A010684 -id:A010684 - cp's OEIS frontend

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

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

A006516 a(n) = 2^(n-1)*(2^n - 1), n >= 0.

Original entry on oeis.org

0, 1, 6, 28, 120, 496, 2016, 8128, 32640, 130816, 523776, 2096128, 8386560, 33550336, 134209536, 536854528, 2147450880, 8589869056, 34359607296, 137438691328, 549755289600, 2199022206976, 8796090925056, 35184367894528, 140737479966720, 562949936644096

Offset: 0

N. J. A. Sloane

a(n) is also the number of different lines determined by pair of vertices in an n-dimensional hypercube. The number of these lines modulo being parallel is in A003462. - Ola Veshta (olaveshta(AT)my-deja.com), Feb 15 2001
Let G_n be the elementary Abelian group G_n = (C_2)^n for n >= 1: A006516 is the number of times the number -1 appears in the character table of G_n and A007582 is the number of times the number 1. Together the two sequences cover all the values in the table, i.e., A006516(n) + A007582(n) = 2^(2n). - Ahmed Fares (ahmedfares(AT)my-deja.com), Jun 01 2001
a(n) is the number of n-letter words formed using four distinct letters, one of which appears an odd number of times. - Lekraj Beedassy, Jul 22 2003 [See, e.g., the Balakrishnan reference, problems 2.67 and 2.68, p. 69. - Wolfdieter Lang, Jul 16 2017]
Number of 0's making up the central triangle in a Pascal's triangle mod 2 gasket. - Lekraj Beedassy, May 14 2004
m-th triangular number, where m is the n-th Mersenne number, i.e., a(n)=A000217(A000225(n)). - Lekraj Beedassy, May 25 2004
Number of walks of length 2n+1 between two nodes at distance 3 in the cycle graph C_8. - Herbert Kociemba, Jul 02 2004
The sequence of fractions a(n+1)/(n+1) is the 3rd binomial transform of (1, 0, 1/3, 0, 1/5, 0, 1/7, ...). - Paul Barry, Aug 05 2005
Number of monic irreducible polynomials of degree 2 in GF(2^n)[x]. - Max Alekseyev, Jan 23 2006
(A007582(n))^2 + a(n)^2 = A007582(2n). E.g., A007582(3) = 36, a(3) = 28; A007582(6) = 2080. 36^2 + 28^2 = 2080. - Gary W. Adamson, Jun 17 2006
The sequence 6*a(n), n>=1, gives the number of edges of the Hanoi graph H_4^{n} with 4 pegs and n>=1 discs. - Daniele Parisse, Jul 28 2006
8*a(n) is the total border length of the 4*n masks used when making an order n regular DNA chip, using the bidimensional Gray code suggested by Pevzner in the book "Computational Molecular Biology." - Bruno Petazzoni (bruno(AT)enix.org), Apr 05 2007
If we start with 1 in binary and at each step we prepend 1 and append 0, we construct this sequence: 1 110 11100 1111000 etc.; see A109241(n-1). - Artur Jasinski, Nov 26 2007
Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A) for which x does not equal y. - Ross La Haye, Jan 02 2008
Wieder calls these "conjoint usual 2-combinations." The set of "conjoint strict k-combinations" is the subset of conjoint usual k-combinations where the empty set and the set itself are excluded from possible selection. These numbers C(2^n - 2,k), which for k = 2 (i.e., {x,y} of the power set of a set) give {1, 0, 1, 15, 91, 435, 1891, 7875, 32131, 129795, 521731, ...}. - Ross La Haye, Jan 15 2008
If n is a member of A000043 then a(n) is also a perfect number (A000396). - Omar E. Pol, Aug 30 2008
a(n) is also the number whose binary representation is A109241(n-1), for n>0. - Omar E. Pol, Aug 31 2008
From Daniel Forgues, Nov 10 2009: (Start)
If we define a spoof-perfect number as:
A spoof-perfect number is a number that would be perfect if some (one or more) of its odd composite factors were wrongly assumed to be prime, i.e., taken as a spoof prime.
And if we define a "strong" spoof-perfect number as:
A "strong" spoof-perfect number is a spoof-perfect number where sigma(n) does not reveal the compositeness of the odd composite factors of n which are wrongly assumed to be prime, i.e., taken as a spoof prime.
The odd composite factors of n which are wrongly assumed to be prime then have to be obtained additively in sigma(n) and not multiplicatively.
Then:
If 2^n-1 is odd composite but taken as a spoof prime then 2^(n-1)*(2^n - 1) is an even spoof perfect number (and moreover "strong" spoof-perfect).
For example:
a(8) = 2^(8-1)*(2^8 - 1) = 128*255 = 32640 (where 255 (with factors 3*5*17) is taken as a spoof prime);
sigma(a(8)) = (2^8 - 1)*(255 + 1) = 255*256 = 2*(128*255) = 2*32640 = 2n is spoof-perfect (and also "strong" spoof-perfect since 255 is obtained additively);
a(11) = 2^(11-1)*(2^11 - 1) = 1024*2047 = 2096128 (where 2047 (with factors 23*89) is taken as a spoof prime);
sigma(a(11)) = (2^11 - 1)*(2047 + 1) = 2047*2048 = 2*(1024*2047) = 2*2096128 = 2n is spoof-perfect (and also "strong" spoof-perfect since 2047 is obtained additively).
I did a Google search and didn't find anything about the distinction between "strong" versus "weak" spoof-perfect numbers. Maybe some other terminology is used.
An example of an even "weak" spoof-perfect number would be:
n = 90 = 2*5*9 (where 9 (with factors 3^2) is taken as a spoof prime);
sigma(n) = (1+2)*(1+5)*(1+9) = 3*(2*3)*(2*5) = 2*(2*5*(3^2)) = 2*90 = 2n is spoof-perfect (but is not "strong" spoof-perfect since 9 is obtained multiplicatively as 3^2 and is thus revealed composite).
Euler proved:
If 2^k - 1 is a prime number, then 2^(k-1)*(2^k - 1) is a perfect number and every even perfect number has this form.
The following seems to be true (is there a proof?):
If 2^k - 1 is an odd composite number taken as a spoof prime, then 2^(k-1)*(2^k - 1) is a "strong" spoof-perfect number and every even "strong" spoof-perfect number has this form?
There is only one known odd spoof-perfect number (found by Rene Descartes) but it is a "weak" spoof-perfect number (cf. 'Descartes numbers' and 'Unsolved problems in number theory' links below). (End)
a(n+1) = A173787(2*n+1,n); cf. A020522, A059153. - Reinhard Zumkeller, Feb 28 2010
Also, row sums of triangle A139251. - Omar E. Pol, May 25 2010
Starting with "1" = (1, 1, 2, 4, 8, ...) convolved with A002450: (1, 5, 21, 85, 341, ...); and (1, 3, 7, 15, 31, ...) convolved with A002001: (1, 3, 12, 48, 192, ...). - Gary W. Adamson, Oct 26 2010
a(n) is also the number of toothpicks in the corner toothpick structure of A153006 after 2^n - 1 stages. - Omar E. Pol, Nov 20 2010
The number of n-dimensional odd theta functions of half-integral characteristic. (Gunning, p.22) - Michael Somos, Jan 03 2014
a(n) = A000217((2^n)-1) = 2^(2n-1) - 2^(n-1) is the nearest triangular number below 2^(2n-1); cf. A007582, A233327. - Antti Karttunen, Feb 26 2014
a(n) is the sum of all the remainders when all the odd numbers < 2^n are divided by each of the powers 2,4,8,...,2^n. - J. M. Bergot, May 07 2014
Let b(m,k) = number of ways to form a sequence of m selections, without replacement, from a circular array of m labeled cells, such that the first selection of a cell whose adjacent cells have already been selected (a "first connect") occurs on the k-th selection. b(m,k) is defined for m >=3, and for 3 <= k <= m. Then b(m,k)/2m ignores rotations and reflection. Let m=n+2, then a(n) = b(m,m-1)/2m. Reiterated, a(n) is the (m-1)th column of the triangle b(m,k)/2m, whose initial rows are (1), (1 2), (2 6 4), (6 18 28 8), (24 72 128 120 16), (120 360 672 840 496 32), (720 2160 4128 5760 5312 2016 64); see A249796. Note also that b(m,3)/2m = n!, and b(m,m)/2m = 2^n. Proofs are easy. - Tony Bartoletti, Oct 30 2014
Beginning at a(1) = 1, this sequence is the sum of the first 2^(n-1) numbers of the form 4*k + 1 = A016813(k). For example, a(4) = 120 = 1 + 5 + 9 + 13 + 17 + 21 + 25 + 29. - J. M. Bergot, Dec 07 2014
a(n) is the number of edges in the (2^n - 1)-dimensional simplex. - Dimitri Boscainos, Oct 05 2015
a(n) is the number of linear elements in a complete plane graph in 2^n points. - Dimitri Boscainos, Oct 05 2015
a(n) is the number of linear elements in a complete parallelotope graph in n dimensions. - Dimitri Boscainos, Oct 05 2015
a(n) is the number of lattices L in Z^n such that the quotient group Z^n / L is C_4. - Álvar Ibeas, Nov 26 2015
a(n) gives the quadratic coefficient of the polynomial ((x + 1)^(2^n) + (x - 1)^(2^n))/2, cf. A201461. - Martin Renner, Jan 14 2017
Let f(x)=x+2*sqrt(x) and g(x)=x-2*sqrt(x). Then f(4^n*x)=b(n)*f(x)+a(n)*g(x) and g(4^n*x)=a(n)*f(x)+b(n)*g(x), where b is A007582. - Luc Rousseau, Dec 06 2018
For n>=1, a(n) is the covering radius of the first order Reed-Muller code RM(1,2n). - Christof Beierle, Dec 22 2021
a(n) =

			G.f. = x + 6*x^2 + 28*x^3 + 120*x^4 + 496*x^5 + 2016*x^6 + 8128*x^7 + 32640*x^8 + ...

V. K. Balakrishnan, Theory and problems of Combinatorics, "Schaum's Outline Series", McGraw-Hill, 1995, p. 69.
Martin Gardner, Mathematical Carnival, "Pascal's Triangle", p. 201, Alfred A. Knopf NY, 1975.
Richard K. Guy, Unsolved problems in number theory, (p. 72).
Ross Honsberger, Mathematical Gems, M.A.A., 1973, p. 113.
Clifford A. Pickover, Wonders of Numbers, Chap. 55, Oxford Univ. Press NY 2000.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n=0..200
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]
M. Archibald, A. Blecher, A. Knopfmacher and M. E. Mays, Inversions and Parity in Compositions of Integers, J. Int. Seq., Vol. 23 (2020), Article 20.4.1.
William Banks, Ahmet Güloğlu, Wesley Nevans and Filip Saidak, Descartes numbers, Anatomy of integers, 167-173, CRM Proc. Lecture Notes, 46, Amer. Math. Soc., Providence, RI, 2008. MathSciNet review (subscription required).
Taylor Brysiewicz, Holger Eble, and Lukas Kühne, Enumerating chambers of hyperplane arrangements with symmetry, arXiv:2105.14542 [math.CO], 2021.
John Elias, Illustration of initial terms: Mersenne-Sierpinski Triangles
Farideh Firoozbakht and M. F. Hasler, Variations on Euclid's formula for Perfect Numbers, JIS 13 (2010) #10.3.1.
R. C. Gunning, Riemann Surfaces and Second-Order Theta Functions, Springer-Verlag, 1976. See page 22.
Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6.
T. Helleseth, T. Klove and J.Mykkeltveit, On the covering radius of binary codes (Corresp.), IEEE Transactions on Information Theory, Vol. 24 (1978).
V. Meally, Letter to N. J. A. Sloane, May 1975
Axel Muller, Metod Saniga, Alain Giorgetti, Henri de Boutray, and Frédéric Holweck, New and improved bounds on the contextuality degree of multi-qubit configurations, arXiv:2305.10225 [quant-ph], 2023.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
O. S. Rothaus, On "bent" functions, Journal of Combinatorial Theory, Series A, Vol. 20 (1976).
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
Thomas Wieder, The number of certain k-combinations of an n-set, Applied Mathematics Electronic Notes, vol. 8 (2008).
Index entries for linear recurrences with constant coefficients, signature (6,-8).

Equals A006095(n+1) - A006095(n). In other words, A006095 gives the partial sums.
Cf. A000108, A007582, A010036, A016290, A003462, A079598, A134346, A048473, A151821, A010684.
Cf. A000043, A000396. - Omar E. Pol, Aug 30 2008
Cf. A109241, A139251, A153006. - Omar E. Pol, Aug 31 2008, May 25 2010, Nov 20 2010
Cf. A002450, A002001. - Gary W. Adamson, Oct 26 2010
Cf. A049072, A000384, A201461, A005059 (binomial transform, and special 5-letter words), A065442, A211705.
Cf. A171476.

List([0..25],n->2^(n-1)*(2^n-1)); # Muniru A Asiru, Dec 06 2018

a006516 n = a006516_list !! n
a006516_list = 0 : 1 :
    zipWith (-) (map (* 6) $ tail a006516_list) (map (* 8) a006516_list)
-- Reinhard Zumkeller, Oct 25 2013

[2^(n-1)*(2^n - 1): n in [0..30]]; // Vincenzo Librandi, Oct 31 2014

GBC := proc(n,k,q) local i; mul( (q^(n-i)-1)/(q^(k-i)-1),i=0..k-1); end; # define q-ary Gaussian binomial coefficient [ n,k ]_q
[ seq(GBC(n+1,2,2)-GBC(n,2,2), n=0..30) ]; # produces A006516
A006516:=1/(4*z-1)/(2*z-1); # Simon Plouffe in his 1992 dissertation
seq(binomial(2^n, 2), n=0..19); # Zerinvary Lajos, Feb 22 2008

Table[2^(n - 1)(2^n - 1), {n, 0, 30}] (* or *) LinearRecurrence[{6, -8}, {0, 1}, 30] (* Harvey P. Dale, Jul 15 2011 *)

A006516(n):=2^(n-1)*(2^n - 1)$ makelist(A006516(n),n,0,30); /* Martin Ettl, Nov 15 2012 */

a(n)=(1<Charles R Greathouse IV, Jun 10 2011

vector(100, n, n--; 2^(n-1)*(2^n-1)) \\ Altug Alkan, Oct 06 2015

for n in range(0, 30): print(2**(n-1)*(2**n - 1), end=', ') # Stefano Spezia, Dec 06 2018

[lucas_number1(n,6,8) for n in range(24)]  # Zerinvary Lajos, Apr 22 2009

[(4**n - 2**n) / 2 for n in range(24)]  # Zerinvary Lajos, Jun 05 2009

G.f.: x/((1 - 2*x)*(1 - 4*x)).
E.g.f. for a(n+1), n>=0: 2*exp(4*x) - exp(2*x).
a(n) = 2^(n-1)*Stirling2(n+1,2), n>=0, with Stirling2(n,m)=A008277(n,m).
Second column of triangle A075497.
a(n) = Stirling2(2^n,2^n-1) = binomial(2^n,2). - Ross La Haye, Jan 12 2008
a(n+1) = 4*a(n) + 2^n. - Philippe Deléham, Feb 20 2004
Convolution of 4^n and 2^n. - Ross La Haye, Oct 29 2004
a(n+1) = Sum_{k=0..n} Sum_{j=0..n} 4^(n-j)*binomial(j,k). - Paul Barry, Aug 05 2005
a(n+2) = 6*a(n+1) - 8*a(n), a(1) = 1, a(2) = 6. - Daniele Parisse, Jul 28 2006 [Typo corrected by Yosu Yurramendi, Aug 06 2008]
Row sums of triangle A134346. Also, binomial transform of A048473: (1, 5, 17, 53, 161, ...); double bt of A151821: (1, 4, 8, 16, 32, 64, ...) and triple bt of A010684: (1, 3, 1, 3, 1, 3, ...). - Gary W. Adamson, Oct 21 2007
a(n) = 3*Stirling2(n+1,4) + Stirling2(n+2,3). - Ross La Haye, Jun 01 2008
a(n) = (4^n - 2^n)/2.
a(n) = A153006(2^n-1). - Omar E. Pol, Nov 20 2010
Sum_{n>=1} 1/a(n) = 2 * (A065442 - 1) = A211705 - 2. - Amiram Eldar, Dec 24 2020
a(n) = binomial(2*n+2, n+1) - Catalan(n+2). - N. J. A. Sloane, Apr 01 2021
a(n) = A171476(n-1), for n >= 1, and a(0) = 0. - Wolfdieter Lang, Jul 27 2022

A020988 a(n) = (2/3)*(4^n-1).

Original entry on oeis.org

0, 2, 10, 42, 170, 682, 2730, 10922, 43690, 174762, 699050, 2796202, 11184810, 44739242, 178956970, 715827882, 2863311530, 11453246122, 45812984490, 183251937962, 733007751850, 2932031007402, 11728124029610, 46912496118442, 187649984473770, 750599937895082

Offset: 0

N. J. A. Sloane

Numbers whose binary representation is 10, n times (see A163662(n) for n >= 1). - Alexandre Wajnberg, May 31 2005
Numbers whose base-4 representation consists entirely of 2's; twice base-4 repunits. - Franklin T. Adams-Watters, Mar 29 2006
Expected time to finish a random Tower of Hanoi problem with 2n disks using optimal moves, so (since 2n is even and A010684(2n) = 1) a(n) = A060590(2n). - Henry Bottomley, Apr 05 2001
a(n) is the number of derangements of [2n + 3] with runs consisting of consecutive integers. E.g., a(1) = 10 because the derangements of {1, 2, 3, 4, 5} with runs consisting of consecutive integers are 5|1234, 45|123, 345|12, 2345|1, 5|4|123, 5|34|12, 45|23|1, 345|2|1, 5|4|23|1, 5|34|2|1 (the bars delimit the runs). - Emeric Deutsch, May 26 2003
For n > 0, also smallest numbers having in binary representation exactly n + 1 maximal groups of consecutive zeros: A087120(n) = a(n-1), see A087116. - Reinhard Zumkeller, Aug 14 2003
Number of walks of length 2n + 3 between any two diametrically opposite vertices of the cycle graph C_6. Example: a(0) = 2 because in the cycle ABCDEF we have two walks of length 3 between A and D: ABCD and AFED. - Emeric Deutsch, Apr 01 2004
From Paul Barry, May 18 2003: (Start)
Row sums of triangle using cumulative sums of odd-indexed rows of Pascal's triangle (start with zeros for completeness):
            0  0
            1  1
         1  4  4  1
      1  6 14 14  6  1
   1  8 27 49 49 27  8  1 (End)
a(n) gives the position of the n-th zero in A173732, i.e., A173732(a(n)) = 0 for all n and this gives all the zeros in A173732. - Howard A. Landman, Mar 14 2010
Smallest number having alternating bit sum -n. Cf. A065359. For n = 0, 1, ..., the last digit of a(n) is 0, 2, 0, 2, ... . - Washington Bomfim, Jan 22 2011
Number of toothpicks minus 1 in the toothpick structure of A139250 after 2^n stages. - Omar E. Pol, Mar 15 2012
For n > 0 also partial sums of the odd powers of 2 (A004171). - K. G. Stier, Nov 04 2013
Values of m such that binomial(4*m + 2, m) is odd. Cf. A002450. - Peter Bala, Oct 06 2015
For a(n) > 2, values of m such that m is two steps away from a power of 2 under the Collatz iteration. - Roderick MacPhee, Nov 10 2016
a(n) is the position of the first occurrence of 2^(n+1)-1 in A020986. See the Brillhart and Morton link, pp. 856-857. - John Keith, Jan 12 2021
a(n) is the number of monotone paths in the n-dimensional cross-polytope for a generic linear orientation. See the Black and De Loera link. - Alexander E. Black, Feb 15 2023

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..170 from Vincenzo Librandi)
Andrei Asinowski, Cyril Banderier, and Benjamin Hackl, On extremal cases of pop-stack sorting, Permutation Patterns (Zürich, Switzerland, 2019) [link is not very stable].
Andrei Asinowski, Cyril Banderier, and Benjamin Hackl, Flip-sort and combinatorial aspects of pop-stack sorting, arXiv:2003.04912 [math.CO], 2020.
Peter Bala, A characterization of A002450, A020988 and A080674.
Alexander E. Black, Monotone Paths on Polytopes: Combinatorics and Optimization, Ph. D. Dissertation, Univ. Calif. Davis (2024). See p. 59.
Alexander Black and Jesús De Loera, Monotone paths on cross-polytopes, arXiv:2102.01237 [math.CO], Feb 2021
John Brillhart and Peter Morton, A case study in mathematical research: the Golay-Rudin-Shapiro sequence, Amer. Math. Monthly, 103 (1996) 854-869.
Nobushige Kurokawa, Zeta functions over F_1, Proc. Japan Acad., 81, Ser. A (2005), 180-184. See Theorem 3 (3).
Jonathan L. Merzel, Ján Minač, Tung T. Nguyen, and Nguyên Duy Tân, On divisibility relation graphs, 2025. See p. 14.
Andrei K. Svinin, Tuenter polynomials and a Catalan triangle, arXiv:1603.05748 [math.CO], 2016. See p.3.
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A020989, A047849, A108019, A295521, A020522.

[(2/3)*(4^n-1): n in [0..40] ]; // Vincenzo Librandi, Apr 28 2011

A020988 := proc(n)
    2*(4^n-1)/3 ;
end proc: # R. J. Mathar, Feb 19 2015

(2(4^Range[0, 30] - 1))/3 (* or *) LinearRecurrence[{5, -4}, {0, 2}, 30] (* Harvey P. Dale, Sep 25 2013 *)

vector(100, n, n--; (2/3)*(4^n-1)) \\ Altug Alkan, Oct 06 2015

Vec(2*z/((1-z)*(1-4*z)) + O(z^30)) \\ Altug Alkan, Oct 11 2015

def A020988(n): return (2 * ((1 << (2 * n)) - 1)) // 3 # John Reimer Morales, Aug 05 2025

(((List.fill(20)(4: BigInt)).scanLeft(1: BigInt)( * )).map(2 * )).scanLeft(0: BigInt)( + ) // _Alonso del Arte, Sep 12 2019

a(n) = 4*a(n-1) + 2, a(0) = 0.
a(n) = A026644(2*n).
a(n) = A007583(n) - 1 = A039301(n+1) - 2 = A083584(n-1) + 1.
E.g.f. : (2/3)*(exp(4*x)-exp(x)). - Paul Barry, May 18 2003
a(n) = A007583(n+1) - 1 = A039301(n+2) - 2 = A083584(n) + 1. - Ralf Stephan, Jun 14 2003
G.f.: 2*x/((1-x)*(1-4*x)). - R. J. Mathar, Sep 17 2008
a(n) = a(n-1) + 2^(2n-1), a(0) = 0. - Washington Bomfim, Jan 22 2011
a(n) = A193652(2*n). - Reinhard Zumkeller, Aug 08 2011
a(n) = 5*a(n-1) - 4*a(n-2) (n > 1), a(0) = 0, a(1) = 2. - L. Edson Jeffery, Mar 02 2012
a(n) = (2/3)*A024036(n). - Omar E. Pol, Mar 15 2012
a(n) = 2*A002450(n). - Yosu Yurramendi, Jan 24 2017
From Seiichi Manyama, Nov 24 2017: (Start)
Zeta_{GL(2)/F_1}(s) = Product_{k = 1..4} (s-k)^(-b(2,k)), where Sum b(2,k)*t^k = t*(t-1)*(t^2-1). That is Zeta_{GL(2)/F_1}(s) = (s-3)*(s-2)/((s-4)*(s-1)).
Zeta_{GL(2)/F_1}(s) = Product_{n > 0} (1 - (1/s)^n)^(-A295521(n)) = Product_{n > 0} (1 - x^n)^(-A295521(n)) = (1-3*x)*(1-2*x)/((1-4*x)*(1-x)) = 1 + Sum_{k > 0} a(k-1)*x^k (x=1/s). (End)
From Oboifeng Dira, May 29 2020: (Start)
a(n) = A078008(2n+1) (second bisection).
a(n) = Sum_{k=0..n} binomial(2n+1, ((n+2) mod 3)+3k). (End)
From John Reimer Morales, Aug 04 2025: (Start)
a(n) = A000302(n) - A047849(n).
a(n) = A020522(n) + A000079(n) - A047849(n). (End)

Edited by N. J. A. Sloane, Sep 06 2006

A003558 Least number m > 0 such that 2^m == +-1 (mod 2n + 1).

Original entry on oeis.org

1, 1, 2, 3, 3, 5, 6, 4, 4, 9, 6, 11, 10, 9, 14, 5, 5, 12, 18, 12, 10, 7, 12, 23, 21, 8, 26, 20, 9, 29, 30, 6, 6, 33, 22, 35, 9, 20, 30, 39, 27, 41, 8, 28, 11, 12, 10, 36, 24, 15, 50, 51, 12, 53, 18, 36, 14, 44, 12, 24, 55, 20, 50, 7, 7, 65, 18, 36, 34, 69, 46

Offset: 0

N. J. A. Sloane

nonn

Multiplicative suborder of 2 (mod 2n+1) (or sord(2, 2n+1)).
This is called quasi-order of 2 mod b, with b = 2*n+1, for n >= 1, in the Hilton/Pederson reference.
For the complexity of computing this, see A002326.
Also, the order of the so-called "milk shuffle" of a deck of n cards, which maps cards (1,2,...,n) to (1,n,2,n-1,3,n-2,...).  See the paper of Lévy. - Jeffrey Shallit, Jun 09 2019
It appears that under iteration of the base-n Kaprekar map, for even n > 2 (A165012, A165051, A165090, A151949 in bases 4, 6, 8, 10), almost all cycles are of length a(n/2 - 1); proved under the additional constraint that the cycle contains at least one element satisfying "number of digits (n-1) - number of digits 0 = o(total number of digits)". - Joseph Myers, Sep 05 2009
From Gary W. Adamson, Sep 20 2011: (Start)
a(n) can be determined by the cycle lengths of iterates using x^2 - 2, seed 2*cos(2*Pi/N); as shown in the A065941 comment of Sep 06 2011. The iterative map of the logistic equation 4x*(1-x) is likewise chaotic with the same cycle lengths but initiating the trajectory with sin^2(2*Pi/N), N = 2n+1 [Kappraff & Adamson, 2004]. Chaotic terms with the identical cycle lengths can be obtained by applying Newton's method to i = sqrt(-1) [Strang, also Kappraff and Adamson, 2003], resulting in the morphism for the cot(2*Pi/N) trajectory: (x^2-1)/2x. (End)
From Gary W. Adamson, Sep 11 2019: (Start)
Using x^2 - 2 with seed 2*cos(Pi/7), we obtain the period-three trajectory 1.8019377...-> 1.24697...-> -0.445041... For an odd prime N, the trajectory terms represent diagonal lengths of regular star 2N-gons, with edge the shortest value (0.445... in this case.) (Cf. "Polygons and Chaos", p. 9, Fig 4.) We can normalize such lengths by dividing through with the lowest value, giving 3 diagonals of the 14-gon: (1, 2.801937..., 4.048917...). Label the terms ranked in magnitude with odd integers (1, 3, 5), and we find that the diagonal lengths are in agreement with the diagonal formula (sin(j*Pi)/14)/(sin(Pi/14)), with j = (1,3,5). (End)
Roots of signed n-th row A054142 polynomials are chaotic with respect to the operation (-2, x^2), with cycle lengths a(n). Example: starting with a root to x^3 - 5x^2 + 6x - 1 = 0; (2 + 2*cos(2*Pi/N) = 3.24697...); we obtain the trajectory (3.24697...-> 1.55495...-> 0.198062...); the roots to the polynomial with cycle length 3 matching a(3) = 3. - Gary W. Adamson, Sep 21 2011
From Juhani Heino, Oct 26 2015: (Start)
Start a sequence with numbers 1 and n. For next numbers, add previous numbers going backwards until the sum is even. Then the new number is sum/2. I conjecture that the sequence returns to 1,n and a(n) is the cycle length.
For example:
  1,7,4,2,1,7,... so a(7) = 4.
  1,6,3,5,4,2,1,6,... so a(6) = 6. (End)
From Juhani Heino, Nov 06 2015: (Start)
Proof of the above conjecture: Let n = -1/2; thus 2n + 1 = 0, so operations are performed mod (2n + 1). When the member is even, it is divided by 2. When it is odd, multiply by n, so effectively divide by -2. This is all well-defined in the sense that new members m are 1 <= m <= n. Now see what happens starting from an odd member m. The next member is -m/2. As long as there are even members, divide by 2 and end up with an odd -m/(2^k). Now add all the members starting with m. The sum is m/(2^k). It's divided by 2, so the next member is m/(2^(k+1)). That is the same as (-m/(2^k))/(-2), as with the definition.
So actually start from 1 and always divide by 2, although the sign sometimes changes. Eventually 1 is reached again. The chain can be traversed backwards and then 2^(cycle length) == +-1 (mod 2n + 1).
To conclude, we take care of a(0): sequence 1,0 continues with zeros and never returns to 1. So let us declare that cycle length 0 means unavailable. (End)
From Gary W. Adamson, Aug 20 2019: (Start)
Terms in the sequence can be obtained by applying the doubling sequence mod (2n + 1), then counting the terms until the next term is == +1 (mod 2n + 1).  Example: given 25, the trajectory is (1, 2, 4, 8, 16, 7, 14, 3, 6, 12).
The cycle ends since the next term is 24 == -1 (mod 25) and has a period of 10. (End)
From Gary W. Adamson, Sep 04 2019: (Start)
Conjecture of Kappraff and Adamson in "Polygons and Chaos", p. 13 Section 7, "Chaos and Number": Given the cycle length for N = 2n + 1, the same cycle length is present in bases 4, 9, 16, 25, ..., m^2, for the expansion of 1/N.
Examples: The cycle length for 7 is 3, likewise for 1/7 in base 4: 0.021021021.... In base 9 the expansion of 1/7 is 0.125125125... Check: The first few terms are 1/9 + 2/81 + 5/729 = 104/279 = 0.1426611... (close to 1/7 = 0.142857...). (End)
From Gary W. Adamson, Sep 24 2019: (Start)
An exception to the rule for 1/N in bases m^2: (when N divides m^2 as in 1/7 in base 49, = 7/49, rational). When all terms in the cycle are the same, the identity reduces to 1/N in (some bases) = .a, a, a, .... The minimal values of "a" for 1/N are provided as examples, with the generalization 1/N in base (N-1)^2 = .a, a, a, ... for N odd:
   1/3  in base  4  =  .1, 1, 1, ...
   1/5  in base 16  =  .3, 3, 3, ...
   1/7  in base 36  =  .5, 5, 5, ...
   1/9  in base 64  =  .7, 7, 7, ...
   1/11 in base 100 =  .9, 9, 9, ... (Check: the first three terms are 9/100 +9/(100^2) + 9/(100^3) = 0.090909 where 1/11 = 0.09090909...). (End)
For N = 2n+1, the corresponding entry is equal to the degree of the polynomial for N shown in (Lang, Table 2, p. 46). As shown, x^3 - 3x - 1 is the minimal polynomial for N = 9, with roots (1.87938..., -1.53208..., 0.347296...); matching the (abs) values of the 2*cos(Pi/9) trajectory using x^2 - 2. Thus, a(4) = 3. If N is prime, the polynomials shown in Table 2 are the same as those for the same N in A065941. If different, the minimal polynomials shown in Table 2 are factors of those in A065941. - Gary W. Adamson, Oct 01 2019
The terms in the 2*cos(Pi/N) trajectory (roots to the minimal polynomials in A187360 and (Lang)), are quickly obtained from the doubling trajectory (mod N) by using the operation L(m) 2*cos(x)--> 2*cos(m*x), where L(2), the second degree Lucas polynomial (A034807) is x^2 - 2. Relating to the heptagon and using seed 2*cos(Pi/7), we obtain the trajectory 1.8019..., 1.24697..., and 0.445041....; cyclic with period 3. All such roots can be derived from the N-th roots of Unity and can be mapped on the Vesica Piscis. Given the roots of Unity (Polar 1Angle(k*2*Pi/N), k = 1, 2, ..., (N-1)/2) the Vesica Piscis maps these points on the left (L) circle to the (R) circle by adding 1A(0) or (a + b*I) = (1 + 0i). But this operation is the same as vector addition in which the resultant vector is 1 + 1A(k*(2*Pi/N)). Example: given the radius at 2*Pi/7 on the left circle, this maps to (1 + 1A(2*Pi/7)) on the right circle; or 1A(2*Pi/7) --> (1.8019377...A(Pi/7).  Similarly, 1 + 1A((2)*2*Pi/7)) maps to (1.24697...A (2*Pi/7); and 1 + 1A(3*2*Pi/7) maps to (.0445041...A(3*Pi/7). - Gary W. Adamson, Oct 23 2019
From Gary W. Adamson, Dec 01 2021: (Start)
As to segregating the two sets: (A014659 terms are those N = (2*n+1), N divides (2^m - 1), and (A014657 terms are those N that divide (2^m + 1)); it appears that the following criteria apply: Given IcoS(N, 1) (cf. Lang link "On the Equivalence...", p. 16, Definition 20), if the number of odd terms is odd, then N belongs to A014659, otherwise A014657. In IcoS(11, 1): (1, 2, 4, 3, 5), three odd terms indicate that 11 is a term in A014657. IcoS(15, 1) has the orbit (1, 2, 4, 7) with two odd terms indicating that 15 is a term in A014659.
It appears that if sin(2^m * Pi/N) has a negative sign, then N is in A014659; otherwise N is in A014657. With N = 15, m is 4 and sin(16 * Pi/15) is -0.2079116... If N is 11, m is 5 and sin(32 * Pi/11) is 0.2817325. (End)
On the iterative map using x^2 - 2, (Devaney, p. 126) states that we must find the function that takes 2*cos(Pi) -> 2*cos(2*Pi). "However, we may write 2*cos(2*Pi) = 2*(2*cos^2(Pi) - 1) = (2*cos(Pi))^2 - 2. So the required function is x^2 - 2." On the period 3 implies chaos theorem of James Yorke and T.Y. Li, proved in 1975; Devaney (p. 133) states that if F is continuous and we find a cycle of period 3, there are infinitely many other cycles for this map with every possible period. Check: The x^2 - 2 orbit for 7 has a period of 3, so this entry has periodic points of all other periods. - Gary W. Adamson, Jan 04 2023
It appears that a(n) is the length of the cycle starting at 2/(2*n+1) for the map x->1 - abs(2*x-1). - Michel Marcus, Jul 16 2025

			a(3) = 3 since f(x) = x^2 - 2 has a period of 3 using seed 2*cos(2*Pi/7), where 7 = 2*3 + 1.
a(15) = 5 since the iterative map of the logistic equation 4x*(1-x) has a period 5 using seed sin^2(2*Pi)/N; N = 31 = 2*15 + 1.

Peter Hilton and Jean Pedersen, A Mathematical Tapestry: Demonstrating the Beautiful Unity of Mathematics, Cambridge University Press, 2010, pp. 261-264.
Carl Schick, Trigonometrie und unterhaltsame Zahlentheorie, Bokos Druck, Zürich, 2003 (ISBN 3-9522917-0-6).
Robert L. Devaney, A First Course in Chaotic Dynamical Systems, Theory and Experiment; Perseus Books Publishing, 1992, pp. 121-126.

T. D. Noe, Table of n, a(n) for n = 0..1000
R. Bekes, J. Pedersen, and B. Shao, Mad tea party cyclic partitions, Coll. Math. J. 43 (1) (2012) 25-36, Jstor.
William Q. Erickson, Daniel Herden, Jonathan Meddaugh, Mark R. Sepanski, Mitchell Minyard, and Kyle Rosengartner, Limits and Periodicity of Metamour 2-Distance Graphs, arXiv:2409.02306 [math.CO], 2024. See p. 9.
Daniel Gabric and Jeffrey Shallit, Borders, Palindrome Prefixes, and Square Prefixes, arXiv:1906.03689 [cs.DM], 2019.
P. Hilton and J. Pedersen, On Factoring 2^k+-1, The Math. Educ. 5 (1) (1994) 29-31.
Jay Kappraff and Gary W. Adamson, Polygons and Chaos, Bridges: Mathematical Connections in Art, Music, and Science, 2001, pages 67-80.
Anthony Kay and Katrina Downes-Ward, Fixed Points and Cycles of the Kaprekar Transformation: 1. Odd Bases, J. Int. Seq., Vol. 25 (2022), Article 22.6.7.
Anthony Kay and Katrina Downes-Ward, Fixed Points and Cycles of the Kaprekar Transformation: 2. Even bases, arXiv:2408.12257 [math.CO], 2024. See p. 6.
Torleiv Klove, On covering sets for limited-magnitude errors, Cryptogr. Commun. 8 (3) (2016) 415-433
Wolfdieter Lang, The Field Q(2Cos Pi/N), its Galois group and length ratios in the regular n-gon, arXiv:1210.1018 [math.GR], 2012-2017.
Wolfdieter Lang, On the Equivalence of Three Complete Cyclic Systems of Integers, arXiv:2008.04300 [math.NT], 2020.
Paul Lévy, Sur quelques classes de permutations, Compositio Mathematica, 8 (1951), 1-48.
Olivier Martin, Andrew M. Odlyzko, and Stephen Wolfram, Algebraic Properties of Cellular Automata, Communications in Mathematical Physics, volume 93, number 2, June 1984, pages 219-258, appendix B part C sord_N(2). Also third author's copy (and text).
Harry J. Smith, XICalc - Extra Precision Integer Calculator. [Broken link]
Gilbert Strang, A Chaotic Search for i, College Mathematics Journal 22, 3-12, (1991) [JSTOR].
Eric Weisstein's World of Mathematics, Multiplicative Order.
Eric Weisstein's World of Mathematics, Suborder Function.

Cf. A054142, A065941, A085478, A160657, A179480, A135303 (coach numbers), A216371 (odd primes with one coach), A000215 (Fermat numbers).
A216066 is an essentially identical sequence apart from the offset.
Cf. A329593, A332433 (signs).
Cf. A014657, A014659.

A003558 := proc(n)
    local m,mo ;
    if n = 0 then
        return 0 ;
    end if;
    for m from 1 do
        mo := modp(2^m,2*n+1) ;
        if mo in {1,2*n} then
            return m;
        end if;
    end do:
end proc:
seq(A003558(n),n=0..20) ; # R. J. Mathar, Dec 01 2014
f:= proc(n) local t;
      t:= numtheory:-mlog(-1,2,n);
      if t = FAIL then numtheory:-order(2,n) else t fi
end proc:
0, seq(f(2*k+1),k=1..1000); # Robert Israel, Oct 26 2015

Suborder[a_,n_]:=If[n>1&&GCD[a,n]==1,Min[MultiplicativeOrder[a,n,{-1,1}]],0];
Join[{1},Table[Suborder[2,2n+1],{n,100}]] (* T. D. Noe, Aug 02 2006 *) (* revised by Vincenzo Librandi, Apr 11 2020 *)

a(n) = {m=1; while(m, if( (2^m) % (2*n+1) == 1 || (2^m) % (2*n+1) == 2*n, return(m)); m++)} \\ Altug Alkan, Nov 06 2015

isok(m, n) = my(md = Mod(2, 2*n+1)^m); (md==1) || (md==-1);
A003558(n) = my(m=1); while(!isok(m,n) , m++); m; \\ Michel Marcus, May 06 2020

def A003558(n):
    m, k = 1, 2 % (c:=(r:=n<<1)+1)
    while not (k==1 or k==r):
        k = 2*k%c
        m += 1
    return m # Chai Wah Wu, Oct 09 2023

a(n) = log_2(A160657(n) + 2) - 1. - Nathaniel Johnston, May 22 2009
a(n-1) = card {cos((2^k)*Pi/(2*n-1)): k in N} for n >= 1 (see A216066, an essentially identical sequence, for more information). - Roman Witula, Sep 01 2012
a(n) <= n. - Charles R Greathouse IV, Sep 15 2012 [For n >= 1]
a(n) = min{k > 0 | q_k = q_0} where q_0 = 1 and q_k = |2*n+1 - 2*q_{k-1}| (cf. [Schick, p. 4]; q_k=1 for n=1; q_k=A010684(k) for n=2; q_k=A130794(k) for n=3; q_k=|A154870(k-1)| for n=4; q_k=|A135449(k)| for n=5.) - Jonathan Skowera, Jun 29 2013
2^(a(n)) == A332433(n) (mod (2*n+1)), and (2^(a(n)) - A332433(n))/(2*n+1) = A329593(n), for n >= 0. - Wolfdieter Lang, Apr 09 2020

More terms from Harry J. Smith, Feb 11 2005

Entry revised by N. J. A. Sloane, Aug 02 2006 and again Dec 10 2017

A048473 a(0)=1, a(n) = 3a(n-1) + 2; a(n) = 23^n - 1.

Original entry on oeis.org

1, 5, 17, 53, 161, 485, 1457, 4373, 13121, 39365, 118097, 354293, 1062881, 3188645, 9565937, 28697813, 86093441, 258280325, 774840977, 2324522933, 6973568801, 20920706405, 62762119217, 188286357653, 564859072961, 1694577218885, 5083731656657, 15251194969973, 45753584909921

Offset: 0

Clark Kimberling

The number of triangles (of all sizes, including holes) in Sierpiński's triangle after n inscriptions. - Lee Reeves, May 10 2004
The sequence is not only related to Sierpiński's triangle, but also to "Floret's cube" and the quaternion factor space Q X Q / {(1,1), (-1,-1)}. It can be written as a_n = ves((A+1)x)^n) as described at the Math Forum Discussions link. - Creighton Dement, Jul 28 2004
Relation to C(n) = Collatz function iteration using only odd steps: If we look for record subsequences where C(n) > n, this subsequence starts at 2^n - 1 and stops at the local maximum of 2*3^n - 1. Examples: [3,5], [7,11,17], [15,23,35,53], ..., [127,191,287,431,647,971,1457]. - Lambert Klasen, Mar 11 2005
Group the natural numbers so that the (2n-1)-th group sum is a multiple of the (2n)-th group containing one term. (1,2),(3),(4,5,6,7,8,9,10,11),(12),(13,14,15,16,17,18,19,...,38),(39),(40,41,...,118,119),(120), (121,122,123,...) ... a(n) = {the sum of the terms of (2n-1)-th group}/{the term of (2n)th group}. The first term of the odd numbered group is given by A003462. The only term of even numbered group is given by A029858. - Amarnath Murthy, Aug 01 2005
a(n)+1 = A008776(n); it appears that this gives the number of terms in the (n+1)-th "gap" of numbers missing in A171884. - M. F. Hasler, May 09 2013
Sum of n-th row of triangle of powers of 3: 1; 1 3 1; 1 3 9 3 1; 1 3 9 27 9 3 1; ... - Philippe Deléham, Feb 23 2014
For n >= 3, also the number of dominating sets in the n-helm graph. - Eric W. Weisstein, May 28 2017
The number of elements of length <= n in the free group on two generators. - Anton Mellit, Aug 10 2017
In general, a first order inhomogeneous recurrence of the form s(0) = a, s(n) = m*s(n-1) + k, n>0, will have a closed form of a*m^n + ((m^n-1)/(m-1))*k. - Gary Detlefs, Jun 07 2024

			a(0) = 1;
a(1) = 1 + 3 + 1 = 5;
a(2) = 1 + 3 + 9 + 3 + 1 = 17;
a(3) = 1 + 3 + 9 + 27 + 9 + 3 + 1 = 53; etc. - _Philippe Deléham_, Feb 23 2014

Theoni Pappas, Math Stuff, Wide World Publ/Tetra, San Carlos CA, page 15, 2002.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Creighton Dement, A paper on floretions and quaternions, some questions, The Math Forum.
Eric Weisstein's World of Mathematics, Dominating Set.
Eric Weisstein's World of Mathematics, Helm Graph.
Index entries for linear recurrences with constant coefficients, signature (4,-3).

a(n)=T(2,n), array T given by A048471.
Cf. A003462, A029858. A column of A119725.
Cf. A008776, A010684, A046055, A112468, A112739, A134347, A171884.

[2*3^n - 1: n in [0..30]]; // Vincenzo Librandi, Sep 23 2011

g:= ((1+x)/(1-3*x)/(1-x)): gser:=series(g, x=0, 43): seq(coeff(gser, x, n), n=0..30); # Zerinvary Lajos, Jan 11 2009; typo fixed by Marko Mihaily, Mar 07 2009

NestList[3 # + 2 &, 1, 30] (* Harvey P. Dale, Mar 06 2012 *)
LinearRecurrence[{4, -3}, {1, 5}, 30] (* Harvey P. Dale, Mar 06 2012 *)
Table[2 3^n - 1, {n, 20}] (* Eric W. Weisstein, May 28 2017 *)
2 3^Range[20] - 1 (* Eric W. Weisstein, May 28 2017 *)

first(m)=vector(m,n,n--;2*3^n - 1) \\ Anders Hellström, Dec 11 2015

n-th difference of a(n), a(n-1), ..., a(0) is 2^(n+1) for n=1, 2, 3, ...
a(0)=1, a(n) = a(n-1) + 3^n + 3^(n-1). - Lee Reeves, May 10 2004
a(n) = (3^n + 3^(n+1) - 2)/2. - Creighton Dement, Jul 31 2004
(1, 5, 17, 53, 161, ...) = Ternary (1, 12, 122, 1222, 12222, ...). - Gary W. Adamson, May 02 2005
Row sums of triangle A134347. Also, binomial transform of A046055: (1, 4, 8, 16, 32, 64, ...); and double binomial transform of A010684: (1, 3, 1, 3, 1, 3, ...). - Gary W. Adamson, Oct 21 2007
G.f.: (1+x)/((1-3*x)*(1-x)). - Zerinvary Lajos, Jan 11 2009; corrected by R. J. Mathar, Jan 21 2009
a(0)=1, a(1)=5, a(n) = 4*a(n-1) - 3*a(n-2). - Harvey P. Dale, Mar 06 2012
a(n) = Sum_{k=0..n} A112468(n,k)*4^k. - Philippe Deléham, Feb 23 2014
E.g.f.: exp(x)*(2*exp(2*x) - 1). - Elmo R. Oliveira, Mar 08 2025

Better description from Amarnath Murthy, May 27 2001

A007413 A squarefree (or Thue-Morse) ternary sequence: closed under 1->123, 2->13, 3->2. Start with 1.

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane

a(n)=2 if and only if n-1 is in A079523. - Benoit Cloitre, Mar 10 2003
Partial sums modulo 4 of the sequence 1, a(1), a(1), a(2), a(2), a(3), a(3), a(4), a(4), a(5), a(5), a(6), a(6), ... - Philippe Deléham, Mar 04 2004
To construct the sequence: start with 1 and concatenate 4 -1 = 3: 1, 3, then change the last term (2 -> 1, 3 ->2 ) gives 1, 2. Concatenate 1, 2 with 4 -1 = 3, 4 - 2 = 2: 1, 2, 3, 2 and change the last term: 1, 2, 3, 1. Concatenate 1, 2, 3, 1 with 4 - 1 = 3, 4 - 2 = 2, 4 - 3 = 1, 4 - 1 = 3: 1, 2, 3, 1, 3, 2, 1, 3 and change the last term: 1, 2, 3, 1, 3, 2, 1, 2 etc. - Philippe Deléham, Mar 04 2004
To construct the sequence: start with the Thue-Morse sequence A010060 = 0, 1, 1, 0, 1, 0, 0, 1, ... Then change 0 -> 1, 2, 3,  and 1 -> 3, 2, 1,  gives: 1, 2, 3, , 3, 2, 1, ,3, 2, 1, , 1, 2, 3, , 3, 2, 1, , ... and fill in the successive holes with the successive terms of the sequence itself. - _Philippe Deléham, Mar 04 2004
To construct the sequence: to insert the number 2 between the A003156(k)-th term and the (1 + A003156(k))-th term of the sequence 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, ... - Philippe Deléham, Mar 04 2004
Conjecture. The sequence is formed by the numbers of 1's between every pair of consecutive 2's in A076826. - Vladimir Shevelev, May 31 2009

			Here are the first 5 stages in the construction of this sequence, together with Mma code, taken from Keranen's article. His alphabet is a,b,c rather than 1,2,3.
productions = {"a" -> "abc ", "b" -> "ac ", "c" -> "b ", " " -> ""};
NestList[g, "a", 5] // TableForm
a
abc
abc ac b
abc ac b abc b ac
abc ac b abc b ac abc ac b ac abc b
abc ac b abc b ac abc ac b ac abc b abc ac b abc b ac abc b abc ac b ac

Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 18.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
A. Thue, Über unendliche Zeichenreihen, Norske Vid. Selsk. Skr. I. Mat. Nat. Kl. Christiania, No. 7 (1906), 1-22.

Paolo Xausa, Table of n, a(n) for n = 1..10000
Roger L. Bagula, Description of sequence as successive rows of a triangle
James D. Currie, Non-repetitive words: Ages and essences, Combinatorica 16.1 (1996): 19-40. See p. 20.
James D. Currie, Palindrome positions in ternary square-free words, Theoretical Computer Science, 396 (2008) 254-257.
F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
V. Keranen, New Abelian Square-Free DT0L-Languages over 4 Letters, Theoretical Computer Science, Volume 410, Issues 38-40, 6 September 2009, Pages 3893-3900.
S. Kitaev and T. Mansour, Counting the occurrences of generalized patterns in words generated by a morphism, arXiv:math/0210170 [math.CO], 2002.
Andrzej Tomski and Maciej Zakarczemny, A note on Browkin's and Cao's cancellation algorithm, Technical Transections 7/2018.

Cf. A001285, A010060.
First differences of A000069.
Equals A036580(n-1) + 1.
Cf. A115384, A159481, A079523, A000120.

Nest[ Flatten[ # /. {1 -> {1, 2, 3}, 2 -> {1, 3}, 3 -> {2}}] &, {1}, 7] (* Robert G. Wilson v, May 07 2005 *)
2 - Differences[ThueMorse[Range[0, 100]]] (* Paolo Xausa, Oct 25 2024 *)

{a(n) = if( n<1 || valuation(n, 2)%2, 2, 2 + (-1)^subst( Pol(binary(n)), x,1))};

def A007413(n): return 2-(n.bit_count()&1)+((n-1).bit_count()&1) # Chai Wah Wu, Mar 03 2023

a(n) modulo 2 = A035263(n). a(A036554(n)) = 2. a(A003159(n)) = 1 if n odd. a(A003159(n)) = 3 if n even. a(n) = A033485(n) mod 4. a(n) = 4 - A036585(n-1). - Philippe Deléham, Mar 04 2004
a(n) = 2 - A029883(n) = 3 - A036577(n). - Philippe Deléham, Mar 20 2004
For n>=1, we have: 1) a(A108269(n))=A010684(n-1); 2) a(A079523(n))=A010684(n-1); 3) a(A081706(2n))=A010684(n). - Vladimir Shevelev, Jun 22 2009

cp's OEIS Frontend

A203230 (n-1)-st elementary symmetric function of the first n terms of A010684.

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A112034 1 / (A010684(n)/A016116(n+5) - 1/A112033(n)).

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A141317 a(n) = A000244(n) - A010684(n).

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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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A006516 a(n) = 2^(n-1)*(2^n - 1), n >= 0.

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A020988 a(n) = (2/3)*(4^n-1).

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A048473 a(0)=1, a(n) = 3a(n-1) + 2; a(n) = 23^n - 1.