A160562 -id:A160562 - 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).
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Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
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Paul Barry, On the Hurwitz Transform of Sequences, Journal of Integer Sequences, Vol. 15 (2012), #12.8.7.
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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.
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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

A071951 Triangle of Legendre-Stirling numbers of the second kind T(n,j), n >= 1, 1 <= j <= n, read by rows.

Original entry on oeis.org

1, 2, 1, 4, 8, 1, 8, 52, 20, 1, 16, 320, 292, 40, 1, 32, 1936, 3824, 1092, 70, 1, 64, 11648, 47824, 25664, 3192, 112, 1, 128, 69952, 585536, 561104, 121424, 7896, 168, 1, 256, 419840, 7096384, 11807616, 4203824, 453056, 17304, 240, 1, 512, 2519296, 85576448, 243248704, 137922336, 23232176, 1422080, 34584, 330, 1

Offset: 1

N. J. A. Sloane, Jun 16 2002

Removing a factor of 2^m from the m-th subdiagonal (the main diagonal corresponds to m = 0) gives the triangle A080248. - Peter Bala, Oct 15 2023

			The triangle begins:
n\j   1      2       3        4       5      6     7   8 9 ...
1:    1
2:    2      1
3:    4      8       1
4:    8     52      20        1
5:   16    320     292       40       1
6:   32   1936    3824     1092      70      1
7:   64  11648   47824    25664    3192    112     1
8:  128  69952  585536   561104  121424   7896   168   1
9:  256 419840 7096384 11807616 4203824 453056 17304 240 1
...
Row 10: 512 2519296 85576448 243248704 137922336 23232176 1422080 34584 330 1. Reformatted by _Wolfdieter Lang_, Apr 10 2013

Robert Israel, Table of n, a(n) for n = 1..10011 (first 141 rows, flattened)
G. E. Andrews, W. Gawronski and L. L. Littlejohn, The Legendre-Stirling Numbers, Discrete Mathematics, Volume 311, Issue 14, 28 July 2011, Pages 1255-1272.
M. W. Coffey, M. C. Lettington, On Fibonacci Polynomial Expressions for Sums of mth Powers, their implications for Faulhaber's Formula and some Theorems of Fermat, arXiv:1510.05402 [math.NT], 2015.
R. B. Corcino, K. J. M. Gonzales, M. J. C. Loquias and E. L. Tan, Dually weighted Stirling-type sequences, arXiv:1302.4694 [math.CO], 2013.
R. B. Corcino, K. J. M. Gonzales, M. J. C. Loquias and E. L. Tan, Dually weighted Stirling-type sequences, Europ. J. Combin., 43, 2015, 55-67.
José L. Cereceda, A refinement of Lang's formula for the sum of powers of integers, arXiv:2301.02141 [math.NT], 2023.
E. S. Egge, Legendre-Stirling permutations, Eur. J. Combin. 13 (2010) 1735-1750.
W. N. Everitt, L. L. Littlejohn and R. Wellman, Legendre polynomials, Legendre-Stirling numbers and the left-definite spectral analysis of the Legendre differential expression, J. Comput. Appl. Math. 148, 2002, 213-238.
H. Li, T. MacHenry, Permanents and Determinants, Weighted Isobaric Polynomials, and Integer Sequences, J. Int. Seq. 16 (2013) #13.3.5, Theorem 43.
L. L. Littlejohn and R. Wellman, A general left-definite theory for certain self-adjoint operators with applications to differential equations, J. Differential Equations, 181(2), 2002, 280-339.

Diagonals give A007290, A000079, A016129, A016309.
The column sequences are A000079 (powers of 2), A016129, A016309, A071952, A089274, A089277.
Cf. A080248, A089278, A089500, A160562.

[[(&+[(-1)^(r+j)*(2*r+1)*(r^2+r)^n/(Factorial(r+j+1)*Factorial(j-r)): r in [1..j]]): j in [1..n]]: n in [1..12]]; // G. C. Greubel, Mar 16 2019

N:= 20: # to get the first N rows, flattened
for j from 1 to N do S[j]:= series(x^j/mul(1-r*(r+1)*x, r=1..j), x, N+1) od:
seq(seq(coeff(S[j],x,i),j=1..i),i=1..N); # Robert Israel, Dec 03 2015
# alternative
A071951 := proc(n,k)
    option remember;
    if k =0 then
        if n = 0 then
            1;
        else
            0;
        end if;
    elif n = 0 then
        if k =0 then
            1;
        else
            0;
        end if;
    else
        procname(n-1,k-1)+k*(k+1)*procname(n-1,k) ;
    end if;
end proc: # R. J. Mathar, Jun 30 2018

Flatten[ Table[ Sum[(-1)^{r + j}(2r + 1)(r^2 + r)^n/((r + j + 1)!(j - r)!), {r, j}], {n, 10}, {j, n}]]

{T(n, k) = sum( i=0, k, (-1)^(i+k) * (2*i + 1) * (i*i + i)^n / (k-i)! / (k+i+1)! )} /* Michael Somos, Feb 25 2012 */

[[sum( (-1)^(r+j)*(2*r+1)*(r^2+r)^n/(factorial(r+j+1)*factorial(j-r)) for r in (1..j)) for j in (1..n)] for n in (1..12)] # G. C. Greubel, Mar 16 2019

T(n, j) = Sum_{r=1..j} (-1)^(r+j)*(2*r+1)*(r^2+r)^n/((r+j+1)!*(j-r)!).
G.f. for j-th column (without leading zeros): 1/Product_{r=1..j} (1 - r*(r+1)*x), j >= 1. From eq.(4.5) of the Everitt et al. paper.
A135921(n+1) = row sums. - Michael Somos, Feb 25 2012
Sum_{n=j..m} binomial(m,n)*T(n,j)*4^(n-j) = A160562(m,j) for 1 <= j <= m. - Werner Schulte, Dec 03 2015

A381512 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where A(n,k) = (2n+k)!/k! [x^(2*n+k)] sinh(x)^k.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 4, 1, 0, 1, 10, 16, 1, 0, 1, 20, 91, 64, 1, 0, 1, 35, 336, 820, 256, 1, 0, 1, 56, 966, 5440, 7381, 1024, 1, 0, 1, 84, 2352, 24970, 87296, 66430, 4096, 1, 0, 1, 120, 5082, 90112, 631631, 1397760, 597871, 16384, 1, 0, 1, 165, 10032, 273988, 3331328, 15857205, 22368256, 5380840, 65536, 1, 0

Offset: 0

Seiichi Manyama, May 11 2025

			Square array begins:
  1, 1,    1,     1,       1,        1, ...
  0, 1,    4,    10,      20,       35, ...
  0, 1,   16,    91,     336,      966, ...
  0, 1,   64,   820,    5440,    24970, ...
  0, 1,  256,  7381,   87296,   631631, ...
  0, 1, 1024, 66430, 1397760, 15857205, ...

Columns k=0..7 give A000007, A000012, A000302, A002452(n+1), A166984, A002453, 4^n * A002451(n), A381513.
Main diagonal gives A383837.
Cf. A136630, A160562, A286899.

a(n, k) = (2*n+k)!/k!*polcoef(sinh(x+x*O(x^(2*n+k)))^k, 2*n+k);

G.f. of column k: 1/Product_{j=0..floor(k/2)} (1 - (k-2*j)^2*x).
A(n,k) = k^2 * A(n-1,k) + A(n,k-2) for k > 1.
A(n,k) = (1/(2^k*k!)) * Sum_{j=0..k} (-1)^j * (k-2*j)^(2*n+k) * binomial(k,j).

A160563 Table of the number of (n,k)-Riordan complexes, read by rows.

Original entry on oeis.org

1, 1, 1, 9, 10, 1, 225, 259, 35, 1, 11025, 12916, 1974, 84, 1, 893025, 1057221, 172810, 8778, 165, 1, 108056025, 128816766, 21967231, 1234948, 28743, 286, 1, 18261468225, 21878089479, 3841278805, 230673443, 6092515, 77077, 455, 1, 4108830350625, 4940831601000

Offset: 0

Jonathan Vos Post, May 19 2009

From Table 4, right-hand side, of Gelineau and Zeng.

Essentially a row-reversal of A008956. - R. J. Mathar, May 20 2009

			Triangle starts:
  [0]         1;
  [1]         1,          1;
  [2]         9,         10,        1;
  [3]       225,        259,       35,        1;
  [4]     11025,      12916,     1974,       84,     1;
  [5]    893025,    1057221,   172810,     8778,   165,    1;
  [6] 108056025,  128816766, 21967231,  1234948, 28743,  286, 1;
.
For row 3: F(x) := 1/cos(x). Then 225*F(x) + 259*(d/dx)^2(F(x)) + 35*(d/dx)^4(F(x)) + (d/dx)^6(F(x)) = 720*(1/cos(x))^7, where F^(r) denotes the r-th derivative of F(x).

Yoann Gelineau and Jiang Zeng, Combinatorial Interpretations of the Jacobi-Stirling Numbers, arXiv:0905.2899 [math.CO], May 2009.
W. Zhang, Some identities involving the Euler and the central factorial numbers, The Fibonacci Quarterly, Vol. 36, Number 2, May 1998.

Cf. A001819, A008275, A008277, A160562, A091885, A182867.

t := proc(n,k) option remember ; expand(x*mul(x+n/2-i,i=1..n-1)) ; coeftayl(%,x=0,k) ; end:
v := proc(n,k) option remember ; 4^(n-k)*t(2*n+1,2*k+1) ; end:
A160563 := proc(n,k) abs(v(n,k)) ; end: for n from 0 to 10 do for k from 0 to n do printf("%d,",A160563(n,k)) ; od: od: # R. J. Mathar, May 20 2009
# Using a bivariate generating function (albeit generating signed terms):
gf := (t + sqrt(1 + t^2))^x: ser := series(gf, t, 20):
ct := n -> coeff(ser, t, n): T := (n, k) -> n!*coeff(ct(n), x, k):
OddPart := (T, len) -> local n, k;
seq(print(seq(T(n, k), k = 1..n, 2)), n = 1..2*len, 2):
OddPart(T, 6);  # Peter Luschny, Mar 03 2024

t[, 0] = 1; t[n, n_] := t[n, n] = ((2*n - 1)!!)^2; t[n_, k_] := t[n, k] = (2*n - 1)^2*t[n - 1, k - 1] + t[n - 1, k];
T[n_, k_] := t[n, n - k];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 28 2017, after R. J. Mathar's comment *)

a(n,k) = |v(n,k)| where v(n,k) = v(n-1,k-1) - (2n-1)^2*v(n-1,k); eq (4.2).
Let F(x) = 1/cos(x). Then (2*n)!*(1/cos(x))^(2*n+1) = Sum_{k=0..n} T(n,k)*F^(2*k)(x), where F^(r) denotes the r-th derivative of F(x) (Zhang 1998). An example is given below. - Peter Bala, Feb 06 2012
Given a (0, 0)-based triangle U we call the triangle [U(n, k), k=1..n step 2, n=1..len step 2] the 'odd subtriangle' of U. This triangle is the odd subtriangle of U(n, k) = n! * [x^(n-k)] [t^n] (t + sqrt(1 + t^2))^x, albeit with signed terms. See A182867 for the even subtriangle. - Peter Luschny, Mar 03 2024

Extended by R. J. Mathar, May 20 2009

A366834 Square array read by descending antidiagonals: (-1)^nT(n,k)/n! is the coefficient of x^(2n+1) in the Taylor expansion of the k-th iteration of sin(x).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 12, 1, 0, 1, 4, 33, 128, 1, 0, 1, 5, 64, 731, 1872, 1, 0, 1, 6, 105, 2160, 25857, 37600, 1, 0, 1, 7, 156, 4765, 121600, 1311379, 990784, 1, 0, 1, 8, 217, 8896, 368145, 10138880, 89060065, 32333824, 1, 0, 1, 9, 288, 14903, 873936, 42807605, 1162426880, 7778778091, 1272660224, 1, 0

Offset: 0

Jianing Song, Oct 25 2023

T(n,k)/n! is the coefficient of x^(2*n+1) in the Taylor expansion of the k-th iteration of sinh(x). This is most easily seen from the relation -i*sin(...sin(sin(sin(i*x)))...) = -i*sin(...sin(sin(i*sinh(x)))...) = -i*sin(...sin(i*sinh(sinh(x)))...) = ... = sinh(...sinh(sinh(sinh(x)))...).

			G.f.s of the first few rows:
n = 0: 1/(1-x);
n = 1: x/(1-x)^2;
n = 2: x/(1-x)^2 + 10*x^2/(1-x)^3;
n = 3: x/(1-x)^2 + 126*x^2/(1-x)^3 + 350*x^3/(1-x)^4:
n = 4: x/(1-x)^2 + 1870*x^2/(1-x)^3 + 20244*x^3/(1-x)^4 + 29400*x^4/(1-x)^5;
n = 5: x/(1-x)^2 + 37598*x^2/(1-x)^3 + 1198582*x^3/(1-x)^4 + 5118960*x^4/(1-x)^5 + 4851000*x^5/(1-x)^6.
The explicit formulas for the first few rows:
T(0,k) = binomial(k,0) = 1 for k = 0, 0 for k > 0;
T(1,k) = binomial(k,1) = k;
T(2,k) = binomial(k,1) + 10*binomial(k,2) = 5*k^2 - 4*k;
T(3,k) = binomial(k,1) + 126*binomial(k,2) + 350*binomial(k,3) = (175/3)*k^3 - 112*k^2 + (164/3)*k;
T(4,k) = binomial(k,1) + 1870*binomial(k,2) + 20244*binomial(k,3) + 29400*binomial(k,4) = 1225*k^4 - 3976*k^3 + 4288*k^2 - 1536*k;
T(5,k) = binomial(k,1) + 37598*binomial(k,2) + 1198582*binomial(k,3) + 5118960*binomial(k,4) + 4851000*binomial(k,5) = 40425*k^5 - 190960*k^4 + (1004696/3)*k^3 - 255552*k^2 + (213568/3)*k.
Table of terms:
Row 0: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Row 1: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Row 2: 0, 1, 12, 33, 64, 105, 156, 217, 288, 369, 460
Row 3: 0, 1, 128, 731, 2160, 4765, 8896, 14903, 23136, 33945, 47680
Row 4: 0, 1, 1872, 25857, 121600, 368145, 873936, 1776817, 3244032, 5472225, 8687440
Row 5: 0, 1, 37600, 1311379, 10138880, 42807605, 130426016, 323774535, 698156544, 1358249385, 2442955360
Row 6: 0, 1, 990784, 89060065, 1162426880, 6937805945, 27344158016, 83303826249, 212901058560, 478937915985, 977877567040
Row 7: 0, 1, 32333824, 7778778091, 174394695680, 1487589904205, 7634965431296, 28668866786679, 87104014381056, 227079171721785, 527214112015360
Row 8: 0, 1, 1272660224, 849264442881, 33044097597440, 406373544070945, 2731282112246016, 12688038285458529, 45949019179646976, 139088689115885505, 367745105831952640
Row 9: 0, 1, 59527313920, 113234181108643, 7701145601933312, 137461463840219237, 1215573962763120128, 7008667055272520967, 30322784763588771840, 106757902382656031049, 321859857651846029824
Row 10: 0, 1, 3252626013184, 18073465545032353, 2162675327569362944, 56311245536706922889, 657730167421332884480, 4719958813316934631353, 24445625744089126797312, 100254353682662263787313, 345053755346367061654528
Demonstration of terms:
sin(sin(x)) = x - 2*x^3/3! + 12*x^5/5! - 128*x^7/7! + 1872*x^9/9! - 37600*x^11/11! + ...;
sin(sin(sin(x))) = x - 3*x^3/3! + 33*x^5/5! - 731*x^7/7! + 25857*x^9/9! - 1311379*x^11/11! + ...;
sin(sin(sin(sin(x)))) = x - 4*x^3/3! + 64*x^5/5! - 2160*x^7/7! + 121600*x^9/9! - 10138880*x^11/11! + ....

Jianing Song, Table of n, a(n) for n = 0..5150 (diagonals 0..100)
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Jianing Song, Formula for A366834.

Cf. A051624 (row n=2), A366827 (row n=3), A003712 (column k=2 signed), A003715 (column k=3 signed).

Cf. A000447, A285018, A285019.

A160562(n,k) = (-1)^k / (2*k+1)! * sum(j=0, k, (-1)^j * binomial(2*k+1, k-j) * (2*j+1)^(2*n+1)) / 2^(2*k)
coeff_of_n_gfs(n) = my(M=matid(1)); for(k=1, n, M = matconcat([concat(M, matrix(k, 1)); concat(0, matrix(1, k, i, j, A160562(k, j-1))*M)])); M \\ The lower triangle matrix (C(i,j))_{0<=j<=i<=n}
T_mat(n,k) = coeff_of_n_gfs(n)*matrix(n+1, k+1, i, j, binomial(j-1,i-1)) \\ gives T(i,j) for i=0..n and j=0..k

T(0,0) = 1, T(n,0) = 0 for n >= 1; T(n,k) = Sum_{i=0..n} A160562(n,i)*T(i,k-1) for k >= 1, where A160562(n,k) = ((-1)^(n-k)*(2*n+1)!/(2*k+1)!) * [x^(2*n+1)]sin(x)^(2*k+1). Note that this is not very efficient to calculate the terms.
A more efficient way would be to calculate the g.f. for each row: the g.f. of the n-th row is C(n,0)/(1-x) + C(n,1)*x/(1-x)^2 + ... + C(n,n)*x^n/(1-x)^(n+1), where C(0,0) = 1, C(n,0) = 0 for n >= 1; C(n,k) = A160562(n,k-1)*C(k-1,k-1) + ... + A160562(n,n-1)*C(n-1,k-1) for n >= k >= 1, so we have T(n,k) = C(n,0)*binomial(k,0) + C(n,1)*binomial(k,1) + ... + C(n,n)*binomial(k,n). See my pdf in the link section for the proof.
From the formula above we see that the n-th row is a degree-n polynomial of k with leading coefficient C(n,n)/n!. We have C(n,n) = A160562(n,n-1)*C(n-1,n-1) = A000447(n)*C(n-1,n-1) for n >= 1, so it can be shown that C(n,n)/n! = n! * A285018(n)/A285019(n).

A370705 Triangle read by rows: T(n, k) = numerator(CF(n, k)) where CF(n, k) = n! * [x^k] [t^n] (t/2 + sqrt(1 + (t/2)^2))^(2*x).

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, -1, 0, 1, 0, 0, -1, 0, 1, 0, 9, 0, -5, 0, 1, 0, 0, 4, 0, -5, 0, 1, 0, -225, 0, 259, 0, -35, 0, 1, 0, 0, -36, 0, 49, 0, -14, 0, 1, 0, 11025, 0, -3229, 0, 987, 0, -21, 0, 1, 0, 0, 576, 0, -820, 0, 273, 0, -30, 0, 1

Offset: 0

Peter Luschny, Mar 02 2024

The rational triangle R(n, k) contains the central factorial numbers. The central factorial of the first kind is the even subtriangle of R, while the central factorial of the second kind is the odd subtriangle. Since the terms of the even subtriangle can be seen as integers, the rational nature of these numbers is generally disregarded. The denominators of the central factorial of second kind are powers of 4; therefore, they are often studied as integers in the form 4^(n-k)*R(n, k). We will refer to the subtriangles by CF1(n, k) and CF2(n, k).
We recall that if T(n, k) is a number triangle (0 <= k <= n) then
   Teven(n, k) = [T(n, k), k=0..n step 2), n=0..len step 2]
   is the even subtriangle of T, and the odd subtriangle of T is
   Todd(n, k) = [T(n, k), k=1..n step 2), n=1..len step 2], where
   'k=a..o step s' denotes the subrange [a, a+s, a+2*s, ..., a+s*floor((o-a)/s)].
The central factorial numbers have their origins in approximation theory and numerical mathematics. They were undoubtedly used for a long time when J. F. Steffensen used them to construct quadrature formulas and presented them in 1924 at the 7th ICM. Four decades later, Carlitz and Riordan adopted the idea for use in combinatorics. While Steffensen originally referred to the numbers as "central differences of nothing," the second part of the name was later omitted.

			Triangle starts:
[0] 1;
[1] 0,     1;
[2] 0,     0,   1;
[3] 0,    -1,   0,     1;
[4] 0,     0,  -1,     0,  1;
[5] 0,     9,   0,    -5,  0,   1;
[6] 0,     0,   4,     0, -5,   0,   1;
[7] 0,  -225,   0,   259,  0, -35,   0,   1;
[8] 0,     0, -36,     0, 49,   0, -14,   0, 1;
[9] 0, 11025,   0, -3229,  0, 987,   0, -21, 0, 1;

Johan Frederik Steffensen, On a class of quadrature formulae. Proceedings of the International Mathematical Congress Toronto 1924, Vol 2, pp. 837-844.

P. L. Butzer, M. Schmidt, E. L. Stark and L. Vogt. Central factorial numbers; their main properties and some applications, Num. Funct. Anal. Optim., 10 (1989) 419-488.
Leonard Carlitz and John Riordan, The Divided Central Differences of Zero, Canadian Journal of Mathematics, Volume 15, 1963, pp. 94-100.
P. A. MacMahon, Divisors of numbers and their continuations in the theory of partitions, Proc. London Math. Soc., (2) 19 (1919), 75-113; Coll. Papers II, pp. 303-341.
Johan Frederik Steffensen, On the Definition of the Central Factorial, Journal of the Institute of Actuaries, Volume 64, Issue 2, July 1933, pp. 165-168.

See the discussion by Sloane in A008955 of Riordan's notation. In particular, the notation 'T' below does not refer to the present triangle.
Central factorials (rational, general case): (this triangle) / A370703;
t(2n, 2k) (first kind, 'even case') A204579; (signed, T(n, 0) missing)
|t(2n, 2k)|                         A269944; (unsigned, T(n, 0) = 0^n)
|t(2n, 2n-2k)|                      A008955;
|t(2n+1, 2n+1-2k)|*4^k              A008956;
T(2n, 2k) (second kind, 'odd case') A269945, A036969;
T(2n+1, 2k+1)*4^(n-k)               A160562.

gf := (t/2 + sqrt(1 + (t/2)^2))^(2*x): ser := series(gf, t, 20):
ct := n -> n!*coeff(ser, t, n):
T := (n, k) -> numer(coeff(ct(n), x, k)):
seq(seq(T(n, k), k = 0..n), n = 0..10);
# Filtering the central factorials of the first resp. second kind:
CF1 := (T,len) -> local n,k; seq(print(seq(T(n,k), k=0..n, 2)), n = 0..len, 2);
CF2 := (T,len) -> local n,k; seq(print(seq(T(n,k), k=1..n, 2)), n = 1..len, 2);

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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A071951 Triangle of Legendre-Stirling numbers of the second kind T(n,j), n >= 1, 1 <= j <= n, read by rows.

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A381512 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where A(n,k) = (2*n+k)!/k! * [x^(2*n+k)] sinh(x)^k.

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A160563 Table of the number of (n,k)-Riordan complexes, read by rows.

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A366834 Square array read by descending antidiagonals: (-1)^n*T(n,k)/n! is the coefficient of x^(2*n+1) in the Taylor expansion of the k-th iteration of sin(x).

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A370705 Triangle read by rows: T(n, k) = numerator(CF(n, k)) where CF(n, k) = n! * [x^k] [t^n] (t/2 + sqrt(1 + (t/2)^2))^(2*x).

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A381512 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where A(n,k) = (2n+k)!/k! [x^(2*n+k)] sinh(x)^k.

A366834 Square array read by descending antidiagonals: (-1)^nT(n,k)/n! is the coefficient of x^(2n+1) in the Taylor expansion of the k-th iteration of sin(x).