A029651 -id:A029651 - 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.
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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
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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

A145460 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where sequence a_k of column k is the exponential transform of C(n,k).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 0, 3, 5, 1, 0, 1, 10, 15, 1, 0, 0, 3, 41, 52, 1, 0, 0, 1, 9, 196, 203, 1, 0, 0, 0, 4, 40, 1057, 877, 1, 0, 0, 0, 1, 10, 210, 6322, 4140, 1, 0, 0, 0, 0, 5, 30, 1176, 41393, 21147, 1, 0, 0, 0, 0, 1, 15, 175, 7273, 293608, 115975, 1, 0, 0, 0, 0, 0, 6, 35, 1176, 49932, 2237921, 678570

Offset: 0

Alois P. Heinz, Oct 10 2008

A(n,k) is also the number of ways of placing n labeled balls into indistinguishable boxes, where in each filled box k balls are seen at the top. E.g. A(3,1)=10:
  |1.| |2.| |3.| |1|2| |1|2| |1|3| |1|3| |2|3| |2|3| |1|2|3|
  |23| |13| |12| |3|.| |.|3| |2|.| |.|2| |1|.| |.|1| |.|.|.|
  +--+ +--+ +--+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+-+

			Square array A(n,k) begins:
   1,   1,  1,  1,  1,  1,  ...
   1,   1,  0,  0,  0,  0,  ...
   2,   3,  1,  0,  0,  0,  ...
   5,  10,  3,  1,  0,  0,  ...
  15,  41,  9,  4,  1,  0,  ...
  52, 196, 40, 10,  5,  1,  ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened
N. J. A. Sloane, Transforms

Columns k=0-9 give: A000110, A000248, A133189, A145453, A145454, A145455, A145456, A145457, A145458, A145459.
A(2n,n) gives A029651.
Cf.: A007318, A143398, A292948.

exptr:= proc(p) local g; g:=
          proc(n) option remember; `if`(n=0, 1,
             add(binomial(n-1, j-1) *p(j) *g(n-j), j=1..n))
        end: end:
A:= (n,k)-> exptr(i-> binomial(i, k))(n):
seq(seq(A(n, d-n), n=0..d), d=0..12);

Exptr[p_] := Module[{g}, g[n_] := g[n] = If[n == 0, 1, Sum[Binomial[n-1, j-1] *p[j]*g[n-j], {j, 1, n}]]; g]; A[n_, k_] := Exptr[Function[i, Binomial[i, k]]][n]; Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Jan 15 2014, translated from Maple *)

def ncr(n, r)
  return 1 if r == 0
  (n - r + 1..n).inject(:*) / (1..r).inject(:*)
end
def A(k, n)
  ary = [1]
  (1..n).each{|i| ary << (0..i - 1).inject(0){|s, j| s + ncr(i - 1, j) * ncr(j + 1, k) * ary[i - 1 - j]}}
  ary
end
def A145460(n)
  a = []
  (0..n).each{|i| a << A(i, n - i)}
  ary = []
  (0..n).each{|i|
    (0..i).each{|j|
      ary << a[i - j][j]
    }
  }
  ary
end
p A145460(20) # Seiichi Manyama, Sep 28 2017

A(0,k) = 1 and A(n,k) = Sum_{i=0..n-1} binomial(n-1,i) * binomial(i+1,k) * A(n-1-i,k) for n > 0. - Seiichi Manyama, Sep 28 2017

A143398 Triangle T(n,k) = number of forests of labeled rooted trees of height at most 1, with n labels, where each root contains k labels, n>=0, 0<=k<=n.

Original entry on oeis.org

1, 0, 1, 0, 3, 1, 0, 10, 3, 1, 0, 41, 9, 4, 1, 0, 196, 40, 10, 5, 1, 0, 1057, 210, 30, 15, 6, 1, 0, 6322, 1176, 175, 35, 21, 7, 1, 0, 41393, 7273, 1176, 105, 56, 28, 8, 1, 0, 293608, 49932, 7084, 756, 126, 84, 36, 9, 1, 0, 2237921, 372060, 42120, 6510, 378, 210, 120, 45, 10, 1

Offset: 0

Alois P. Heinz, Aug 12 2008

			T(4,2) = 9: 3->{1,2}<-4, 2->{1,3}<-4, 2->{1,4}<-3, 1->{2,3}<-4, 1->{2,4}<-3, 1->{3,4}<-2, {1,2}{3,4}, {1,3}{2,4}, {1,4}{2,3}.
Triangle begins:
  1;
  0,   1;
  0,   3,  1;
  0,  10,  3,  1;
  0,  41,  9,  4,  1;
  0, 196, 40, 10,  5,  1;
  ...

Alois P. Heinz, Rows n = 0..140, flattened
Index entries for sequences related to rooted trees

Columns k=0-9 give: A000007, A000248, A133189, A145453, A145454, A145455, A145456, A145457, A145458, A145459.
Main diagonal gives A000012.
Row sums give A143406.
T(2n,n) gives A029651.
Cf. A000142, A145460, A351703.

u:= (n, k)-> `if`(k=0, 0, floor(n/k)):
T:= (n, k)-> n! *add(i^(n-k*i)/ ((n-k*i)! *i! *k!^i), i=0..u(n, k)):
seq(seq(T(n, k), k=0..n), n=0..12);

t[n_, n_] = 1; t[, 0] = 0; t[n, k_] := n!*Sum[i^(n-k*i)/((n-k*i)!*i!*k!^i), {i, 0, n/k}]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 19 2013 *)

u(n,k) = if(k==0, 0, n\k);
T(n, k) = n!*sum(j=0, u(n, k), j^(n-k*j)/(k!^j*j!*(n-k*j)!)); \\ Seiichi Manyama, May 13 2022

T(n,k) = n! * Sum_{i=0..u(n,k)} i^(n-k*i)/((n-k*i)!*i!*k!^i) with u(n,k) = 0 if k=0 and u(n,k) = floor(n/k) else.

A265612 a(n) = CatalanNumber(n+1)n(1+3n)/(6+2n).

Original entry on oeis.org

0, 1, 7, 35, 156, 660, 2717, 11011, 44200, 176358, 700910, 2778446, 10994920, 43459650, 171655785, 677688675, 2674776720, 10555815270, 41656918050, 164401379610, 648887951400, 2561511781920, 10113397410402, 39937416869070, 157743149913776, 623178050662300

Offset: 0

Peter Luschny, Dec 15 2015

nonn

This is row n=7 in the array A(n,k) = (rf(k+n-2,k-1)-(k-1)*(k-2)*rf(k+n-2, k-3))/ (k-1)! if n>=3 and A(n,0)=0, A(n,1)=1, A(n,2)=n; rf(n,k) denotes the rising factorial. See the cross-references for other values of n and the table in A264357.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000108, A007946, A097613, A051960, A029651, A051924, A129869, A220101, A264357, A265613.

A265612 := n -> 2*4^n*GAMMA(3/2+n)*n*(1+3*n)/(sqrt(Pi)*GAMMA(4+n)):
seq(simplify(A265612(n)), n=0..25);

Table[SeriesCoefficient[(5 x + (I (x - 1) (7 x - 2))/Sqrt[4 x - 1] - 2 - x^2)/(2 x^3), {x, 0, n}], {n, 0, 25}] (* or *)
Table[2*4^n Gamma[3/2 + n] n (1 + 3 n)/(Sqrt[Pi] Gamma[4 + n]), {n, 0, 25}] (* or *)
Table[CatalanNumber[n + 1] n ((1 + 3 n)/(6 + 2 n)), {n, 0, 25}] (* Michael De Vlieger, Dec 15 2015 *)

for(n=0,25, print1(round(2*4^n*gamma(3/2+n)*n*(1+3*n)/(sqrt(Pi)*gamma(4+n))), ", ")) \\ G. C. Greubel, Feb 06 2017

a = lambda n: catalan_number(n+1)*n*(1+3*n)/(6+2*n)
[a(n) for n in range(26)]

G.f.: (5*x+(I*(x-1)*(7*x-2))/sqrt(4*x-1)-2-x^2)/(2*x^3).
a(n) = 2*4^n*Gamma(3/2+n)*n*(1+3*n)/(sqrt(Pi)*Gamma(4+n)).
a(n) = (rf(5+n, n-1)-(n-1)*(n-2)*rf(5+n, n-3))/(n-1)! for n>=3, rf(n,k) the rising factorial.
a(n) = a(n-1)*(2*n*(1+3*n)*(1+2*n)/((n-1)*(3*n-2)*(3+n))) for n>=2.
a(n) ~  4^n*(6-(127/4)/n+(7995/64)/n^2-(223405/512)/n^3+(23501457/16384)/n^4-...) /sqrt(n*Pi).
a(n) = [x^n] x*(1 + x)/(1 - x)^(n+4). - Ilya Gutkovskiy, Oct 09 2017

A191662 a(n) = n! / A000034(n-1).

Original entry on oeis.org

1, 1, 6, 12, 120, 360, 5040, 20160, 362880, 1814400, 39916800, 239500800, 6227020800, 43589145600, 1307674368000, 10461394944000, 355687428096000, 3201186852864000, 121645100408832000, 1216451004088320000, 51090942171709440000, 562000363888803840000

Offset: 1

Paul Curtz, Jun 10 2011

The a(n) are the denominators in the formulas of the k-dimensional square pyramidal numbers:
A005408 = (2*n+1)/1                       = 1, 3,  5,   7,   9, ... (k=1)
A000290 = (n^2)/1                         = 1, 4,  9,  16,  25, ... (k=2)
A000330 = n*(n+1)*(2*n+1)/6               = 1, 5, 14,  30,  55, ... (k=3)
A002415 = (n^2)*(n^2-1)/12                = 1, 6, 20,  50, 105, ... (k=4)
A005585 = n*(n+1)*(n+2)*(n+3)*(2*n+3)/120 = 1, 7, 27,  77, 182, ... (k=5)
A040977 = (n^2)*(n^2-1)*(n^2-4)/360       = 1, 8, 35, 112, 294, ... (k=6)
A050486 (k=7), A053347 (k=8), A054333 (k=9), A054334 (k=10), A057788 (k=11).
The first superdiagonal of this array appears in A029651. - Paul Curtz, Jul 04 2011
The general formula for the k-dimensional square pyramidal numbers is (2*n+k)*binomial(n+k-1,k-1)/k, k >= 1, n >= 0, see A097207. - Johannes W. Meijer, Jun 22 2011

Cf. A000142, A009445, A002674, A064680.
Cf. A000290, A000330, A002415, A005408, A005585, A029651, A040977, A050486, A053347, A054333, A054334, A057788.
Cf. A052612.

A191662:= proc(n): n!/A000034(n-1) end: A000034 := proc(n) op((n mod 2)+1, [1, 2]) ; end proc: seq(A191662(n),n=1..17); # Johannes W. Meijer, Jun 22 2011

Array[If[EvenQ[#],#!/2,#!]&,20] (* Harvey P. Dale, Mar 14 2014 *)

a(2*n-1) = (2*n-1)!, a(2*n) = (2*n)!/2.
a(n+1) = A064680(n+1) * a(n).
From Amiram Eldar, Jul 06 2022: (Start)
Sum_{n>=1} 1/a(n) = sinh(1) + 2*cosh(1) - 2.
Sum_{n>=1} (-1)^(n+1)/a(n) = sinh(1) - 2*cosh(1) + 2. (End)
D-finite with recurrence: a(n) - (n-1)*n*a(n-2) = 0 for n >= 3 with a(1)=a(2)=1. - Georg Fischer, Nov 25 2022
a(n) = A052612(n)/2 for n >= 1. - Alois P. Heinz, Sep 05 2023

More terms from Harvey P. Dale, Mar 14 2014

A264357 Array A(r, n) of number of independent components of a symmetric traceless tensor of rank r and dimension n, written as triangle T(n, r) = A(r, n-r+2), n >= 1, r = 2..n+1.

Original entry on oeis.org

0, 2, 0, 5, 2, 0, 9, 7, 2, 0, 14, 16, 9, 2, 0, 20, 30, 25, 11, 2, 0, 27, 50, 55, 36, 13, 2, 0, 35, 77, 105, 91, 49, 15, 2, 0, 44, 112, 182, 196, 140, 64, 17, 2, 0, 54, 156, 294, 378, 336, 204, 81, 19, 2, 0

Offset: 1

Wolfdieter Lang, Dec 10 2015

A (totally) symmetric traceless tensor of rank r >= 2 and dimension n >= 1 is irreducible.
The array of the number of independent components of a rank r symmetric traceless tensor A(r, n), for r >= 2 and n >=1, is given by risefac(n,r)/r! - risefac(n,r-2)/(r-2)!, where the first term gives the number of independent components of a symmetric tensors of rank r (see a Dec 10 2015 comment under A135278) and the second term is the number of constraints from the tracelessness requirement. The tensor has to be traceless in each pair of indices.
The first rows of the array A, or the first columns (without the first r-2 zeros) of the triangle T are for r = 2..6: A000096, A005581, A005582, A005583, A005584.
Equals A115241 with the first column of positive integers removed. - Georg Fischer, Jul 26 2023

			The array A(r, n) starts:
   r\n 1 2  3   4   5    6    7     8     9    10 ...
   2:  0 2  5   9  14   20   27    35    44    54
   3:  0 2  7  16  30   50   77   112   156   210
   4:  0 2  9  25  55  105  182   294   450   660
   5:  0 2 11  36  91  196  378   672  1122  1782
   6:  0 2 13  49 140  336  714  1386  2508  4290
   7:  0 2 15  64 204  540 1254  2640  5148  9438
   8:  0 2 17  81 285  825 2079  4719  9867 19305
   9:  0 2 19 100 385 1210 3289  8008 17875 37180
  10:  0 2 21 121 506 1716 5005 13013 30888 68068
  ...
The triangle T(n, r) starts:
   n\r  2   3   4   5   6   7  8  9 10 11 ...
   1:   0
   2:   2   0
   3:   5   2   0
   4:   9   7   2   0
   5:  14  16   9   2   0
   6:  20  30  25  11   2   0
   7:  27  50  55  36  13   2  0
   8:  35  77 105  91  49  15  2  0
   9:  44 112 182 196 140  64 17  2  0
  10:  54 156 294 378 336 204 81 19  2  0
  ...
A(r, 1) = 0 , r >= 2, because a symmetric rank r tensor t of dimension one has one component t(1,1,...,1) (r 1's) and if the traces vanish then t vanishes.
A(3, 2) = 2 because a symmetric rank 3 tensor t with three indices taking values from 1 or 2 (n=2) has the four independent components t(1,1,1), t(1,1,2), t(1,2,2), t(2,2,2), and (invoking symmetry) the vanishing traces are Sum_{j=1..2} t(j,j,1) = 0 and Sum_{j=1..2} t(j,j,2) = 0. These are two constraints, which can be used to eliminate, say, t(1,1,1) and t(2,2,2), leaving 2 = A(3, 2) independent components, say, t(1,1,2) and t(1,2,2).
From _Peter Luschny_, Dec 14 2015: (Start)
The diagonals diag(n, k) start:
   k\n  0       1       2       3       4      5       6
   0:   0,      2,      9,     36,    140,   540,   2079, ... A007946
   1:   2,      7,     25,     91,    336,  1254,   4719, ... A097613
   2:   5,     16,     55,    196,    714,  2640,   9867, ... A051960
   3:   9,     30,    105,    378,   1386,  5148,  19305, ... A029651
   4:  14,     50,    182,    672,   2508,  9438,  35750, ... A051924
   5:  20,     77,    294,   1122,   4290, 16445,  63206, ... A129869
   6:  27,    112,    450,   1782,   7007, 27456, 107406, ... A220101
   7:  35,    156,    660,   2717,  11011, 44200, 176358, ... A265612
   8:  44,    210,    935,    4004, 16744, 68952, 281010, ... A265613
  A000096,A005581,A005582,A005583,A005584.
(End)

Cf. A000096, A005581, A005582, A005583, A005584, A007946, A024482, A029651, A051924, A051960, A097613, A115241, A129869, A135278, A220101, A265612, A265613.

A[r_, n_] := Pochhammer[n, r]/r! - Pochhammer[n, r-2]/(r-2)!;
T[n_, r_] := A[r, n-r+2];
Table[T[n, r], {n, 1, 10}, {r, 2, n+1}] (* Jean-François Alcover, Jun 28 2019 *)

A = lambda r, n: rising_factorial(n,r)/factorial(r) - rising_factorial(n,r-2)/factorial(r-2)
for r in (2..10): [A(r,n) for n in (1..10)] # Peter Luschny, Dec 13 2015

T(n, r) = A(r, n-r+2) with the array A(r, n) = risefac(n,r)/r! - risefac(n,r-2)/(r-2)! where the rising factorial risefac(n,k) = Product_{j=0..k-1} (n+j) and risefac(n,0) = 1.
From Peter Luschny, Dec 14 2015: (Start)
A(n+2, n+1) = A007946(n-1) = CatalanNumber(n)*3*n*(n+1)/(n+2) for n>=0.
A(n+2, n+2) = A024482(n+2) = A097613(n+2) = CatalanNumber(n+1)*(3*n+4)/2 for n>=0.
A(n+2, n+3) = A051960(n+1) = CatalanNumber(n+1)*(3*n+5) for n>=0.
A(n+2, n+4) = A029651(n+2) = CatalanNumber(n+1)*(6*n+9) for n>=0.
A(n+2, n+5) = A051924(n+3) = CatalanNumber(n+2)*(3*n+7) for n>=0.
A(n+2, n+6) = A129869(n+4) = CatalanNumber(n+2)*(3*n+8)*(2*n+5)/(n+4) for n>=0.
A(n+2, n+7) = A220101(n+4) = CatalanNumber(n+3)*(3*(n+3)^2)/(n+5) for n>=0.
A(n+2, n+8) = CatalanNumber(n+4)*(n+3)*(3*n+10)/(2*n+12) for n>=0.
Let for n>=0 and k>=0 diag(n,k) = A(k+2,n+k+1) and G(n,k) = 2^(k+2*n)*Gamma((3-(-1)^k+2*k+4*n)/4)/(sqrt(Pi)*Gamma(k+n+0^k)) then
diag(n,0) = G(n,0)*(n*3)/(n+2),
diag(n,1) = G(n,1)*(3*n+4)/((n+1)*(n+2)),
diag(n,2) = G(n,2)*(3*n+5)/(n+2),
diag(n,3) = G(n,3)*3,
diag(n,4) = G(n,4)*(3*n+7),
diag(n,5) = G(n,5)*(3*n+8),
diag(n,6) = G(n,6)*3*(3+n)^2,
diag(n,7) = G(n,7)*(3+n)*(10+3*n). (End)

A265613 a(n) = CatalanNumber(n+1)n(3n^2+5n+2)/((4+n)*(3+n)).

Original entry on oeis.org

0, 1, 8, 44, 210, 935, 4004, 16744, 68952, 281010, 1136960, 4576264, 18349630, 73370115, 292746300, 1166182800, 4639918800, 18443677230, 73261092240, 290845019400, 1154169552900, 4578702310182, 18159992594568, 72014135814704, 285542883894800, 1132125641947300

Offset: 0

Peter Luschny, Dec 15 2015

nonn

This is row n=8 in the array A(n,k) = (rf(k+n-2,k-1)-(k-1)*(k-2)*rf(k+n-2, k-3))/ (k-1)! if n>=3 and A(n,0)=0, A(n,1)=1, A(n,2)=n; rf(n,k) denotes the rising factorial. See the cross-references for other values of n and the table in A264357.

Robert Israel, Table of n, a(n) for n = 0..1630

Cf. A000108, A007946, A097613, A051960, A029651, A051924, A129869, A220101, A264357, A265612.

A265613 := n -> (4*4^n*n*(n+1)*(3*n+2)*GAMMA(n+3/2))/(sqrt(Pi)*GAMMA(n+5)):
seq(simplify(A265613(n)), n=0..25);

Table[SeriesCoefficient[I (14 x^2 + I Sqrt[4 x - 1] (4 x^2 - 7 x + 2) - 11 x + 2 (1 - x^3))/(2 x^4 Sqrt[4 x - 1]), {x, 0, n}], {n, 0, 25}]
(* or *)
Table[(4^(n + 1) n (n + 1) (3 n + 2) Gamma[n + 3/2])/(Sqrt[Pi] Gamma[n + 5]), {n, 0, 25}] (* or *)
Table[CatalanNumber(n+1) n (3 n^2 + 5 n + 2)/((4 + n) (3 + n)), {n, 0, 25}] (* Michael De Vlieger, Dec 15 2015 *)

a = lambda n: catalan_number(n+1)*n*(3*n^2+5*n+2)/((4+n)*(3+n))
[a(n) for n in range(26)]

G.f.: I*(14*x^2+I*sqrt(4*x-1)*(4*x^2-7*x+2)-11*x+2*(1-x^3))/(2*x^4*sqrt(4*x-1)).
a(n) = (4^(n+1)*n*(n+1)*(3*n+2)*Gamma(n+3/2))/(sqrt(Pi)*Gamma(n+5)).
a(n) = (rf(n+6, n-1)-(n-1)*(n-2)*rf(n+6, n-3))/(n-1)! for n>=3, rf(n,k) the rising factorial.
a(n) = a(n-1)*((2*(n+1))*(3*n+2)*(1+2*n)/((n-1)*(3*n-1)*(4+n))) for n>=2.
a(n) ~ 4^n*(12-(191/2)/n+(17595/32)/n^2-(705005/256)/n^3+(104705937/8192)/ n^4-...)/sqrt(n*Pi).
a(n) = [x^n] x*(1 + x)/(1 - x)^(n+5). - Ilya Gutkovskiy, Oct 09 2017

A146533 Catalan transform of A135092.

Original entry on oeis.org

1, 7, 21, 70, 245, 882, 3234, 12012, 45045, 170170, 646646, 2469012, 9464546, 36402100, 140408100, 542911320, 2103781365, 8167621770, 31762973550, 123708423300, 482462850870, 1883902560540, 7364346373020, 28817007546600, 112866612890850, 442437122532132, 1735714865318364

Offset: 0

Philippe Deléham, Oct 31 2008

nonn

Cf. A088218, A000984, A029651, A100320, A029609, A144706.

a[n_]:=(7*Binomial[2n,n]-5*KroneckerDelta[n,0])/2; Array[a,27,0] (* Stefano Spezia, Feb 14 2025 *)

a(n) = Sum_{k=0..n} A039599(n,k)*A010687(k) = Sum_{k=0..n} A106566(n,k)*A135092(k).
a(n) = (7*C(2n,n) - 5*0^n)/2.
From Stefano Spezia, Feb 14 2025: (Start)
G.f.: (7/sqrt(1 - 4*x) - 5)/2.
E.g.f.: (7*exp(2*x)*BesselI(0, 2*x) - 5)/2. (End)

a(23)-a(26) from Stefano Spezia, Feb 14 2025

A146534 a(n) = 4C(2n,n) - 30^n.

Original entry on oeis.org

1, 8, 24, 80, 280, 1008, 3696, 13728, 51480, 194480, 739024, 2821728, 10816624, 41602400, 160466400, 620470080, 2404321560, 9334424880, 36300541200, 141381055200, 551386115280, 2153031497760, 8416395854880, 32933722910400, 128990414732400, 505642425751008, 1983674131792416

Offset: 0

Philippe Deléham, Oct 31 2008

nonn

Cf. A000984, A010688, A029609, A029651, A039599, A088218, A100320, A144706, A146533, A240530.

a[n_]:=4*Binomial[2n,n]-3*KroneckerDelta[n,0]; Array[a,27,0] (* Stefano Spezia, Feb 14 2025 *)

a(n) = Sum_{k=0..n} A039599(n,k)*A010688(k).
From Stefano Spezia, Feb 14 2025: (Start)
G.f.: 4/sqrt(1 - 4*x) - 3.
E.g.f.: 4*exp(2*x)*BesselI(0, 2*x) - 3. (End)

a(22)-a(26) from Stefano Spezia, Feb 14 2025

A370487 Number of partitions of [3n] into 3 sets of size n having at least one set of consecutive numbers whose maximum (if n>0) is a multiple of n.

Original entry on oeis.org

1, 1, 7, 28, 103, 376, 1384, 5146, 19303, 72928, 277132, 1058146, 4056232, 15600898, 60174898, 232676278, 901620583, 3500409328, 13612702948, 53017895698, 206769793228, 807386811658, 3156148445578, 12350146091398, 48371405524648, 189615909656626, 743877799422154

Offset: 0

Alois P. Heinz, Feb 19 2024

nonn

			a(0) = 1: {}|{}|{}.
a(1) = 1: 1|2|3.
a(2) = 7: 12|34|56, 12|35|46, 12|36|45, 13|24|56, 14|23|56, 15|26|34, 16|25|34.

Alois P. Heinz, Table of n, a(n) for n = 0..1663
Wikipedia, Partition of a set

Row n=3 of A370363.

Cf. A000079, A001700, A003409, A029651, A131577, A255738.

a:= n-> `if`(n=0, 1, 3*binomial(2*n-1,n)-2):
seq(a(n), n=0..27);
# second Maple program:
a:= proc(n) option remember; `if`(n<3, 3*n*(n-1)+1, ((15*n^2
      -31*n+12)*a(n-1)-(3*n-2)*(4*n-6)*a(n-2))/((3*n-5)*n))
    end:
seq(a(n), n=0..27);

a(n) = 3*binomial(2*n-1,n) - 2 for n >= 1, a(0) = 1.
G.f.: ((3*x+1)*(4*x-1)+3*(1-x)*sqrt(1-4*x))/(2*(1-x)*(1-4*x)).
a(n) = A003409(n) - 2 = A029651(n) - 2 = 3*A001700(n-1) - 2 for n >= 1.
a(n) mod 2 = A255738(n+1).
a(n) mod 2 = 1 <=> n in { A131577 }.

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).

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A145460 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where sequence a_k of column k is the exponential transform of C(n,k).

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A143398 Triangle T(n,k) = number of forests of labeled rooted trees of height at most 1, with n labels, where each root contains k labels, n>=0, 0<=k<=n.

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A265612 a(n) = CatalanNumber(n+1)*n*(1+3*n)/(6+2*n).

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A191662 a(n) = n! / A000034(n-1).

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A264357 Array A(r, n) of number of independent components of a symmetric traceless tensor of rank r and dimension n, written as triangle T(n, r) = A(r, n-r+2), n >= 1, r = 2..n+1.

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A265613 a(n) = CatalanNumber(n+1)*n*(3*n^2+5*n+2)/((4+n)*(3+n)).

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A265612 a(n) = CatalanNumber(n+1)n(1+3n)/(6+2n).

A265613 a(n) = CatalanNumber(n+1)n(3n^2+5n+2)/((4+n)*(3+n)).

A146534 a(n) = 4C(2n,n) - 30^n.