*cos(2*Pi/9) = 3.5320888.... The equations are (1,0,0,0) = 1; (1,1,0,0) = c; (2,3,1,0) = c^2; (5,9,5,1) = c^3. The solutions are 1, 2.53208..., 2.87938..., and 1.87938...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. (Cf. comment in A089942 which uses the analogous operations but with c = 1 + 2*cos(2*Pi/9).) - Gary W. Adamson, Sep 21 2011
Also called the Lobb numbers, after Andrew Lobb, are a natural generalization of the Catalan numbers, given by L(m,n)=(2m+1)*Binomial(2n,m+n)/(m+n+1), where n >= m >= 0. For m=0, we get the n-th Catalan number. See added reference. - Jayanta Basu, Apr 30 2013
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n, 2*k). T(n, k) appears in the formula for the (2*n)-th power of the algebraic number rho(N):= 2*cos(Pi/N) = R(N, 2) in terms of the odd-indexed diagonal/side length ratios R(N, 2*k+1) = S(2*k, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310):
  rho(N)^(2*n) = Sum_{k=0..n} T(n, k)*R(N, 2*k+1), n >= 0, identical in N > = 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears.
  For the odd powers of rho(n) see A039598. (End)
Unsigned coefficients of polynomial numerators of Eqn. 2.1 of the Chakravarty and Kodama paper, defining the polynomials of A067311. - Tom Copeland, May 26 2016
The triangle is the Riordan square of the Catalan numbers in the sense of A321620. - Peter Luschny, Feb 14 2023

			Triangle T(n, k) begins:
  n\k     0     1     2     3     4     5    6   7   8  9
  0:      1
  1:      1     1
  2:      2     3     1
  3:      5     9     5     1
  4:     14    28    20     7     1
  5:     42    90    75    35     9     1
  6:    132   297   275   154    54    11    1
  7:    429  1001  1001   637   273    77   13   1
  8:   1430  3432  3640  2548  1260   440  104  15   1
  9:   4862 11934 13260  9996  5508  2244  663 135  17  1
  ... Reformatted by _Wolfdieter Lang_, Dec 21 2015
From _Paul Barry_, Feb 17 2011: (Start)
Production matrix begins
  1, 1,
  1, 2, 1,
  0, 1, 2, 1,
  0, 0, 1, 2, 1,
  0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 0, 1, 2, 1 (End)
From _Wolfdieter Lang_, Sep 20 2013: (Start)
Example for rho(N) = 2*cos(Pi/N) powers:
n=2: rho(N)^4 = 2*R(N,1) + 3*R(N,3) + 1*R(N, 5) =
  2 + 3*S(2, rho(N)) + 1*S(4, rho(N)), identical in N >= 1. For N=4 (the square with only one distinct diagonal), the degree delta(4) = 2, hence R(4, 3) and R(4, 5) can be reduced, namely to R(4, 1) = 1 and R(4, 5) = -R(4,1) = -1, respectively. Therefore, rho(4)^4 =(2*cos(Pi/4))^4 = 2 + 3 -1 = 4. (End)

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.
T. Myers and L. Shapiro, Some applications of the sequence 1, 5, 22, 93, 386, ... to Dyck paths and ordered trees, Congressus Numerant., 204 (2010), 93-104.

T. D. Noe, Rows n=0..50 of triangle, flattened
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Quang T. Bach and Jeffrey B. Remmel, Generating functions for descents over permutations which avoid sets of consecutive patterns, arXiv:1510.04319 [math.CO], 2015 (see p.25).
M. Barnabei, F. Bonetti and M. Silimbani, Two permutation classes enumerated by the central binomial coefficients, arXiv preprint arXiv:1301.1790 [math.CO], 2013 and J. Int. Seq. 16 (2013) #13.3.8
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Paul Barry and A. Hennessy, The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences, J. Int. Seq. 13 (2010), Article 10.8.2, example 15.
Paul Barry, On the Hurwitz Transform of Sequences, Journal of Integer Sequences, Vol. 15 (2012), #12.8.7.
Paul Barry, Comparing two matrices of generalized moments defined by continued fraction expansions, arXiv preprint arXiv:1311.7161 [math.CO], 2013 and J. Int. Seq. 17 (2014) # 14.5.1.
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Paul Barry, Notes on the Hankel transform of linear combinations of consecutive pairs of Catalan numbers, arXiv:2011.10827 [math.CO], 2020.
Paul Barry, Notes on Riordan arrays and lattice paths, arXiv:2504.09719 [math.CO], 2025. See pp. 15, 29.
Paul Barry, d-orthogonal polynomials, Fuss-Catalan matrices and lattice paths, arXiv:2505.16718 [math.CO], 2025. See p. 12.
Jonathan E. Beagley and Paul Drube, Combinatorics of Tableau Inversions, Electron. J. Combin., 22 (2015), #P2.44.
S. Chakravarty and Y. Kodama, A generating function for the N-soliton solutions of the Kadomtsev-Petviashvili II equation, arXiv preprint arXiv:0802.0524v2 [nlin.SI], 2008.
Wun-Seng Chou, Tian-Xiao He, and Peter J.-S. Shiue, On the Primality of the Generalized Fuss-Catalan Numbers, J. Int. Seqs., Vol. 21 (2018), #18.2.1.
Johann Cigler, Some elementary observations on Narayana polynomials and related topics, arXiv:1611.05252 [math.CO], 2016. See p. 11.
Paul Drube, Generating Functions for Inverted Semistandard Young Tableaux and Generalized Ballot Numbers, arXiv:1606.04869 [math.CO], 2016.
Paul Drube, Generalized Path Pairs and Fuss-Catalan Triangles, arXiv:2007.01892 [math.CO], 2020. See Figure 4 p. 8.
T.-X. He and L. W. Shapiro, Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group, Lin. Alg. Applic. 532 (2017) 25-41, example p 32.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
Thomas Koshy, Lobb's generalization of Catalan's parenthesization problem, The College Mathematics Journal 40 (2), March 2009, 99-107, DOI:10.1080/07468342.2009.11922344.
Huyile Liang, Jeffrey Remmel, and Sainan Zheng, Stieltjes moment sequences of polynomials, arXiv:1710.05795 [math.CO], 2017, see page 11.
Andrew Lobb, Deriving the n-th Catalan number, Mathematical Gazette, Vol. 83, No. 496 (March 1999), 109-110.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174. See page 161.
Pedro J. Miana, Hideyuki Ohtsuka, and Natalia Romero, Sums of powers of Catalan triangle numbers, arXiv:1602.04347 [math.NT], 2016 (see 2.8).
A. Papoulis, A new method of inversion of the Laplace transform, Quart. Appl. Math 14 (1957), 405-414. [Annotated scan of selected pages]
Athanasios Papoulis, A new method of inversion of the Laplace transform, Quart. Appl. Math., Vol. 14, No. 4 (1957), 405-414: 124. [Note: there is a typo]
J. Riordan, The distribution of crossings of chords joining pairs of 2n points on a circle, Math. Comp. 29 (129) (1975) 215-222
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv preprint arXiv:1305.2017 [math.CO], 2013.
Yidong Sun and Luping Ma, Minors of a class of Riordan arrays related to weighted partial Motzkin paths. Eur. J. Comb. 39, 157-169 (2014), Table 2.2.
Wikipedia, Lobb number
W.-J. Woan, L. Shapiro and D. G. Rogers, The Catalan numbers, the Lebesgue integral and 4^{n-2}, Amer. Math. Monthly, 104 (1997), 926-931.
Sheng-Liang Yang, Yan-Ni Dong, and Tian-Xiao He, Some matrix identities on colored Motzkin paths, Discrete Mathematics 340.12 (2017), 3081-3091.

Columns give: A000108, A000245, A000344, A000588, A001392, A000589, A000590, A000012, A005408, A014107(n>1).
Row sums: A000984.
Triangle sums (see the comments): A000958 (Kn11), A001558 (Kn12), A088218 (Fi1, Fi2).
Cf. A008313, A039598, A084938, A000007, A067311, A321620.

/* As triangle */ [[Binomial(2*n, k+n)*(2*k+1)/(k+n+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 16 2015

T:=(n,k)->(2*k+1)*binomial(2*n,n-k)/(n+k+1): for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form # Emeric Deutsch, May 06 2006
T := proc(n, k) option remember; if k = n then 1 elif k > n then 0 elif k = 0 then T(n-1, 0) + T(n-1,1) else T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1) fi end:
seq(seq(T(n, k), k = 0..n), n = 0..9) od; # Peter Luschny, Feb 14 2023

Table[Abs[Differences[Table[Binomial[2 n, n + i], {i, 0, n + 1}]]], {n, 0,7}] // Flatten (* Geoffrey Critzer, Dec 18 2011 *)
Join[{1},Flatten[Table[Binomial[2n-1,n-k]-Binomial[2n-1,n-k-2],{n,10},{k,0,n}]]] (* Harvey P. Dale, Dec 18 2011 *)
Flatten[Table[Binomial[2*n,m+n]*(2*m+1)/(m+n+1),{n,0,9},{m,0,n}]] (* Jayanta Basu, Apr 30 2013 *)

a(n, k) = (2*n+1)/(n+k+1)*binomial(2*k, n+k)
trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(a(y, x), ", ")); print(""))
trianglerows(10) \\ Felix Fröhlich, Jun 24 2016

# Algorithm of L. Seidel (1877)
# Prints the first n rows of the triangle
def A039599_triangle(n) :
    D = [0]*(n+2); D[1] = 1
    b = True ; h = 1
    for i in range(2*n-1) :
        if b :
            for k in range(h,0,-1) : D[k] += D[k-1]
            h += 1
        else :
            for k in range(1,h, 1) : D[k] += D[k+1]
        if b : print([D[z] for z in (1..h-1)])
        b = not b
A039599_triangle(10)  # Peter Luschny, May 01 2012

T(n,k) = C(2*n-1, n-k) - C(2*n-1, n-k-2), n >= 1, T(0,0) = 1.
From Emeric Deutsch, May 06 2006: (Start)
T(n,k) = (2*k+1)*binomial(2*n,n-k)/(n+k+1).
G.f.: G(t,z)=1/(1-(1+t)*z*C), where C=(1-sqrt(1-4*z))/(2*z) is the Catalan function. (End)
The following formulas were added by Philippe Deléham during 2003 to 2009: (Start)
Triangle T(n, k) read by rows; given by A000012 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
T(n, k) = C(2*n, n-k)*(2*k+1)/(n+k+1). Sum(k>=0; T(n, k)*T(m, k) = A000108(n+m)); A000108: numbers of Catalan.
T(n, 0) = A000108(n); T(n, k) = 0 if k>n; for k>0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
T(n, k) = A009766(n+k, n-k) = A033184(n+k+1, 2k+1).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+1) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
T(0, 0) = 1, T(n, k) = 0 if n<0 or n=1, T(n, k) = T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1).
a(n) + a(n+1) = 1 + A000108(m+1) if n = m*(m+3)/2; a(n) + a(n+1) = A039598(n) otherwise.
T(n, k) = A050165(n, n-k).
Sum_{j>=0} T(n-k, j)*A039598(k, j) = A028364(n, k).
Matrix inverse of the triangle T(n, k) = (-1)^(n+k)*binomial(n+k, 2*k) = (-1)^(n+k)*A085478(n, k).
Sum_{k=0..n} T(n, k)*x^k = A000108(n), A000984(n), A007854(n), A076035(n), A076036(n) for x = 0, 1, 2, 3, 4.
Sum_{k=0..n} (2*k+1)*T(n, k) = 4^n.
T(n, k)*(-2)^(n-k) = A114193(n, k).
Sum_{k>=h} T(n,k) = binomial(2n,n-h).
Sum_{k=0..n} T(n,k)*5^k = A127628(n).
Sum_{k=0..n} T(n,k)*7^k = A115970(n).
T(n,k) = Sum_{j=0..n-k} A106566(n+k,2*k+j).
Sum_{k=0..n} T(n,k)*6^k = A126694(n).
Sum_{k=0..n} T(n,k)*A000108(k) = A007852(n+1).
Sum_{k=0..floor(n/2)} T(n-k,k) = A000958(n+1).
Sum_{k=0..n} T(n,k)*(-1)^k = A000007(n).
Sum_{k=0..n} T(n,k)*(-2)^k = (-1)^n*A064310(n).
T(2*n,n) = A126596(n).
Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x=1,2,3,4,5,6,7,8,9,10 respectively.
Sum_{j>=0} T(n,j)*binomial(j,k) = A116395(n,k).
T(n,k) = Sum_{j>=0} A106566(n,j)*binomial(j,k).
T(n,k) = Sum_{j>=0} A127543(n,j)*A038207(j,k).
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A101490(n+1).
T(n,k) = A053121(2*n,2*k).
Sum_{k=0..n} T(n,k)*sin((2*k+1)*x) = sin(x)*(2*cos(x))^(2*n).
T(n,n-k) = Sum_{j>=0} (-1)^(n-j)*A094385(n,j)*binomial(j,k).
Sum_{j>=0} A110506(n,j)*binomial(j,k) = Sum_{j>=0} A110510(n,j)*A038207(j,k) = T(n,k)*2^(n-k).
Sum_{j>=0} A110518(n,j)*A027465(j,k) = Sum_{j>=0} A110519(n,j)*A038207(j,k) = T(n,k)*3^(n-k).
Sum_{k=0..n} T(n,k)*A001045(k) = A049027(n), for n>=1.
Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n if Sum_{k>=0} a(k)*x^k = (1+x)/(x^2-m*x+1).
Sum_{k=0..n} T(n,k)*A040000(k) = A001700(n).
Sum_{k=0..n} T(n,k)*A122553(k) = A051924(n+1).
Sum_{k=0..n} T(n,k)*A123932(k) = A051944(n).
Sum_{k=0..n} T(n,k)*k^2 = A000531(n), for n>=1.
Sum_{k=0..n} T(n,k)*A000217(k) = A002457(n-1), for n>=1.
Sum{j>=0} binomial(n,j)*T(j,k)= A124733(n,k).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
Sum_{k=0..n} T(n,k)*A005043(k) = A127632(n).
Sum_{k=0..n} T(n,k)*A132262(k) = A089022(n).
T(n,k) + T(n,k+1) = A039598(n,k).
T(n,k) = A128899(n,k)+A128899(n,k+1).
Sum_{k=0..n} T(n,k)*A015518(k) = A076025(n), for n>=1. Also Sum_{k=0..n} T(n,k)*A015521(k) = A076026(n), for n>=1.
Sum_{k=0..n} T(n,k)*(-1)^k*x^(n-k) = A033999(n), A000007(n), A064062(n), A110520(n), A132863(n), A132864(n), A132865(n), A132866(n), A132867(n), A132869(n), A132897(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively.
Sum_{k=0..n} T(n,k)*(-1)^(k+1)*A000045(k) = A109262(n), A000045:= Fibonacci numbers.
Sum_{k=0..n} T(n,k)*A000035(k)*A016116(k) = A143464(n).
Sum_{k=0..n} T(n,k)*A016116(k) = A101850(n).
Sum_{k=0..n} T(n,k)*A010684(k) = A100320(n).
Sum_{k=0..n} T(n,k)*A000034(k) = A029651(n).
Sum_{k=0..n} T(n,k)*A010686(k) = A144706(n).
Sum_{k=0..n} T(n,k)*A006130(k-1) = A143646(n), with A006130(-1)=0.
T(n,2*k)+T(n,2*k+1) = A118919(n,k).
Sum_{k=0..j} T(n,k) = A050157(n,j).
Sum_{k=0..2} T(n,k) = A026012(n); Sum_{k=0..3} T(n,k)=A026029(n).
Sum_{k=0..n} T(n,k)*A000045(k+2) = A026671(n).
Sum_{k=0..n} T(n,k)*A000045(k+1) = A026726(n).
Sum_{k=0..n} T(n,k)*A057078(k) = A000012(n).
Sum_{k=0..n} T(n,k)*A108411(k) = A155084(n).
Sum_{k=0..n} T(n,k)*A057077(k) = 2^n = A000079(n).
Sum_{k=0..n} T(n,k)*A057079(k) = 3^n = A000244(n).
Sum_{k=0..n} T(n,k)*(-1)^k*A011782(k) = A000957(n+1).
(End)
T(n,k) = Sum_{j=0..k} binomial(k+j,2j)*(-1)^(k-j)*A000108(n+j). - Paul Barry, Feb 17 2011
Sum_{k=0..n} T(n,k)*A071679(k+1) = A026674(n+1). - Philippe Deléham, Feb 01 2014
Sum_{k=0..n} T(n,k)*(2*k+1)^2 = (4*n+1)*binomial(2*n,n). - Werner Schulte, Jul 22 2015
Sum_{k=0..n} T(n,k)*(2*k+1)^3 = (6*n+1)*4^n. - Werner Schulte, Jul 22 2015
Sum_{k=0..n} (-1)^k*T(n,k)*(2*k+1)^(2*m) = 0 for 0 <= m < n (see also A160562). - Werner Schulte, Dec 03 2015
T(n,k) = GegenbauerC(n-k,-n+1,-1) - GegenbauerC(n-k-1,-n+1,-1). - Peter Luschny, May 13 2016
T(n,n-2) = A014107(n). - R. J. Mathar, Jan 30 2019
T(n,n-3) = n*(2*n-1)*(2*n-5)/3. - R. J. Mathar, Jan 30 2019
T(n,n-4) = n*(n-1)*(2*n-1)*(2*n-7)/6. - R. J. Mathar, Jan 30 2019
T(n,n-5) = n*(n-1)*(2*n-1)*(2*n-3)*(2*n-9)/30. - R. J. Mathar, Jan 30 2019

Corrected by Philippe Deléham, Nov 26 2009, Dec 14 2009

A228351 Triangle read by rows in which row n lists the compositions (ordered partitions) of n (see Comments lines for definition).

Original entry on oeis.org

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

Offset: 1

Omar E. Pol, Aug 30 2013

The representation of the compositions (for fixed n) is as lists of parts, the order between individual compositions (for the same n) is (list-)reversed co-lexicographic. - Joerg Arndt, Sep 02 2013
Dropping the "(list-)reversed" in the comment above gives A228525.
The equivalent sequence for partitions is A026792.
This sequence lists (without repetitions) all finite compositions, in such a way that, if [P_1, ..., P_r] denotes the composition occupying the n-th position in the list, then (((2*n/2^(P_1)-1)/2^(P_2)-1)/...)/2^(P_r)-1 = 0. - Lorenzo Sauras Altuzarra, Jan 22 2020
The k-th composition in the list is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, and taking first differences. Reversing again gives A066099, which is described as the standard ordering. Both sequences define a bijective correspondence between nonnegative integers and integer compositions. - Gus Wiseman, Apr 01 2020
It follows from the previous comment that A000120(k) is the length of the k-th composition that is listed by this sequence (recall that A000120(k) is the number of 1's in the binary expansion of k). - Lorenzo Sauras Altuzarra, Sep 29 2020

			Illustration of initial terms:
-----------------------------------
n  j     Diagram     Composition j
-----------------------------------
.         _
1  1     |_|         1;
.         _ _
2  1     |_  |       2,
2  2     |_|_|       1, 1;
.         _ _ _
3  1     |_    |     3,
3  2     |_|_  |     1, 2,
3  3     |_  | |     2, 1,
3  4     |_|_|_|     1, 1, 1;
.         _ _ _ _
4  1     |_      |   4,
4  2     |_|_    |   1, 3,
4  3     |_  |   |   2, 2,
4  4     |_|_|_  |   1, 1, 2,
4  5     |_    | |   3, 1,
4  6     |_|_  | |   1, 2, 1,
4  7     |_  | | |   2, 1, 1,
4  8     |_|_|_|_|   1, 1, 1, 1;
.
Triangle begins:
[1];
[2],[1,1];
[3],[1,2],[2,1],[1,1,1];
[4],[1,3],[2,2],[1,1,2],[3,1],[1,2,1],[2,1,1],[1,1,1,1];
[5],[1,4],[2,3],[1,1,3],[3,2],[1,2,2],[2,1,2],[1,1,1,2],[4,1],[1,3,1],[2,2,1],[1,1,2,1],[3,1,1],[1,2,1,1],[2,1,1,1],[1,1,1,1,1];
...
For example [1,2] occupies the 5th position in the corresponding list of compositions and indeed (2*5/2^1-1)/2^2-1 = 0. - _Lorenzo Sauras Altuzarra_, Jan 22 2020
12 --binary expansion--> [1,1,0,0] --reverse--> [0,0,1,1] --positions of 1's--> [3,4] --prepend 0--> [0,3,4] --first differences--> [3,1]. - _Lorenzo Sauras Altuzarra_, Sep 29 2020

Peter Kagey, Table of n, a(n) for n = 1..10000
Mikhail Kurkov, Comments on A228351
Index entries for sequences that are related to compositions

Row n has length A001792(n-1). Row sums give A001787, n >= 1.
Cf. A000120 (binary weight), A001511, A006519, A011782, A026792, A065120.
A related ranking of finite sets is A048793/A272020.
All of the following consider the k-th row to be the k-th composition, ignoring the coarser grouping by sum.
- Indices of weakly increasing rows are A114994.
- Indices of weakly decreasing rows are A225620.
- Indices of strictly decreasing rows are A333255.
- Indices of strictly increasing rows are A333256.
- Indices of reversed interval rows A164894.
- Indices of interval rows are A246534.
- Indices of strict rows are A233564.
- Indices of constant rows are A272919.
- Indices of anti-run rows are A333489.
- Row k has A124767(k) runs and A333381(k) anti-runs.
- Row k has GCD A326674(k) and LCM A333226(k).
- Row k has Heinz number A333219(k).
Cf. A000120, A029931, A035327, A070939, A233249, A333217, A333218, A333220, A333227, A333627, A333628.
Equals A163510+1, termwise.
Cf. A124734 (increasing length, then lexicographic).
Cf. A296774 (increasing length, then reverse lexicographic).
Cf. A337243 (increasing length, then colexicographic).
Cf. A337259 (increasing length, then reverse colexicographic).
Cf. A296773 (decreasing length, then lexicographic).
Cf. A296772 (decreasing length, then reverse lexicographic).
Cf. A337260 (decreasing length, then colexicographic).
Cf. A108244 (decreasing length, then reverse colexicographic).
Cf. A228369 (lexicographic).
Cf. A066099 (reverse lexicographic).
Cf. A228525 (colexicographic).

a228351 n = a228351_list !! (n - 1)
a228351_list = concatMap a228351_row [1..]
a228351_row 0 = []
a228351_row n = a001511 n : a228351_row (n `div` 2^(a001511 n))
-- Peter Kagey, Jun 27 2016

# Program computing the sequence:
A228351 := proc(n) local c, k, L, N: L, N := [], [seq(2*r, r = 1 .. n)]: for k in N do c := 0: while k != 0 do if gcd(k, 2) = 2 then k := k/2: c := c+1: else L := [op(L), op(c)]: k := k-1: c := 0: fi: od: od: L[n]: end: # Lorenzo Sauras Altuzarra, Jan 22 2020
# Program computing the list of compositions:
List := proc(n) local c, k, L, M, N: L, M, N := [], [], [seq(2*r, r = 1 .. 2^n-1)]: for k in N do c := 0: while k != 0 do if gcd(k, 2) = 2 then k := k/2: c := c+1: else L := [op(L), c]: k := k-1: c := 0: fi: od: M := [op(M), L]: L := []: od: M: end: # Lorenzo Sauras Altuzarra, Jan 22 2020

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Table[Differences[Prepend[bpe[n],0]],{n,0,30}] (* Gus Wiseman, Apr 01 2020 *)

from itertools import count, islice
def A228351_gen(): # generator of terms
    for n in count(1):
        k = n
        while k:
            yield (s:=(~k&k-1).bit_length()+1)
            k >>= s
A228351_list = list(islice(A228351_gen(),30)) # Chai Wah Wu, Jul 17 2023

A131577 Zero followed by powers of 2 (cf. A000079).

Original entry on oeis.org

0, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824, 2147483648, 4294967296, 8589934592

Offset: 0

Paul Curtz, Aug 29 2007, Dec 06 2007

A000079 is the main entry for this sequence.
Binomial transform of A000035.
Essentially the same as A034008 and A000079.
a(n) = a(n-1)-th even natural numbers (A005846) for n > 1. - Jaroslav Krizek, Apr 25 2009
Where record values greater than 1 occur in A083662: A000045(n)=A083662(a(n)). - Reinhard Zumkeller, Sep 26 2009
Number of compositions of natural number n into parts >0.
The signed sequence 0, 1, -2, 4, -8, 16, -32, 64, -128, 256, -512, 1024, ... is the Lucas U(-2,0) sequence. - R. J. Mathar, Jan 08 2013
In computer programming, these are the only unsigned numbers such that k&(k-1)=0, where & is the bitwise AND operator and numbers are expressed in binary. - Stanislav Sykora, Nov 29 2013
Also the 0-additive sequence: a(n) is the smallest number larger than a(n-1) which is not the sum of any subset of earlier terms, with initial values {0, 1, 2}. - Robert G. Wilson v, Jul 12 2014
Also the smallest nonnegative superincreasing sequence: each term is larger than the sum of all preceding terms. Indeed, an equivalent definition is a(0)=0, a(n+1)=1+sum_{k=0..n} a(k). - M. F. Hasler, Jan 13 2015

Mohammad K. Azarian, A Generalization of the Climbing Stairs Problem, Mathematics and Computer Education Journal, Vol. 31, No. 1, pp. 24-28, Winter 1997.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Adi Dani, Compositions of natural numbers over arithmetic progressions
Jimmy Devillet, Bisymmetric and quasitrivial operations: characterizations and enumerations, arXiv:1712.07856 [math.RA], 2017.
J. T. Rowell, Solution Sequences for the Keyboard Problem and its Generalizations, Journal of Integer Sequences, 18 (2015), #15.10.7.
Wikipedia, Lucas sequence
Index entries for linear recurrences with constant coefficients, signature (2).
Index entries for Lucas sequences

Cf. A000079, A003945, A042950, A020406, A046045, A011782.

int is (unsigned long n) { return !(n & (n-1)); } /* Charles R Greathouse IV, Sep 15 2012 */

a131577 = (`div` 2) . a000079
a131577_list = 0 : a000079_list  -- Reinhard Zumkeller, Dec 09 2012

[(2^n-0^n)/2: n in [0..50]]; // Vincenzo Librandi, Aug 10 2011

A131577 := proc(n)
    if n =0 then
        0;
    else
        2^(n-1) ;
    end if;
end proc: # R. J. Mathar, Jul 22 2012

Floor[2^Range[-1, 33]] (* Robert G. Wilson v, Sep 02 2007 *)
Join[{0}, 2^Range[0, 60]] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)

a(n)=1<Charles R Greathouse IV, Sep 15 2012

def A131577(n): return 1<Chai Wah Wu, Sep 09 2023

a(n) = floor(2^(n-1)). - Robert G. Wilson v, Sep 02 2007
G.f.: x/(1-2*x); a(n) = (2^n-0^n)/2. - Paul Barry, Jan 05 2009
E.g.f.: exp(x)*sinh(x). - Geoffrey Critzer, Oct 28 2012
E.g.f.: x/T(0) where T(k) = 4*k+1 - x/(1 + x/(4*k+3 - x/(1 + x/T(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Mar 17 2013
a(n) = Sum_{k=0..floor((n+1)/2)} binomial(n, 2*k-1). - Taras Goy, Jan 02 2025

More terms from Robert G. Wilson v, Sep 02 2007
Edited by N. J. A. Sloane, Sep 13 2007
Edited by M. F. Hasler, Jan 13 2015

A238343 Triangle T(n,k) read by rows: T(n,k) is the number of compositions of n with k descents, n>=0, 0<=k<=n.

Original entry on oeis.org

1, 1, 0, 2, 0, 0, 3, 1, 0, 0, 5, 3, 0, 0, 0, 7, 9, 0, 0, 0, 0, 11, 19, 2, 0, 0, 0, 0, 15, 41, 8, 0, 0, 0, 0, 0, 22, 77, 29, 0, 0, 0, 0, 0, 0, 30, 142, 81, 3, 0, 0, 0, 0, 0, 0, 42, 247, 205, 18, 0, 0, 0, 0, 0, 0, 0, 56, 421, 469, 78, 0, 0, 0, 0, 0, 0, 0, 0, 77, 689, 1013, 264, 5, 0, 0, 0, 0, 0, 0, 0, 0, 101, 1113, 2059, 786, 37, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Joerg Arndt and Alois P. Heinz, Feb 25 2014

Counting ascents gives the same triangle.

For n > 0, also the number of compositions of n with k + 1 maximal weakly increasing runs. - Gus Wiseman, Mar 23 2020

			Triangle starts:
00:    1;
01:    1,    0;
02:    2,    0,    0;
03:    3,    1,    0,    0;
04:    5,    3,    0,    0,   0;
05:    7,    9,    0,    0,   0, 0;
06:   11,   19,    2,    0,   0, 0, 0;
07:   15,   41,    8,    0,   0, 0, 0, 0;
08:   22,   77,   29,    0,   0, 0, 0, 0, 0;
09:   30,  142,   81,    3,   0, 0, 0, 0, 0, 0;
10:   42,  247,  205,   18,   0, 0, 0, 0, 0, 0, 0;
11:   56,  421,  469,   78,   0, 0, 0, 0, 0, 0, 0, 0;
12:   77,  689, 1013,  264,   5, 0, 0, 0, 0, 0, 0, 0, 0;
13:  101, 1113, 2059,  786,  37, 0, 0, 0, 0, 0, 0, 0, 0, 0;
14:  135, 1750, 4021, 2097, 189, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
15:  176, 2712, 7558, 5179, 751, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
...
From _Gus Wiseman_, Mar 23 2020: (Start)
Row n = 5 counts the following compositions:
  (5)          (3,2)
  (1,4)        (4,1)
  (2,3)        (1,3,1)
  (1,1,3)      (2,1,2)
  (1,2,2)      (2,2,1)
  (1,1,1,2)    (3,1,1)
  (1,1,1,1,1)  (1,1,2,1)
               (1,2,1,1)
               (2,1,1,1)
(End)

Joerg Arndt and Alois P. Heinz, Rows n = 0..140, flattened

Columns k=0-10 give: A000041, A241626, A241627, A241628, A241629, A241630, A241631, A241632, A241633, A241634, A241635.
T(3n,n) gives A000045(n+1).
T(3n+1,n) = A136376(n+1).
Row sums are A011782.
Compositions by length are A007318.
The version for co-runs or levels is A106356.
The case of partitions (instead of compositions) is A133121.
The version for runs is A238279.
The version without zeros is A238344.
The version for weak ascents is A333213.
Cf. A008284, A045883, A124765, A124766, A332875, A333215.

b:= proc(n, i) option remember; `if`(n=0, 1, expand(
       add(b(n-j, j)*`if`(j (p-> seq(coeff(p, x, i), i=0..n))(b(n, 0)):
seq(T(n), n=0..20);

b[n_, i_] := b[n, i] = If[n == 0, 1, Sum[b[n-j, j]*If[jJean-François Alcover, Jan 08 2015, translated from Maple *)
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],n==0||Length[Split[#,LessEqual]]==k+1&]],{n,0,9},{k,0,n}] (* Gus Wiseman, Mar 23 2020 *)

Sum_{k=0..n} k * T(n,k) = A045883(n-2) for n>=2.

A027383 a(2n) = 32^n - 2; a(2*n+1) = 2^(n+2) - 2.

Original entry on oeis.org

1, 2, 4, 6, 10, 14, 22, 30, 46, 62, 94, 126, 190, 254, 382, 510, 766, 1022, 1534, 2046, 3070, 4094, 6142, 8190, 12286, 16382, 24574, 32766, 49150, 65534, 98302, 131070, 196606, 262142, 393214, 524286, 786430, 1048574, 1572862, 2097150, 3145726, 4194302, 6291454

Offset: 0

R. K. Guy

Number of balanced strings of length n: let d(S) = #(1's) - #(0's), # == count in S, then S is balanced if every substring T of S has -2 <= d(T) <= 2.
Number of "fold lines" seen when a rectangular piece of paper is folded n+1 times along alternate orthogonal directions and then unfolded. - Quim Castellsaguer (qcastell(AT)pie.xtec.es), Dec 30 1999
Also the number of binary strings with the property that, when scanning from left to right, once the first 1 is seen in position j, there must be a 1 in positions j+2, j+4, ... until the end of the string. (Positions j+1, j+3, ... can be occupied by 0 or 1.) - Jeffrey Shallit, Sep 02 2002
a(n-1) is also the Moore lower bound on the order of a (3,n)-cage. - Eric W. Weisstein, May 20 2003 and Jason Kimberley, Oct 30 2011
Partial sums of A016116. - Hieronymus Fischer, Sep 15 2007
Equals row sums of triangle A152201. - Gary W. Adamson, Nov 29 2008
From John P. McSorley, Sep 28 2010: (Start)
a(n) = DPE(n+1) is the total number of k-double-palindromes of n up to cyclic equivalence. See sequence A180918 for the definitions of a k-double-palindrome of n and of cyclic equivalence. Sequence A180918 is the 'DPE(n,k)' triangle read by rows where DPE(n,k) is the number of k-double-palindromes of n up to cyclic equivalence. For example, we have a(4) = DPE(5) = DPE(5,1) + DPE(5,2) + DPE(5,3) + DPE(5,4) + DPE(5,5) = 0 + 2 + 2 + 1 + 1 = 6.
The 6 double-palindromes of 5 up to cyclic equivalence are 14, 23, 113, 122, 1112, 11111. They come from cyclic equivalence classes {14,41}, {23,32}, {113,311,131}, {122,212,221}, {1112,2111,1211,1121}, and {11111}. Hence a(n)=DPE(n+1) is the total number of cyclic equivalence classes of n containing at least one double-palindrome.
(End)
From Herbert Eberle, Oct 02 2015: (Start)
For n > 0, there is a red-black tree of height n with a(n-1) internal nodes and none with less.
In order a red-black tree of given height has minimal number of nodes, it has exactly 1 path with strictly alternating red and black nodes. All nodes outside this height defining path are black.
Consider:
mrbt5                R
                    / \
                   /   \
                  /     \
                 /       B
                /       / \
mrbt4          B       /   B
              / \     B   E E
             /   B   E E
mrbt3       R   E E
           / \
          /   B
mrbt2    B   E E
        / E
mrbt1  R
      E E
(Red nodes shown as R, blacks as B, externals as E.)
Red-black trees mrbt1, mrbt2, mrbt3, mrbt4, mrbt5 of respective heights h = 1, 2, 3, 4, 5; all minimal in the number of internal nodes, namely 1, 2, 4, 6, 10.
Recursion (let n = h-1): a(-1) = 0, a(n) = a(n-1) + 2^floor(n/2), n >= 0.
(End)
Also the number of strings of length n with the digits 1 and 2 with the property that the sum of the digits of all substrings of uneven length is not divisible by 3. An example with length 8 is 21221121. - Herbert Kociemba, Apr 29 2017
a(n-2) is the number of achiral n-bead necklaces or bracelets using exactly two colors.  For n=4, the four arrangements are AAAB, AABB, ABAB, and ABBB. - Robert A. Russell, Sep 26 2018
Partial sums of powers of 2 repeated 2 times, like A200672 where is 3 times. - Yuchun Ji, Nov 16 2018
Also the number of binary words of length n with cuts-resistance <= 2, where, for the operation of shortening all runs by one, cuts-resistance is the number of applications required to reach an empty word. Explicitly, these are words whose sequence of run-lengths, all of which are 1 or 2, has no odd-length run of 1's sandwiched between two 2's. - Gus Wiseman, Nov 28 2019
Also the number of up-down paths with n steps such that the height difference between the highest and lowest points is at most 2. - Jeremy Dover, Jun 17 2020
Also the number of non-singleton integer compositions of n + 2 with no odd part other than the first or last. Including singletons gives A052955. This is an unsorted (or ordered) version of A351003. The version without even (instead of odd) interior parts is A001911, complement A232580. Note that A000045(n-1) counts compositions without odd parts, with non-singleton case A077896, and A052952/A074331 count non-singleton compositions without even parts. Also the number of compositions y of n + 1 such that y_i = y_{i+1} for all even i. - Gus Wiseman, Feb 19 2022

			After 3 folds one sees 4 fold lines.
Example: a(3) = 6 because the strings 001, 010, 100, 011, 101, 110 have the property.
Binary: 1, 10, 100, 110, 1010, 1110, 10110, 11110, 101110, 111110, 1011110, 1111110, 10111110, 11111110, 101111110, 111111110, 1011111110, 1111111110, 10111111110, ... - _Jason Kimberley_, Nov 02 2011
Example: Partial sums of powers of 2 repeated 2 times:
a(3) = 1+1+2 = 4;
a(4) = 1+1+2+2 = 6;
a(5) = 1+1+2+2+4 = 10.
_Yuchun Ji_, Nov 16 2018

John P. McSorley: Counting k-compositions of n with palindromic and related structures. Preprint, 2010. [John P. McSorley, Sep 28 2010]

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
J. Jordan and R. Southwell, Further Properties of Reproducing Graphs, Applied Mathematics, Vol. 1 No. 5, 2010, pp. 344-350. - From _N. J. A. Sloane_, Feb 03 2013
Leonard F. Klosinski, Gerald L. Alexanderson and Loren C. Larson, Under misprinted head B3, Amer Math. Monthly, 104(1997) 753-754.
Laurent Noé, Spaced seed design on profile HMMs for precise HTS read-mapping efficient sliding window product on the matrix semi-group, in Rapide Bilan 2012-2013, LIFL, Université Lille 1 - INRIA Journées au vert 11 et 12 juin 2013.
Eric Weisstein's World of Mathematics, Cage Graph
Index entries for sequences obtained by enumerating foldings
Index entries for linear recurrences with constant coefficients, signature (1,2,-2).

Cf. A132666, A152201.
Moore lower bound on the order of a (k,g) cage: A198300 (square); rows: A000027 (k=2), this sequence (k=3), A062318 (k=4), A061547 (k=5), A198306 (k=6), A198307 (k=7), A198308 (k=8), A198309 (k=9), A198310 (k=10), A094626 (k=11); columns: A020725 (g=3), A005843 (g=4), A002522 (g=5), A051890 (g=6), A188377 (g=7). - Jason Kimberley, Oct 30 2011
Cf. A000066 (actual order of a (3,g)-cage).
Bisections are A033484 (even) and A000918 (odd).
a(n) = A305540(n+2,2), the second column of the triangle.
Numbers whose binary expansion is a balanced word are A330029.
Binary words counted by cuts-resistance are A319421 or A329860.
Cf. A000975, A062011, A329862, A329863.
The complementary compositions are counted by A274230(n-1) + 1, with bisections A060867 (even) and A134057 (odd).
Cf. A000346, A000984, A001405, A001700, A011782 (compositions).
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A029744 = {s(n), n>=1}, the numbers 2^k and 3*2^k, as the parent: A029744 (s(n)); A052955 (s(n)-1), A027383 (s(n)-2), A354788 (s(n)-3), A347789 (s(n)-4), A209721 (s(n)+1), A209722 (s(n)+2), A343177 (s(n)+3), A209723 (s(n)+4); A060482, A136252 (minor differences from A354788 at the start); A354785 (3*s(n)), A354789 (3*s(n)-7). The first differences of A029744 are 1,1,1,2,2,4,4,8,8,... which essentially matches eight sequences: A016116, A060546, A117575, A131572, A152166, A158780, A163403, A320770. The bisections of A029744 are A000079 and A007283. - N. J. A. Sloane, Jul 14 2022

import Data.List (transpose)
a027383 n = a027383_list !! n
a027383_list = concat $ transpose [a033484_list, drop 2 a000918_list]
-- Reinhard Zumkeller, Jun 17 2015

[2^Floor((n+2)/2)+2^Floor((n+1)/2)-2: n in [0..50]]; // Vincenzo Librandi, Aug 16 2011

a[0]:=0:a[1]:=1:for n from 2 to 100 do a[n]:=2*a[n-2]+2 od: seq(a[n], n=1..41); # Zerinvary Lajos, Mar 16 2008

a[n_?EvenQ] := 3*2^(n/2)-2; a[n_?OddQ] := 2^(2+(n-1)/2)-2; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Oct 21 2011, after Quim Castellsaguer *)
LinearRecurrence[{1, 2, -2}, {1, 2, 4}, 41] (* Robert G. Wilson v, Oct 06 2014 *)
Table[Length[Select[Tuples[{0,1},n],And[Max@@Length/@Split[#]<=2,!MatchQ[Length/@Split[#],{_,2,ins:1..,2,_}/;OddQ[Plus[ins]]]]&]],{n,0,15}] (* Gus Wiseman, Nov 28 2019 *)

a(n)=2^(n\2+1)+2^((n+1)\2)-2 \\ Charles R Greathouse IV, Oct 21 2011

def a(n): return 2**((n+2)//2) + 2**((n+1)//2) - 2
print([a(n) for n in range(43)]) # Michael S. Branicky, Feb 19 2022

a(0)=1, a(1)=2; thereafter a(n+2) = 2*a(n) + 2.
a(2n) = 3*2^n - 2 = A033484(n);
a(2n-1) = 2^(n+1) - 2 = A000918(n+1).
G.f.: (1 + x)/((1 - x)*(1 - 2*x^2)). - David Callan, Jul 22 2008
a(n) = Sum_{k=0..n} 2^min(k, n-k).
a(n) = 2^floor((n+2)/2) + 2^floor((n+1)/2) - 2. - Quim Castellsaguer (qcastell(AT)pie.xtec.es)
a(n) = 2^(n/2)*(3 + 2*sqrt(2) + (3-2*sqrt(2))*(-1)^n)/2 - 2. - Paul Barry, Apr 23 2004
a(n) = A132340(A052955(n)). - Reinhard Zumkeller, Aug 20 2007
a(n) = A052955(n+1) - 1. - Hieronymus Fischer, Sep 15 2007
a(n) = A132666(a(n+1)) - 1. - Hieronymus Fischer, Sep 15 2007
a(n) = A132666(a(n-1)+1) for n > 0. - Hieronymus Fischer, Sep 15 2007
A132666(a(n)) = a(n-1) + 1 for n > 0. - Hieronymus Fischer, Sep 15 2007
G.f.: (1 + x)/((1 - x)*(1 - 2*x^2)). - David Callan, Jul 22 2008
a(n) = 2*( (a(n-2)+1) mod (a(n-1)+1) ), n > 1. - Pierre Charland, Dec 12 2010
a(n) = A136252(n-1) + 1, for n > 0. - Jason Kimberley, Nov 01 2011
G.f.: (1+x*R(0))/(1-x), where R(k) = 1 + 2*x/( 1 - x/(x + 1/R(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 16 2013
a(n) = 2^((2*n + 3*(1-(-1)^n))/4)*3^((1+(-1)^n)/2) - 2. - Luce ETIENNE, Sep 01 2014
a(n) = a(n-1) + 2^floor((n-1)/2) for n>0, a(0)=1. - Yuchun Ji, Nov 23 2018
E.g.f.: 3*cosh(sqrt(2)*x) - 2*cosh(x) + 2*sqrt(2)*sinh(sqrt(2)*x) - 2*sinh(x). - Stefano Spezia, Apr 06 2022

More terms from Larry Reeves (larryr(AT)acm.org), Mar 24 2000

Replaced definition with a simpler one. - N. J. A. Sloane, Jul 09 2022

A238130 Triangle read by rows: T(n,k) is the number of compositions into nonzero parts with k parts directly followed by a different part, n>=0, 0<=k<=n.

Original entry on oeis.org

1, 1, 0, 2, 0, 0, 2, 2, 0, 0, 3, 4, 1, 0, 0, 2, 10, 4, 0, 0, 0, 4, 12, 14, 2, 0, 0, 0, 2, 22, 29, 10, 1, 0, 0, 0, 4, 26, 56, 36, 6, 0, 0, 0, 0, 3, 34, 100, 86, 31, 2, 0, 0, 0, 0, 4, 44, 148, 200, 99, 16, 1, 0, 0, 0, 0, 2, 54, 230, 374, 278, 78, 8, 0, 0, 0, 0, 0, 6, 58, 322, 680, 654, 274, 52, 2, 0, 0, 0, 0, 0

Offset: 0

Joerg Arndt and Alois P. Heinz, Feb 21 2014

First column (k=0) is A000005, second column (k=1) is 2*A002133.

Row sums are A011782.

			Triangle starts:
00:  1,
01:  1, 0,
02:  2, 0, 0,
03:  2, 2, 0, 0,
04:  3, 4, 1, 0, 0,
05:  2, 10, 4, 0, 0, 0,
06:  4, 12, 14, 2, 0, 0, 0,
07:  2, 22, 29, 10, 1, 0, 0, 0,
08:  4, 26, 56, 36, 6, 0, 0, 0, 0,
09:  3, 34, 100, 86, 31, 2, 0, 0, 0, 0,
10:  4, 44, 148, 200, 99, 16, 1, 0, 0, 0, 0,
11:  2, 54, 230, 374, 278, 78, 8, 0, 0, 0, 0, 0,
12:  6, 58, 322, 680, 654, 274, 52, 2, 0, 0, 0, 0, 0,
13:  2, 74, 446, 1122, 1390, 814, 225, 22, 1, 0, 0, 0, 0, 0,
...
Row 5 is [2, 10, 4, 0, 0, 0] because in the 16 compositions of 5
##:  [composition]  no. of changes
01:  [ 1 1 1 1 1 ]   0
02:  [ 1 1 1 2 ]   1
03:  [ 1 1 2 1 ]   2
04:  [ 1 1 3 ]   1
05:  [ 1 2 1 1 ]   2
06:  [ 1 2 2 ]   1
07:  [ 1 3 1 ]   2
08:  [ 1 4 ]   1
09:  [ 2 1 1 1 ]   1
10:  [ 2 1 2 ]   2
11:  [ 2 2 1 ]   1
12:  [ 2 3 ]   1
13:  [ 3 1 1 ]   1
14:  [ 3 2 ]   1
15:  [ 4 1 ]   1
16:  [ 5 ]   0
there are 2 with no changes, 10 with one change, and 4 with two changes.

Joerg Arndt and Alois P. Heinz, Rows n = 0..140, flattened

Cf. A238279 (same sequence with zeros omitted).
Cf. A106356 (compositions with k successive parts same).
Cf. A225084 (compositions with maximal up-step k).

b:= proc(n, v) option remember; `if`(n=0, 1, expand(
      add(b(n-i, i)*`if`(v=0 or v=i, 1, x), i=1..n)))
    end:
T:= n-> seq(coeff(b(n, 0), x, i), i=0..n):
seq(T(n), n=0..14);

b[n_, v_] := b[n, v] = If[n == 0, 1, Sum[b[n-i, i]*If[v == 0 || v == i, 1, x], {i, 1, n}]]; T[n_] := Table[Coefficient[b[n, 0], x, i], {i, 0, n}]; Table[T[n], {n, 0, 14}] // Flatten (* Jean-François Alcover, Jan 12 2015, translated from Maple *)

A081294 Expansion of (1-2x)/(1-4x).

Original entry on oeis.org

1, 2, 8, 32, 128, 512, 2048, 8192, 32768, 131072, 524288, 2097152, 8388608, 33554432, 134217728, 536870912, 2147483648, 8589934592, 34359738368, 137438953472, 549755813888, 2199023255552, 8796093022208, 35184372088832

Offset: 0

Paul Barry, Mar 17 2003

Binomial transform of A046717. Second binomial transform of A000302 (with interpolated zeros). Partial sums are A007583.
Counts closed walks of length 2n at a vertex of the cyclic graph on 4 nodes C_4. With interpolated zeros, counts closed walks of length n at a vertex of the cyclic graph on 4 nodes C_4. - Paul Barry, Mar 10 2004
In general, Sum_{k=0..n} Sum_{j=0..n} C(2*(n-k), j)*C(2*k, j)*r^j has expansion (1 - (r+1)*x)/(1 - (r+3)*x - (r-1)*(r+3)*x^2 + (r-1)^3*x^3). - Paul Barry, Jun 04 2005 [corrected by Jason Yuen, Jan 20 2025]
a(n) is the number of binary strings of length 2n with an even number of 0's (and hence an even number of 1's). - Toby Gottfried, Mar 22 2010
Number of compositions of n where there are 2 sorts of part 1, 4 sorts of part 2, 8 sorts of part 3, ..., 2^k sorts of part k. - Joerg Arndt, Aug 04 2014
a(n) is also the number of permutations simultaneously avoiding 231 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl Aug 07 2014
INVERT transform of powers of 2 (A000079). - Alois P. Heinz, Feb 11 2021
a(n) is the number of elements in an n-interval of the binomial poset of even-sized subsets of positive integers, cf. Stanley reference and second formula by Paul Barry.  Each multichain 0 = x_0 <= x_1 <= x_2 = 1 in such an n-interval corresponds to a closed walk described above by Paul Barry.  More generally, each multichain 0 = x_0 <= x_1 <= ... <= x_k = 1 corresponds to a closed walk of length 2n on the k-dimensional hypercube, cf. A054879, A092812, A121822. - Geoffrey Critzer, Apr 21 2023

			G.f. = 1 + 2*x + 8*x^2 + 32*x^3 + 128*x^4 + 512*x^5 + 2048*x^6 + 8192*x^7 + ...

Richard P. Stanley, Enumerative Combinatorics, Vol 1, second edition, Example 3.18.3-f, page 323.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
M. Paukner, L. Pepin, M. Riehl, and J. Wieser, Pattern Avoidance in Task-Precedence Posets, arXiv:1511.00080 [math.CO], 2015-2016.
Index to divisibility sequences.
Index entries for linear recurrences with constant coefficients, signature (4).

Row sums of triangle A136158.

Cf. A000079, A081295, A009117, A016742, A054879, A092812, A121822. Essentially the same as A004171.

[(4^n+0^n)/2: n in [0..30]]; // Vincenzo Librandi, Jul 26 2011

R:=PowerSeriesRing(Rationals(), 25); Coefficients(R!( (1-2*x)/(1-4*x))); // Marius A. Burtea, Jan 20 2020

a:= n-> 2^max(0, (2*n-1)):
seq(a(n), n=0..30);  # Alois P. Heinz, Jul 20 2017

CoefficientList[Series[(1-2x)/(1-4x),{x,0,40}],x] (* or *)
Join[{1}, NestList[4 # &, 2, 40]] (* Harvey P. Dale, Apr 22 2011 *)

a(n)=1<Charles R Greathouse IV, Jul 25 2011

x='x+O('x^100); Vec((1-2*x)/(1-4*x)) \\ Altug Alkan, Dec 21 2015

G.f.: (1-2*x)/(1-4*x).
a(n) = 4*a(n-1) n > 1, with a(0)=1, a(1)=2.
a(n) = (4^n+0^n)/2 (i.e., 1 followed by 4^n/2, n > 0).
E.g.f.: exp(2*x)*cosh(2*x) = (exp(4*x)+exp(0))/2. - Paul Barry, May 10 2003
a(n) = Sum_{k=0..n} C(2*n, 2*k). - Paul Barry, May 20 2003
a(n) = A001045(2*n+1) - A001045(2*n-1) + 0^n/2. - Paul Barry, Mar 10 2004
a(n) = 2^n*A011782(n); a(n) = gcd(A011782(2n), A011782(2n+1)). - Paul Barry, Jan 12 2005
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(2*(n-k), j)*C(2*k, j). - Paul Barry, Jun 04 2005
a(n) = Sum_{k=0..n} A038763(n,k). - Philippe Deléham, Sep 22 2006
a(n) = Integral_{x=0..4} p(n,x)^2/(Pi*sqrt(x(4-x))) dx, where p(n,x) is the sequence of orthogonal polynomials defined by C(2*n,n): p(n,x) = (2*x-4)*p(n-1,x) - 4*p(n-2,x), with p(0,x)=1, p(1,x)=-2+x. - Paul Barry, Mar 01 2007
a(n) = ((2+sqrt(4))^n + (2-sqrt(4))^n)/2. - Al Hakanson (hawkuu(AT)gmail.com), Nov 22 2008
a(n) = A000079(n) * A011782(n). - Philippe Deléham, Dec 01 2008
a(n) = A004171(n-1) = A028403(n) - A000079(n) for n >= 1. - Jaroslav Krizek, Jul 27 2009
a(n) = Sum_{k=0..n} A201730(n,k)*3^k. - Philippe Deléham, Dec 06 2011
a(n) = Sum_{k=0..n} A134309(n,k)*2^k = Sum_{k=0..n} A055372(n,k). - Philippe Deléham, Feb 04 2012
G.f.: Q(0), where Q(k) = 1 - 2*x/(1 - 2/(2 - 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Apr 29 2013
E.g.f.: 1/2 + exp(4*x)/2 = (Q(0)+1)/2, where Q(k) = 1 + 4*x/(2*k+1 - 2*x*(2*k+1)/(2*x + (k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Apr 29 2013
a(n) = ceiling( 2^(2n-1) ). - Wesley Ivan Hurt, Jun 30 2013
G.f.: 1 + 2*x/(1 + x)*( 1 + 5*x/(1 + 4*x)*( 1 + 8*x/(1 + 7*x)*( 1 + 11*x/(1 + 10*x)*( 1 + ... )))). - Peter Bala, May 27 2017
Sum_{n>=0} 1/a(n) = 5/3. - Amiram Eldar, Aug 18 2022
Sum_{n>=0} a(n)*x^n/A000680(n) = E(x)^2 where E(x) = Sum_{n>=0} x^n/A000680(n). - Geoffrey Critzer, Apr 21 2023

cp's OEIS Frontend

A006257 Josephus problem: a(2*n) = 2*a(n)-1, a(2*n+1) = 2*a(n)+1.

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A039599 Triangle formed from even-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x).

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A228351 Triangle read by rows in which row n lists the compositions (ordered partitions) of n (see Comments lines for definition).

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A131577 Zero followed by powers of 2 (cf. A000079).

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A238343 Triangle T(n,k) read by rows: T(n,k) is the number of compositions of n with k descents, n>=0, 0<=k<=n.

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Maple

A006257 Josephus problem: a(2n) = 2a(n)-1, a(2n+1) = 2a(n)+1.

A027383 a(2n) = 32^n - 2; a(2*n+1) = 2^(n+2) - 2.

A081294 Expansion of (1-2x)/(1-4x).