A124574 -id:A124574 - 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

A053121 Catalan triangle (with 0's) read by rows.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 2, 0, 1, 2, 0, 3, 0, 1, 0, 5, 0, 4, 0, 1, 5, 0, 9, 0, 5, 0, 1, 0, 14, 0, 14, 0, 6, 0, 1, 14, 0, 28, 0, 20, 0, 7, 0, 1, 0, 42, 0, 48, 0, 27, 0, 8, 0, 1, 42, 0, 90, 0, 75, 0, 35, 0, 9, 0, 1, 0, 132, 0, 165, 0, 110, 0, 44, 0, 10, 0, 1, 132, 0, 297, 0, 275, 0, 154, 0, 54, 0, 11, 0

Offset: 0

Wolfdieter Lang

Inverse lower triangular matrix of A049310(n,m) (coefficients of Chebyshev's S polynomials).
Walks with a wall: triangle of number of n-step walks from (0,0) to (n,m) where each step goes from (a,b) to (a+1,b+1) or (a+1,b-1) and the path stays in the nonnegative quadrant.
T(n,m) is the number of left factors of Dyck paths of length n ending at height m. Example: T(4,2)=3 because we have UDUU, UUDU, and UUUD, where U=(1,1) and D=(1,-1). (This is basically a different formulation of the previous - walks with a wall - property.) - Emeric Deutsch, Jun 16 2011
"The Catalan triangle is formed in the same manner as Pascal's triangle, except that no number may appear on the left of the vertical bar." [Conway and Smith]
G.f. for row polynomials p(n,x) := Sum_{m=0..n} (a(n,m)*x^m): c(z^2)/(1-x*z*c(z^2)). Row sums (x=1): A001405 (central binomial).
In the language of the Shapiro et al. reference such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group. The g.f. Ginv(x) of the m=0 column of the inverse of a given Bell-matrix (here A049310) is obtained from its g.f. of the m=0 column (here G(x)=1/(1+x^2)) by Ginv(x)=(f^{(-1)}(x))/x, with f(x) := x*G(x) and f^{(-1)}is the compositional inverse function of f (here one finds, with Ginv(0)=1, c(x^2)). See the Shapiro et al. reference.
Number of involutions of {1,2,...,n} that avoid the patterns 132 and have exactly k fixed points. Example: T(4,2)=3 because we have 2134, 4231 and 3214. Number of involutions of {1,2,...,n} that avoid the patterns 321 and have exactly k fixed points. Example: T(4,2)=3 because we have 1243, 1324 and 2134. Number of involutions of {1,2,...,n} that avoid the patterns 213 and have exactly k fixed points. Example: T(4,2)=3 because we have 1243, 1432 and 4231. - Emeric Deutsch, Oct 12 2006
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. - Philippe Deléham, Sep 25 2007
Riordan array (c(x^2),xc(x^2)), where c(x) is the g.f. of Catalan numbers A000108. - Philippe Deléham, Nov 25 2007
A053121^2 = triangle A145973. Convolved with A001405 = triangle A153585. - Gary W. Adamson, Dec 28 2008
By columns without the zeros, n-th row = A000108 convolved with itself n times; equivalent to A = (1 + x + 2x^2 + 5x^3 + 14x^4 + ...), then n-th row = coefficients of A^(n+1). - Gary W. Adamson, May 13 2009
Triangle read by rows,product of A130595 and A064189 considered as infinite lower triangular arrays; A053121 = A130595*A064189 = B^(-1)*A097609*B where B = A007318. - Philippe Deléham, Dec 07 2009
From Mark Dols, Aug 17 2010: (Start)
As an upper right triangle, rows represent powers of 5-sqrt(24):
  5 - sqrt(24)^1 = 0.101020514...
  5 - sqrt(24)^2 = 0.010205144...
  5 - sqrt(24)^3 = 0.001030928...
(Divided by sqrt(96) these powers give a decimal representation of the columns of A007318, with 1/sqrt(96) being the middle column.) (End)
T(n,k) is the number of dispersed Dyck paths of length n (i.e., Motzkin paths of length n with no (1,0) steps at positive heights) having k (1,0)-steps. Example: T(5,3)=4 because, denoting U=(1,1), D=(1,-1), H=(1,0), we have HHHUD, HHUDH, HUDHH, and UDHHH. - Emeric Deutsch, Jun 01 2011
Let S(N,x) denote the N-th Chebyshev S-polynomial in x (see A049310, cf. [W. Lang]). Then x^n = sum_{k=0..n} T(n,k)*S(k,x). - L. Edson Jeffery, Sep 06 2012
This triangle a(n,m) appears also in the (unreduced) formula for the powers rho(N)^n for the algebraic number over the rationals rho(N) = 2*cos(Pi/N) = R(N, 2), the smallest diagonal/side ratio R in the regular N-gon:
  rho(N)^n = sum(a(n,m)*R(N,m+1),m=0..n), n>=0, identical in N >= 1. R(N,j) = S(j-1, x=rho(N)) (Chebyshev S (A049310)). See a comment on this under A039599 (even powers) and A039598 (odd powers). Proof: see the Sep 06 2012 comment by L. Edson Jeffery, which follows from T(n,k) (called here a(n,k)) being the inverse of the Riordan triangle A049310. - Wolfdieter Lang, Sep 21 2013
The so-called A-sequence for this Riordan triangle of the Bell type (c(x^2), x*c(x^2)) (see comments above) is A(x) = 1 + x^2. This proves the recurrence given in the formula section by Henry Bottomley for a(n, m) = a(n-1, m-1) + a(n-1, m+1) for n>=1 and m>=1, with inputs. The Z-sequence for this Riordan triangle is Z(x) = x which proves the recurrence a(n,0) = a(n-1,1), n>=1, a(0,0) = 1. For A- and Z-sequences for Riordan triangles see the W. Lang link under A006232. - Wolfdieter Lang, Sep 22 2013
Rows of the triangle describe decompositions of tensor powers of the standard (2-dimensional) representation of the Lie algebra sl(2) into irreducibles. Thus a(n,m) is the multiplicity of the m-th ((m+1)-dimensional) irreducible representation in the n-th tensor power of the standard one. - Mamuka Jibladze, May 26 2015
The Riordan row polynomials p(n, x) belong to the Boas-Buck class (see a comment and references in A046521), hence they satisfy the Boas-Buck identity: (E_x - n*1)*p(n, x) = (E_x + 1)*Sum_{j=0..n-1} (1/2)*(1 - (-1)^j)*binomial(j+1, (j+1)/2)*p(n-1-j, x), for n >= 0, where E_x = x*d/dx (Euler operator). For the triangle a(n, m) this entails a recurrence for the sequence of column m, given in the formula section. - Wolfdieter Lang, Aug 11 2017
From Roger Ford, Jan 22 2018: (Start)
For row n, the nonzero values represent the odd components (loops) formed by n+1 nonintersecting arches above and below the x-axis with the following constraints:  The top has floor((n+3)/2) starting arches at position 1 and the next consecutive odd positions. All other starting top arches are in even positions. The bottom arches are a rainbow of arches.  If the component=1 then the arch configuration is a semimeander solution.
Examples: For row 3 {0, 2, 0, 1} there are 3 arch configurations: 2 arch configurations have a component=1; 1 has a component=3.  c=components, U=top arch starting in odd position, u=top arch starting in an even position, d=ending top arch:
.
  top UuUdUddd c=3   top UdUuUddd c=1     top UdUdUudd c=1
       /\                    /\
      //\\                  /  \
     //  \\                / /\ \                    /\
    //    \\              / /  \ \                  /  \
   ///\  /\\\        /\  / / /\ \ \        /\  /\  / /\ \
   \\\ \/ ///        \ \ \ \/ / / /        \ \ \ \/ / / /
    \\\  ///          \ \ \  / / /          \ \ \  / / /
     \\\///            \ \ \/ / /            \ \ \/ / /
      \\//              \ \  / /              \ \  / /
       \/                \ \/ /                \ \/ /
                          \  /                  \  /
                           \/                    \/
For row 4 {2, 0, 3, 0, 1} there are 6 arch configurations: 2 have a component=1; 3 have a component=3: 1 has a component=1. (End)

			Triangle a(n,m) begins:
  n\m  0   1   2   3   4   5   6  7  8  9 10 ...
  0:   1
  1:   0   1
  2:   1   0   1
  3:   0   2   0   1
  4:   2   0   3   0   1
  5:   0   5   0   4   0   1
  6:   5   0   9   0   5   0   1
  7:   0  14   0  14   0   6   0  1
  8:  14   0  28   0  20   0   7  0  1
  9:   0  42   0  48   0  27   0  8  0  1
  10: 42   0  90   0  75   0  35  0  9  0  1
  ... (Reformatted by _Wolfdieter Lang_, Sep 20 2013)
E.g., the fourth row corresponds to the polynomial p(3,x)= 2*x + x^3.
From _Paul Barry_, May 29 2009: (Start)
Production matrix is
  0, 1,
  1, 0, 1,
  0, 1, 0, 1,
  0, 0, 1, 0, 1,
  0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 0, 0, 1, 0, 1 (End)
Boas-Buck recurrence for column k = 2, n = 6: a(6, 2) = (3/4)*(0 + 2*a(4 ,2) + 0 + 6*a(2, 2)) = (3/4)*(2*3 + 6) = 9. - _Wolfdieter Lang_, Aug 11 2017

J. H. Conway and D. A. Smith, On Quaternions and Octonions, A K Peters, Ltd., Natick, MA, 2003. See p. 60. MR1957212 (2004a:17002)
A. Nkwanta, Lattice paths and RNA secondary structures, in African Americans in Mathematics, ed. N. Dean, Amer. Math. Soc., 1997, pp. 137-147.

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
I. Bajunaid et al., Function series, Catalan numbers and random walks on trees, Amer. Math. Monthly 112 (2005), 765-785.
C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps
Paul Barry, Riordan Arrays, Orthogonal Polynomials as Moments, and Hankel Transforms, J. Int. Seq. 14 (2011) # 11.2.2, example 3.
Paul Barry, On the inversion of Riordan arrays, arXiv:2101.06713 [math.CO], 2021.
Paul Barry and A. Hennessy, Meixner-Type Results for Riordan Arrays and Associated Integer Sequences, J. Int. Seq. 13 (2010) # 10.9.4, example 3.
Xiang-Ke Chang, X.-B. Hu, H. Lei, and Y.-N. Yeh, Combinatorial proofs of addition formulas, The Electronic Journal of Combinatorics, 23(1) (2016), #P1.8.
J. Cigler, Some q-analogues of Fibonacci, Lucas and Chebyshev polynomials with nice moments, 2013.
J. Cigler, Some remarks about q-Chebyshev polynomials and q-Catalan numbers and related results, 2013.
J. Cigler, Some notes on q-Gould polynomials, 2013.
Emeric Deutsch, A. Robertson and D. Saracino, Refined restricted involutions, European Journal of Combinatorics 28 (2007), 481-498 (see pp. 486 and 498).
J. East and R. D. Gray, Idempotent generators in finite partition monoids and related semigroups, arXiv preprint arXiv:1404.2359, 2014
D. Gouyou-Beauchamps, Chemins sous-diagonaux et tableau de Young, pp. 112-125 of "Combinatoire Enumerative (Montreal 1985)", Lect. Notes Math. 1234, 1986 (see |F_{l,p}| on page 114). - _N. J. A. Sloane_, Jan 29 2011
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.
V. E. Hoggatt, Jr. and M. Bicknell, Catalan and related sequences arising from inverses of Pascal's triangle matrices, Fib. Quart., 14 (1976), 395-405.
W. F. Klostermeyer, M. E. Mays, L. Soltes and G. Trapp, A Pascal rhombus, Fibonacci Quarterly, 35 (1997), 318-328.
Wolfdieter Lang, Chebyshev S-polynomials: ten applications.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419; Note 4, pp. 414-415.
MathOverflow, Catalan numbers as sums of squares of numbers in the rows of the Catalan triangle - is there a combinatorial explanation?
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Karim Ritter von Merkl, Computing colored Khovanov homology, arXiv:2505.03916 [math.QA], 2025. See p. 2.
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Index entries for sequences related to Chebyshev polynomials.

Cf. A008315, A049310, A000108, A001405 (row sums), A145973, A153585, A108786, A037952. Another version: A008313. A039598 and A039599 without zeros, and odd and even numbered rows.

Variant without zero-diagonals: A033184 and with rows reversed: A009766.

a053121 n k = a053121_tabl !! n !! k
a053121_row n = a053121_tabl !! n
a053121_tabl = iterate
   (\row -> zipWith (+) ([0] ++ row) (tail row ++ [0,0])) [1]
-- Reinhard Zumkeller, Feb 24 2012

T:=proc(n,k): if n+k mod 2 = 0 then (k+1)*binomial(n+1,(n-k)/2)/(n+1) else 0 fi end: for n from 0 to 13 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form; Emeric Deutsch, Oct 12 2006
F:=proc(l,p) if ((l-p) mod 2) = 1 then 0 else (p+1)*l!/( ( (l-p)/2 )! * ( (l+p)/2 +1)! ); fi; end;
r:=n->[seq( F(n,p),p=0..n)]; [seq(r(n),n=0..15)]; # N. J. A. Sloane, Jan 29 2011
A053121 := proc(n,k) option remember; `if`(k>n or k<0,0,`if`(n=k,1,
procname(n-1,k-1)+procname(n-1,k+1))) end proc:
seq(print(seq(A053121(n,k), k=0..n)),n=0..12); # Peter Luschny, May 01 2011

a[n_, m_] /; n < m || OddQ[n-m] = 0; a[n_, m_] = (m+1) Binomial[n+1, (n-m)/2]/(n+1); Flatten[Table[a[n, m], {n, 0, 12}, {m, 0, n}]] [[1 ;; 90]] (* Jean-François Alcover, May 18 2011 *)
T[0, 0] := 1; T[n_, k_]/;0<=k<=n := T[n, k] = T[n-1, k-1]+T[n-1, k+1]; T[n_, k_] := 0; Flatten@Table[T[n, k], {n, 0, 12}, {k, 0, n}] (* Oliver Seipel, Dec 31 2024 *)

T(n, m)=if(nCharles R Greathouse IV, Mar 09 2016

def A053121_triangle(dim):
    M = matrix(ZZ,dim,dim)
    for n in (0..dim-1): M[n,n] = 1
    for n in (1..dim-1):
        for k in (0..n-1):
            M[n,k] = M[n-1,k-1] + M[n-1,k+1]
    return M
A053121_triangle(13) # Peter Luschny, Sep 19 2012

a(n, m) := 0 if n

a(n, m) = (4*(n-1)*a(n-2, m) + 2*(m+1)*a(n-1, m-1))/(n+m+2), a(n, m)=0 if n

G.f. for m-th column: c(x^2)*(x*c(x^2))^m, where c(x) = g.f. for Catalan numbers A000108.

G.f.: G(t,z) = c(z^2)/(1 - t*z*c(z^2)), where c(z) = (1 - sqrt(1-4*z))/(2*z) is the g.f. for the Catalan numbers (A000108). - Emeric Deutsch, Jun 16 2011

a(n, m) = a(n-1, m-1) + a(n-1, m+1) if n > 0 and m >= 0, a(0, 0)=1, a(0, m)=0 if m > 0, a(n, m)=0 if m < 0. - Henry Bottomley, Jan 25 2001

Sum_{k>=0} T(m,k)^2 = A000108(m). - Paul D. Hanna, Apr 23 2005

Sum_{k>=0} T(m, k)*T(n, k) = 0 if m+n is odd; Sum_{k>=0} T(m, k)*T(n, k) = A000108((m+n)/2) if m+n is even. - Philippe Deléham, May 26 2005

T(n,k)=sum{i=0..n, (-1)^(n-i)*C(n,i)*sum{j=0..i, C(i,j)*(C(i-j,j+k)-C(i-j,j+k+2))}}; Column k has e.g.f. BesselI(k,2x)-BesselI(k+2,2x). - Paul Barry, Feb 16 2006

Sum_{k=0..n} T(n,k)*(k+1) = 2^n. - Philippe Deléham, Mar 22 2007

Sum_{j>=0} T(n,j)*binomial(j,k) = A054336(n,k). - Philippe Deléham, Mar 30 2007

T(2*n+1,2*k+1) = A039598(n,k), T(2*n,2*k) = A039599(n,k). - Philippe Deléham, Apr 16 2007

Sum_{k=0..n} T(n,k)^x = A000027(n+1), A001405(n), A000108(n), A003161(n), A129123(n) for x = 0,1,2,3,4 respectively. - Philippe Deléham, Nov 22 2009

Sum_{k=0..n} T(n,k)*x^k = A126930(n), A126120(n), A001405(n), A054341(n), A126931(n) for x = -1, 0, 1, 2, 3 respectively. - Philippe Deléham, Nov 28 2009

Sum_{k=0..n} T(n,k)*A000045(k+1) = A098615(n). - Philippe Deléham, Feb 03 2012

Recurrence for row polynomials C(n, x) := Sum_{m=0..n} a(n, m)*x^m = x*Sum_{k=0..n} Chat(k)*C(n-1-k, x), n >= 0, with C(-1, 1/x) = 1/x and Chat(k) = A000108(k/2) if n is even and 0 otherwise. From the o.g.f. of the row polynomials: G(z; x) := Sum_{n >= 0} C(n, x)*z^n = c(z^2)*(1 + x*z*G(z, x)), with the o.g.f. c of A000108. - Ahmet Zahid KÜÇÜK and Wolfdieter Lang, Aug 23 2015

The Boas-Buck recurrence (see a comment above) for the sequence of column m is: a(n, m) = ((m+1)/(n-m))*Sum_{j=0..n-1-m} (1/2)*(1 - (-1)^j)*binomial(j+1, (j+1)/2)* a(n-1-j, k), for n > m >= 0 and input a(m, m) = 1. - Wolfdieter Lang, Aug 11 2017

Sum_{m=1..n} a(n,m) = A037952(n). - R. J. Mathar, Sep 23 2021

Edited by N. J. A. Sloane, Jan 29 2011

A039598 Triangle formed from odd-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x). Sometimes called Catalan's triangle.

Original entry on oeis.org

1, 2, 1, 5, 4, 1, 14, 14, 6, 1, 42, 48, 27, 8, 1, 132, 165, 110, 44, 10, 1, 429, 572, 429, 208, 65, 12, 1, 1430, 2002, 1638, 910, 350, 90, 14, 1, 4862, 7072, 6188, 3808, 1700, 544, 119, 16, 1, 16796, 25194, 23256, 15504, 7752, 2907, 798, 152, 18, 1

Offset: 0

N. J. A. Sloane

T(n,k) is the number of leaves at level k+1 in all ordered trees with n+1 edges. - Emeric Deutsch, Jan 15 2005
Riordan array ((1-2x-sqrt(1-4x))/(2x^2),(1-2x-sqrt(1-4x))/(2x)). Inverse array is A053122. - Paul Barry, Mar 17 2005
T(n,k) is the number of walks of n steps, each in direction N, S, W, or E, starting at the origin, remaining in the upper half-plane and ending at height k (see the R. K. Guy reference, p. 5). Example: T(3,2)=6 because we have ENN, WNN, NEN, NWN, NNE and NNW. - Emeric Deutsch, Apr 15 2005
Triangle T(n,k), 0<=k<=n, read by rows given by T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0) = 2*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + 2*T(n-1,k) + T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 30 2007
Number of (2n+1)-step walks from (0,0) to (2n+1,2k+1) and consisting of steps u=(1,1) and d=(1,-1) in which the path stays in the nonnegative quadrant. Examples: T(2,0)=5 because we have uuudd, uudud, uuddu, uduud, ududu; T(2,1)=4 because we have uuuud, uuudu, uuduu, uduuu; T(2,2)=1 because we have uuuuu. - Philippe Deléham, Apr 16 2007, Apr 18 2007
Triangle read by rows: T(n,k)=number of lattice paths from (0,0) to (n,k) that do not go below the line y=0 and consist of steps U=(1,1), D=(1,-1) and two types of steps H=(1,0); example: T(3,1)=14 because we have UDU, UUD, 4 HHU paths, 4 HUH paths and 4 UHH paths. - Philippe Deléham, Sep 25 2007
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. - Philippe Deléham, Sep 25 2007
With offset [1,1] this is the (ordinary) convolution triangle a(n,m) with o.g.f. of column m given by (c(x)-1)^m, where c(x) is the o.g.f. for Catalan numbers A000108. See the Riordan comment by Paul Barry.
T(n, k) is also the number of order-preserving full transformations (of an n-chain) with exactly k fixed points. - Abdullahi Umar, Oct 02 2008
T(n,k)/2^(2n+1) = coefficients of the maximally flat lowpass digital differentiator of the order N=2n+3. - Pavel Holoborodko (pavel(AT)holoborodko.com), Dec 19 2008
The signed triangle S(n,k) := (-1)^(n-k)*T(n,k) provides the transformation matrix between f(n,l) := L(2*l)*5^n*F(2*l)^(2*n+1) (F=Fibonacci numbers A000045, L=Lucas numbers A000032) and F(4*l*(k+1)), k = 0, ..., n, for each l>=0: f(n,l) = Sum_{k=0..n} S(n,k)*F(4*l*(k+1)), n>=0, l>=0. Proof: the o.g.f. of the l.h.s., G(l;x) := Sum_{n>=0} f(n,l)*x^n = F(4*l)/(1 - 5*F(2*l)^2*x) is shown to match the o.g.f. of the r.h.s.: after an interchange of the n- and k-summation, the Riordan property of S = (C(x)/x,C(x)) (compare with the above comments by Paul Barry), with C(x) := 1 - c(-x), with the o.g.f. c(x) of A000108 (Catalan numbers), is used, to obtain, after an index shift, first Sum_{k>=0} F(4*l*(k))*GS(k;x), with the o.g.f of column k of triangle S which is GS(k;x) := Sum_{n>=k} S(n,k)*x^n = C(x)^(k+1)/x. The result is GF(l;C(x))/x with the o.g.f. GF(l,x) := Sum_{k>=0} F(4*l*k)*x^k = x*F(4*l)/(1-L(4*l)*x+x^2) (see a comment on A049670, and A028412). If one uses then the identity L(4*n) - 5*F(2*n)^2 = 2 (in Koshy's book [reference under A065563] this is No. 15, p. 88, attributed to Lucas, 1876), the proof that one recovers the o.g.f. of the l.h.s. from above boils down to a trivial identity on the Catalan o.g.f., namely 1/c^2(-x) = 1 + 2*x - (x*c(-x))^2. - Wolfdieter Lang, Aug 27 2012
O.g.f. for row polynomials R(x) := Sum_{k=0..n} a(n,k)*x^k:
  ((1+x) - C(z))/(x - (1+x)^2*z) with C the o.g.f. of A000108 (Catalan numbers). From Riordan ((C(x)-1)/x,C(x)-1), compare with a Paul Barry comment above. This coincides with the o.g.f. given by Emeric Deutsch in the formula section. - Wolfdieter Lang, Nov 13 2012
The A-sequence for this Riordan triangle is [1,2,1] and the Z-sequence is [2,1]. See a W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 13 2012
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n+1, 2*k+1). T(n, k) appears in the formula for the (2*n+1)-th power of the algebraic number rho(N) := 2*cos(Pi/N) = R(N, 2) in terms of the even-indexed diagonal/side length ratios R(N, 2*(k+1)) = S(2*k+1, 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+1) = 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 even powers of rho(n) see A039599. (End)
The tridiagonal Toeplitz production matrix P in the Example section corresponds to the unsigned Cartan matrix for the simple Lie algebra A_n as n tends to infinity (cf. Damianou ref. in A053122). -  Tom Copeland, Dec 11 2015 (revised Dec 28 2015)
T(n,k) is the number of pairs of non-intersecting walks of n steps, each in direction N or E, starting at the origin, and such that the end points of the two paths are separated by a horizontal distance of k. See Shapiro 1976. - Peter Bala, Apr 12 2017
Also the convolution triangle of the Catalan numbers A000108. - Peter Luschny, Oct 07 2022

			Triangle T(n,k) starts:
n\k     0      1      2      3      4     5    6    7   8  9 10
0:      1
1:      2      1
2:      5      4      1
3:     14     14      6      1
4:     42     48     27      8      1
5:    132    165    110     44     10     1
6:    429    572    429    208     65    12    1
7:   1430   2002   1638    910    350    90   14    1
8:   4862   7072   6188   3808   1700   544  119   16   1
9:  16796  25194  23256  15504   7752  2907  798  152  18  1
10: 58786  90440  87210  62016  33915 14364 4655 1120 189 20  1
... Reformatted and extended by _Wolfdieter Lang_, Nov 13 2012.
Production matrix begins:
2, 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
0, 0, 0, 0, 0, 0, 1, 2, 1
- _Philippe Deléham_, Nov 07 2011
From _Wolfdieter Lang_, Nov 13 2012: (Start)
Recurrence: T(5,1) = 165 = 1*42 + 2*48 +1*27. The Riordan A-sequence is [1,2,1].
Recurrence from Riordan Z-sequence [2,1]: T(5,0) = 132 = 2*42 + 1*48. (End)
From _Wolfdieter Lang_, Sep 20 2013: (Start)
  Example for rho(N) = 2*cos(Pi/N) powers:
  n=2: rho(N)^5 = 5*R(N, 2) + 4*R(N, 4) + 1*R(N, 6) = 5*S(1, rho(N)) + 4*S(3, rho(N)) + 1*S(5, rho(N)), identical in N >= 1. For N=5 (the pentagon with only one distinct diagonal) the degree delta(5) = 2, hence R(5, 4) and R(5, 6) can be reduced, namely to R(5, 1) = 1 and R(5, 6) = -R(5,1) = -1, respectively. Thus rho(5)^5 = 5*R(N, 2) + 4*1  + 1*(-1) = 3 + 5*R(N, 2) = 3 + 5*rho(5), with the golden section rho(5). (End)

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Mirror image of A050166. Row sums are A001700.

Cf. A000108, A008313, A039599, A183134, A094527, A033184, A035324, A053122, A172417.

/* As triangle: */ [[Binomial(2*n,n-k) - Binomial(2*n,n-k-2): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 22 2015

T:=(n,k)->binomial(2*n, n-k) - binomial(2*n, n-k-2); # N. J. A. Sloane, Aug 26 2013
# Uses function PMatrix from A357368. Adds row and column above and to the left.
PMatrix(10, n -> binomial(2*n, n) / (n + 1)); # Peter Luschny, Oct 07 2022

Flatten[Table[Binomial[2n, n-k] - Binomial[2n, n-k-2], {n,0,9}, {k,0,n}]] (* Jean-François Alcover, May 03 2011 *)

T(n,k)=binomial(2*n,n-k) - binomial(2*n,n-k-2) \\ Charles R Greathouse IV, Nov 07 2016

# Algorithm of L. Seidel (1877)
# Prints the first n rows of the triangle.
def A039598_triangle(n) :
    D = [0]*(n+2); D[1] = 1
    b = True; h = 1
    for i in range(2*n) :
        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]
        b = not b
        if b : print([D[z] for z in (1..h-1) ])
A039598_triangle(10)  # Peter Luschny, May 01 2012

Row n: C(2n, n-k) - C(2n, n-k-2).
a(n, k) = C(2n+1, n-k)*2*(k+1)/(n+k+2) = A050166(n, n-k) = a(n-1, k-1) + 2*a(n-1, k)+ a (n-1, k+1) [with a(0, 0) = 1 and a(n, k) = 0 if n<0 or nHenry Bottomley, Sep 24 2001
From Philippe Deléham, Feb 14 2004: (Start)
T(n, 0) = A000108(n+1), T(n, k) = 0 if n0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+2) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
Sum_{k>=0} T(m, k)*T(n, k) = A000108(m+n+1). (End)
T(n, k) = A009766(n+k+1, n-k) = A033184(n+k+2, 2k+2). - Philippe Deléham, Feb 14 2004
Sum_{j>=0} T(k, j)*A039599(n-k, j) = A028364(n, k). - Philippe Deléham, Mar 04 2004
Antidiagonal Sum_{k=0..n} T(n-k, k) = A000957(n+3). - Gerald McGarvey, Jun 05 2005
The triangle may also be generated from M^n * [1,0,0,0,...], where M = an infinite tridiagonal matrix with 1's in the super- and subdiagonals and [2,2,2,...] in the main diagonal. - Gary W. Adamson, Dec 17 2006
G.f.: G(t,x) = C^2/(1-txC^2), where C = (1-sqrt(1-4x))/(2x) is the Catalan function. From here G(-1,x)=C, i.e., the alternating row sums are the Catalan numbers (A000108). - Emeric Deutsch, Jan 20 2007
Sum_{k=0..n} T(n,k)*x^k = A000957(n+1), A000108(n), A000108(n+1), A001700(n), A049027(n+1), A076025(n+1), A076026(n+1) for x=-2,-1,0,1,2,3,4 respectively (see square array in A067345). - Philippe Deléham, Mar 21 2007, Nov 04 2011
Sum_{k=0..n} T(n,k)*(k+1) = 4^n. - Philippe Deléham, Mar 30 2007
Sum_{j>=0} T(n,j)*binomial(j,k) = A035324(n,k), A035324 with offset 0 (0 <= k <= n). - Philippe Deléham, Mar 30 2007
T(n,k) = A053121(2*n+1,2*k+1). - Philippe Deléham, Apr 16 2007, Apr 18 2007
T(n,k) = A039599(n,k) + A039599(n,k+1). - Philippe Deléham, Sep 11 2007
Sum_{k=0..n+1} T(n+1,k)*k^2 = A029760(n). - Philippe Deléham, Dec 16 2007
Sum_{k=0..n} T(n,k)*A059841(k) = A000984(n). - Philippe Deléham, Nov 12 2008
G.f.: 1/(1-xy-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-.... (continued fraction).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A001700(n), A194723(n+1), A194724(n+1), A194725(n+1), A194726(n+1), A194727(n+1), A194728(n+1), A194729(n+1), A194730(n+1) for x = 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Nov 03 2011
From Peter Bala, Dec 21 2014: (Start)
This triangle factorizes in the Riordan group as ( C(x), x*C(x) ) * ( 1/(1 - x), x/(1 - x) ) = A033184 * A007318, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.
Let U denote the lower unit triangular array with 1's on or below the main diagonal and zeros elsewhere. For k = 0,1,2,... define U(k) to be the lower unit triangular block array
/I_k 0\
\ 0  U/ having the k X k identity matrix I_k as the upper left block; in particular, U(0) = U. Then this array equals the bi-infinite product (...*U(2)*U(1)*U(0))*(U(0)*U(1)*U(2)*...). (End)
From Peter Bala, Jul 21 2015: (Start)
O.g.f. G(x,t) = (1/x) * series reversion of ( x/f(x,t) ), where f(x,t) = ( 1 + (1 + t)*x )^2/( 1 + t*x ).
1 + x*d/dx(G(x,t))/G(x,t) = 1 + (2 + t)*x + (6 + 4*t + t^2)*x^2 + ... is the o.g.f for A094527. (End)
Conjecture: Sum_{k=0..n} T(n,k)/(k+1)^2 = H(n+1)*A000108(n)*(2*n+1)/(n+1), where H(n+1) = Sum_{k=0..n} 1/(k+1). - Werner Schulte, Jul 23 2015
From Werner Schulte, Jul 25 2015: (Start)
Sum_{k=0..n} T(n,k)*(k+1)^2 = (2*n+1)*binomial(2*n,n). (A002457)
Sum_{k=0..n} T(n,k)*(k+1)^3 = 4^n*(3*n+2)/2.
Sum_{k=0..n} T(n,k)*(k+1)^4 = (2*n+1)^2*binomial(2*n,n).
Sum_{k=0..n} T(n,k)*(k+1)^5 = 4^n*(15*n^2+15*n+4)/4. (End)
The o.g.f. G(x,t) is such that G(x,t+1) is the o.g.f. for A035324, but with an offset of 0, and G(x,t-1) is the o.g.f. for A033184, again with an offset of 0. - Peter Bala, Sep 20 2015
Denote this lower triangular array by L; then L * transpose(L) is the Cholesky factorization of the Hankel matrix ( 1/(i+j)*binomial(2*i+2*j-2, i+j-1) )A172417%20read%20as%20a%20square%20array.%20See%20Chamberland,%20p.%201669.%20-%20_Peter%20Bala">i,j >= 1 = A172417 read as a square array. See Chamberland, p. 1669. - _Peter Bala, Oct 15 2023

Typo in one entry corrected by Philippe Deléham, Dec 16 2007

A064189 Triangle T(n,k), 0 <= k <= n, read by rows, defined by: T(0,0)=1, T(n,k)=0 if n < k, T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-1,k+1).

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 4, 5, 3, 1, 9, 12, 9, 4, 1, 21, 30, 25, 14, 5, 1, 51, 76, 69, 44, 20, 6, 1, 127, 196, 189, 133, 70, 27, 7, 1, 323, 512, 518, 392, 230, 104, 35, 8, 1, 835, 1353, 1422, 1140, 726, 369, 147, 44, 9, 1, 2188, 3610, 3915, 3288, 2235, 1242, 560, 200, 54, 10, 1

Offset: 0

N. J. A. Sloane, Sep 21 2001

Motzkin triangle read in reverse order.
T(n,k) = number of lattice paths from (0,0) to (n,k), staying weakly above the x-axis and consisting of steps U=(1,1), D=(1,-1) and H=(1,0). Example: T(3,1) = 5 because we have HHU, UDU, HUH, UHH and UUD. Columns 0,1,2 and 3 give A001006 (Motzkin numbers), A002026 (first differences of Motzkin numbers), A005322 and A005323, respectively. - Emeric Deutsch, Feb 29 2004
Riordan array ((1-x-sqrt(1-2x-3x^2))/(2x^2), (1-x-sqrt(1-2x-3x^2))/(2x)). Inverse is the array (1/(1+x+x^2), x/(1+x+x^2)) (A104562). - Paul Barry, Mar 15 2005
Inverse binomial matrix applied to A039598. - Philippe Deléham, Feb 28 2007
Triangle T(n,k), 0 <= k <= n, read by rows given by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-1,k+1) for k >= 1. - Philippe Deléham, Mar 27 2007
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 from 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. - Philippe Deléham, Sep 25 2007
Equals binomial transform of triangle A053121. - Gary W. Adamson, Oct 25 2008
Consider a semi-infinite chessboard with squares labeled (n,k), ranks or rows n >= 0, files or columns k >= 0; the number of king-paths of length n from (0,0) to (n,k), 0 <= k <= n, is T(n,k). The recurrence relation given above relates to the movements of the king. This is essentially the comment made by Harrie Grondijs for the Motzkin triangle A026300. - Johannes W. Meijer, Oct 10 2010

			Triangle begins:
  [0]   1;
  [1]   1,    1;
  [2]   2,    2,    1;
  [3]   4,    5,    3,    1;
  [4]   9,   12,    9,    4,   1;
  [5]  21,   30,   25,   14,   5,   1;
  [6]  51,   76,   69,   44,  20,   6,   1;
  [7] 127,  196,  189,  133,  70,  27,   7,  1;
  [8] 323,  512,  518,  392, 230, 104,  35,  8, 1;
  [9] 835, 1353, 1422, 1140, 726, 369, 147, 44, 9, 1;
  ...
From _Philippe Deléham_, Nov 04 2011: (Start)
Production matrix begins:
  1, 1
  1, 1, 1
  0, 1, 1, 1
  0, 0, 1, 1, 1
  0, 0, 0, 1, 1, 1
  0, 0, 0, 0, 1, 1, 1 (End)

See A026300 for additional references and other information.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
E. Barcucci, R. Pinzani and R. Sprugnoli, The Motzkin family, P.U.M.A. Ser. A, Vol. 2, 1991, No. 3-4, pp. 249-279.
Paul Barry, On Motzkin-Schröder Paths, Riordan Arrays, and Somos-4 Sequences, J. Int. Seq. (2023) Vol. 26, Art. 23.4.7.
Paul Barry, Moment sequences, transformations, and Spidernet graphs, arXiv:2307.00098 [math.CO], 2023.
Paul Barry, Notes on Riordan arrays and lattice paths, arXiv:2504.09719 [math.CO], 2025. See pp. 16, 29.
Xiaomei Chen, Counting humps and peaks in Motzkin paths with height k, arXiv:2412.00668 [math.CO], 2024. See pp. 3, 7.
Igor Dolinka, James East, Athanasios Evangelou, Desmond FitzGerald, Nicholas Ham, James Hyde, Nicholas Loughlin, and James Mitchell, Idempotent Statistics of the Motzkin and Jones Monoids, arXiv preprint arXiv:1507.04838 [math.CO], 2015.
Igor Dolinka, James East, and Robert D. Gray, Motzkin monoids and partial Brauer monoids, arXiv preprint arXiv:1512.02279 [math.GR], 2015.
Robert Donaghey and Louis W. Shapiro, Motzkin numbers, J. Combin. Theory, Series A, 23 (1977), 291-301.
Ivana Đurđev, Igor Dolinka, and James East, Sandwich semigroups in diagram categories, arXiv:1910.10286 [math.GR], 2019.
Samuele Giraudo, Tree series and pattern avoidance in syntax trees, arXiv:1903.00677 [math.CO], 2019.
Tom Halverson and Theodore N. Jacobson, Set-partition tableaux and representations of diagram algebras, arXiv:1808.08118 [math.RT], 2018.
Toufik Mansour and Mark Shattuck, Enumeration of Catalan and smooth words according to capacity, Integers (2025) Vol. 25, Art. No. A5. See p. 29.
Donatella Merlini and Massimo Nocentini, Algebraic Generating Functions for Languages Avoiding Riordan Patterns, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.3.
Robin Pemantle and Mark C. Wilson, Twenty Combinatorial Examples of Asymptotics Derived from Multivariate Generating Functions, SIAM Rev., 50 (2008), no. 2, 199-272. See p. 265.
Sheng-Liang Yang, Yan-Ni Dong, and Tian-Xiao He, Some matrix identities on colored Motzkin paths, Discrete Mathematics 340.12 (2017): 3081-3091.
Sheng-Liang Yang and Yuan-Yuan Gao, The Pascal rhombus and Riordan arrays, Fib. Q., 56:4 (2018), 337-347. See Fig. 3.

A026300 (the main entry for this sequence) with rows reversed.
Cf. A001006, A002026, A005322, A005323, A053121.
Row sums give: A005773(n+1) or A307789(n+2).

alias(C=binomial): A064189 := (n,k) -> add(C(n,j)*(C(n-j,j+k)-C(n-j,j+k+2)), j=0..n): seq(seq(A064189(n,k), k=0..n),n=0..10); # Peter Luschny, Dec 31 2019
# Uses function PMatrix from A357368. Adds a row above and a column to the left.
PMatrix(10, n -> simplify(hypergeom([1 -n/2, -n/2+1/2], [2], 4))); # Peter Luschny, Oct 08 2022

T[0, 0, x_, y_] := 1; T[n_, 0, x_, y_] := x*T[n - 1, 0, x, y] + T[n - 1, 1, x, y]; T[n_, k_, x_, y_] := T[n, k, x, y] = If[k < 0 || k > n, 0, T[n - 1, k - 1, x, y] + y*T[n - 1, k, x, y] + T[n - 1, k + 1, x, y]]; Table[T[n, k, 1, 1], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Apr 21 2017 *)
T[n_, k_] := Binomial[n, k] Hypergeometric2F1[(k - n)/2, (k - n + 1)/2, k + 2, 4];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten  (* Peter Luschny, May 19 2021 *)

{T(n, k) = if( k<0 || k>n, 0, polcoeff( polcoeff( 2 / (1 - x + sqrt(1 - 2*x - 3*x^2) - 2*x*y) + x * O(x^n), n), k))}; /* Michael Somos, Jun 06 2016 */

def A064189_triangel(dim):
    M = matrix(ZZ,dim,dim)
    for n in range(dim): M[n,n] = 1
    for n in (1..dim-1):
        for k in (0..n-1):
            M[n,k] = M[n-1,k-1]+M[n-1,k]+M[n-1,k+1]
    return M
A064189_triangel(9) # Peter Luschny, Sep 20 2012

Sum_{k=0..n} T(n, k)*(k+1) = 3^n.
Sum_{k=0..n} T(n, k)*T(n, n-k) = T(2*n, n) - T(2*n, n+2)
G.f.: M/(1-t*z*M), where M = 1 + z*M + z^2*M^2 is the g.f. of the Motzkin numbers (A001006). - Emeric Deutsch, Feb 29 2004
Sum_{k>=0} T(m, k)*T(n, k) = A001006(m+n). - Philippe Deléham, Mar 05 2004
Sum_{k>=0} T(n-k, k) = A005043(n+2). - Philippe Deléham, May 31 2005
Column k has e.g.f. exp(x)*(BesselI(k,2*x)-BesselI(k+2,2*x)). - Paul Barry, Feb 16 2006
T(n,k) = Sum_{j=0..n} C(n,j)*(C(n-j,j+k) - C(n-j,j+k+2)). - Paul Barry, Feb 16 2006
n-th row is generated from M^n * V, where M = the infinite tridiagonal matrix with all 1's in the super, main and subdiagonals; and V = the infinite vector [1,0,0,0,...]. E.g., Row 3 = (4, 5, 3, 1), since M^3 * V = [4, 5, 3, 1, 0, 0, 0, ...]. - Gary W. Adamson, Nov 04 2006
T(n,k) = A122896(n+1,k+1). - Philippe Deléham, Apr 21 2007
T(n,k) = (k/n)*Sum_{j=0..n} binomial(n,j)*binomial(j,2*j-n-k). - Vladimir Kruchinin, Feb 12 2011
Sum_{k=0..n} T(n,k)*(-1)^k*(k+1) = (-1)^n. - Werner Schulte, Jul 08 2015
Sum_{k=0..n} T(n,k)*(k+1)^3 = (2*n+1)*3^n. - Werner Schulte, Jul 08 2015
G.f.: 2 / (1 - x + sqrt(1 - 2*x - 3*x^2) - 2*x*y) = Sum_{n >= k >= 0} T(n, k) * x^n * y^k. - Michael Somos, Jun 06 2016
T(n,k) = binomial(n, k)*hypergeom([(k-n)/2, (k-n+1)/2], [k+2], 4). - Peter Luschny, May 19 2021
The coefficients of the n-th degree Taylor polynomial of the function (1 - x^2)*(1 + x + x^2)^n expanded about the point x = 0 give the entries in row n in reverse order. - Peter Bala, Sep 06 2022

More terms from Vladeta Jovovic, Sep 23 2001

A061554 Square table read by antidiagonals: a(n,k) = binomial(n+k, floor(k/2)).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 6, 4, 4, 1, 1, 10, 10, 5, 5, 1, 1, 20, 15, 15, 6, 6, 1, 1, 35, 35, 21, 21, 7, 7, 1, 1, 70, 56, 56, 28, 28, 8, 8, 1, 1, 126, 126, 84, 84, 36, 36, 9, 9, 1, 1, 252, 210, 210, 120, 120, 45, 45, 10, 10, 1, 1, 462, 462, 330, 330, 165, 165, 55, 55, 11, 11, 1, 1

Offset: 0

Henry Bottomley, May 17 2001

Equivalently, a triangle read by rows, where the rows are obtained by sorting the elements of the rows of Pascal's triangle (A007318) into descending order. - Philippe Deléham, May 21 2005
Equivalently, as a triangle read by rows, this is T(n,k)=binomial(n,floor((n-k)/2)); column k then has e.g.f. Bessel_I(k,2x)+Bessel_I(k+1,2x). - Paul Barry, Feb 28 2006
Antidiagonal sums are A037952(n+1) = C(n+1,[n/2]). Matrix inverse is the row reversal of triangle A066170. Eigensequence is A125094(n) = Sum_{k=0..n-1} A125093(n-1,k)*A125094(k). - Paul D. Hanna, Nov 20 2006
Riordan array (1/(1-x-x^2*c(x^2)),x*c(x^2)); where c(x)=g.f.for Catalan numbers A000108. - Philippe Deléham, Mar 17 2007
Triangle T(n,k), 0<=k<=n, read by rows given by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 27 2007
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. - Philippe Deléham, Sep 25 2007
T(n,k) is the number of paths from (0,k) to some (n,m) which never dip below y=0, touch y=0 at least once and are made up only of the steps (1,1) and (1,-1). This can be proved using the recurrence supplied by Deléham. - Gerald McGarvey, Oct 15 2008
Triangle read by rows = partial sums of A053121 terms starting from the right. - Gary W. Adamson, Oct 24 2008
As a subset of the "family of triangles" (Deleham comment of Sep 25 2007), beginning with a signed variant of A061554, M = (-1,0) =  (1; -1, 1; 2, -1, 1; -3, 3, -1, 1; ...) successive binomial transforms of M yield (0,1) - A089942; (1,2) - A039599; (2,3) - A124733; (3,4) - A124574; (4,5) - A126331; ... such that the binomial transform of the triangle generated from (n,n+1) = the triangle generated from (n+1,n+2). Similarly, another subset beginning with A053121 - (0,0), and taking successive binomial transforms yields (1,1) - A064189; (2,2) - A039598; (3,3) - A091965, ... By rows, the triangle generated from (n,n) can be obtained by taking pairwise sums from the (n-1,n) triangle starting from the right. For example, row 2 of (1,2) - A039599 = (2, 3, 1); and taking pairwise sums from the right we obtain (5, 4, 1) = row 2 of (2,2) - A039598. - Gary W. Adamson, Aug 04 2011
The triangle by rows (n) with alternating signs (+-+...) from the top as a set of simultaneous equations solves for diagonal lengths of odd N (N = 2n+1) regular polygons. The constants in each case are powers of c = 2*cos(2*Pi/N). By way of example, the first 3 rows relate to the heptagon and the simultaneous equations are (1,0,0) = 1; (-1,1,0) = c = 1.24697...; and (2,-1,1) = c^2. The answers are 1, 2.24697..., and 1.801...; the 3 distinct diagonal lengths of the heptagon with edge = 1. - Gary W. Adamson, Sep 07 2011

			The array starts:
   1, 1,  2,  3,  6, 10,  20,  35,   70,  126, ...
   1, 1,  3,  4, 10, 15,  35,  56,  126,  210, ...
   1, 1,  4,  5, 15, 21,  56,  84,  210,  330, ...
   1, 1,  5,  6, 21, 28,  84, 120,  330,  495, ...
   1, 1,  6,  7, 28, 36, 120, 165,  495,  715, ...
   1, 1,  7,  8, 36, 45, 165, 220,  715, 1001, ...
   1, 1,  8,  9, 45, 55, 220, 286, 1001, 1365, ...
   1, 1,  9, 10, 55, 66, 286, 364, 1365, 1820, ...
   1, 1, 10, 11, 66, 78, 364, 455, 1820, 2380, ...
   1, 1, 11, 12, 78, 91, 455, 560, 2380, 3060, ...
Triangle (antidiagonal) version begins:
    1;
    1,   1;
    2,   1,   1;
    3,   3,   1,   1;
    6,   4,   4,   1,   1;
   10,  10,   5,   5,   1,   1;
   20,  15,  15,   6,   6,   1,  1;
   35,  35,  21,  21,   7,   7,  1,  1;
   70,  56,  56,  28,  28,   8,  8,  1,  1;
  126, 126,  84,  84,  36,  36,  9,  9,  1,  1;
  252, 210, 210, 120, 120,  45, 45, 10, 10,  1, 1;
  462, 462, 330, 330, 165, 165, 55, 55, 11, 11, 1, 1; ...
Matrix inverse begins:
   1;
  -1,  1;
  -1, -1,   1;
   1, -2,  -1,   1;
   1,  2,  -3,  -1,  1;
  -1,  3,   3,  -4, -1,  1;
  -1, -3,   6,   4, -5, -1,  1;
   1, -4,  -6,  10,  5, -6, -1,  1;
   1,  4, -10, -10, 15,  6, -7, -1, 1; ...
From _Paul Barry_, May 21 2009: (Start)
Production matrix is
  1, 1,
  1, 0, 1,
  0, 1, 0, 1,
  0, 0, 1, 0, 1,
  0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 1, 0, 1 (End)

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Reed Acton, T. Kyle Petersen, Blake Shirman, and Bridget Eileen Tenner, The clairvoyant maître d', arXiv:2401.11680 [math.CO], 2024. See p. 15.
Johann Cigler, Some remarks and conjectures related to lattice paths in strips along the x-axis, arXiv:1501.04750 [math.CO], 2015.
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.
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.

Rows are A001405, A037952, A037955, A037951, A037956, A037953, A037957 etc. Columns are truncated pairs of A000012, A000027, A000217, A000292, A000332, A000389, A000579, etc. Main diagonal is alternate values of A051036.

Cf. A007318, A107430, A125094, A037952, A066170.

T := proc(n, k) option remember;
if n = k then 1 elif k < 0 or n < 0 or k > n then 0
elif k = 0 then T(n-1, 0) + T(n-1, 1) else T(n-1, k-1) + T(n-1, k+1) fi end:
for n from 0 to 9 do seq(T(n, k), k = 0..n) od; # Peter Luschny, May 25 2021

t[n_, k_] = Binomial[n, Floor[(n+1)/2 - (-1)^(n-k)*(k+1)/2]]; Flatten[Table[t[n, k], {n, 0, 11}, {k, 0, n}]] (* Jean-François Alcover, May 31 2011 *)

T(n,k)=binomial(n,(n+1)\2-(-1)^(n-k)*((k+1)\2))

As a triangle: T(n,k) = binomial(n,m) where m = floor((n+1)/2 - (-1)^(n-k)*(k+1)/2).
a(0, k) = binomial(k, floor(k/2)) = A001405(k); for n>0 T(n, k) = T(n+1, k-2) + T(n-1, k).
n-th row = M^n * V, where M = the infinite tridiagonal matrix with all 1's in the super and subdiagonals and (1,0,0,0,...) in the main diagonal. V = the infinite vector [1,0,0,0,...]. Example: (3,3,1,1,0,0,0,...) = M^3 * V. - Gary W. Adamson, Nov 04 2006
Sum_{k=0..n} T(m,k)*T(n,k) = T(m+n,0) = A001405(m+n). - Philippe Deléham, Feb 26 2007
Sum_{k=0..n} T(n,k)=2^n. - Philippe Deléham, Mar 27 2007
Sum_{k=0..n} T(n,k)*x^k = A127361(n), A126869(n), A001405(n), A000079(n), A127358(n), A127359(n), A127360(n) for x = -2, -1, 0, 1, 2, 3, 4 respectively. - Philippe Deléham, Dec 04 2009

Entry revised by N. J. A. Sloane, Nov 22 2006

A052179 Triangle of numbers arising in enumeration of walks on cubic lattice.

Original entry on oeis.org

1, 4, 1, 17, 8, 1, 76, 50, 12, 1, 354, 288, 99, 16, 1, 1704, 1605, 700, 164, 20, 1, 8421, 8824, 4569, 1376, 245, 24, 1, 42508, 48286, 28476, 10318, 2380, 342, 28, 1, 218318, 264128, 172508, 72128, 20180, 3776, 455, 32, 1, 1137400, 1447338

Offset: 0

N. J. A. Sloane, Jan 26 2000

Triangle T(n,k), 0 <= k <= n, read by rows given by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = 4*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + 4*T(n-1,k) + T(n-1,k+1) for k >= 1. - Philippe Deléham, Mar 27 2007
Triangle read by rows: T(n,k) = number of lattice paths from (0,0) to (n,k) that do not go below the line y=0 and consist of steps U=(1,1), D=(1,-1) and four types of steps H=(1,0); example: T(3,1)=50 because we have UDU, UUD, 16 HHU paths, 16 HUH paths and 16 UHH paths. - Philippe Deléham, Sep 25 2007
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. - Philippe Deléham, Sep 25 2007
Riordan array ((1-4x-sqrt(1-8x+12x^2))/(2x^2), (1-4x-sqrt(1-8x+12x^2))/(2x)). Inverse of A159764. - Paul Barry, Apr 21 2009
6^n = (n-th row terms) dot (first n+1 terms in (1,2,3,...)). Example: 6^3 =  216 = (76, 50, 12, 1) dot (1, 2, 3, 4) = (76 + 100 + 36 + 4) = 216. - Gary W. Adamson, Jun 15 2011
A subset of the "family of triangles" (Deléham comment of Sep 25 2007) is the succession of binomial transforms beginning with triangle A053121, (0,0); giving -> A064189, (1,1); -> A039598, (2,2); -> A091965, (3,3); -> A052179, (4,4); -> A125906, (5,5) ->, etc.; generally the binomial transform of the triangle generated from (n,n) = that generated from ((n+1),(n+1)). - Gary W. Adamson, Aug 03 2011

			Triangle begins:
    1;
    4,   1;
   17,   8,   1;
   76,  50,  12,   1;
  354, 288,  99,  16,   1;
  ...
Production matrix begins:
  4, 1;
  1, 4, 1;
  0, 1, 4, 1;
  0, 0, 1, 4, 1;
  0, 0, 0, 1, 4, 1;
  0, 0, 0, 0, 1, 4, 1;
  0, 0, 0, 0, 0, 1, 4, 1;
- _Philippe Deléham_, Nov 04 2011

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened
Rigoberto Flórez, Leandro Junes, José L. Ramírez, Further Results on Paths in an n-Dimensional Cubic Lattice, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.2.
R. K. Guy, Catwalks, Sandsteps and Pascal Pyramids, J. Integer Seqs., Vol. 3 (2000), #00.1.6.

Cf. A039598, A053121, A064189, A091965.

T:= proc(n, k) option remember; `if`(min(n, k)<0, 0,
     `if`(max(n, k)=0, 1, T(n-1, k-1)+4*T(n-1, k)+T(n-1, k+1)))
    end:
seq(seq(T(n,k), k=0..n), n=0..10);  # Alois P. Heinz, Oct 28 2021

t[0, 0] = 1; t[n_, k_] /; k < 0 || k > n = 0; t[n_, 0] := t[n, 0] = 4*t[n-1, 0] + t[n-1, 1]; t[n_, k_] := t[n, k] = t[n-1, k-1] + 4*t[n-1, k] + t[n-1, k+1]; Flatten[ Table[t[n, k], {n, 0, 9}, {k, 0, n}]] (* Jean-François Alcover, Oct 10 2011, after Philippe Deleham *)

Sum_{k>=0} T(m, k)*T(n, k) = T(m+n, 0) = A005572(m+n). - Philippe Deléham, Sep 15 2005
n-th row = M^n * V, where M = the infinite tridiagonal matrix with all 1's in the super and subdiagonals and (4,4,4,...) in the main diagonal. E.g., Row 3 = (76, 50, 12, 1) since M^3 * V = [76, 50, 12, 1, 0, 0, 0, ...]. - Gary W. Adamson, Nov 04 2006
Sum_{k=0..n} T(n,k) = A005573(n). - Philippe Deléham, Feb 04 2007
Sum_{k=0..n} T(n,k)*(k+1) = 6^n. - Philippe Deléham, Mar 27 2007
Sum_{k=0..n} T(n,k)*x^k = A033543(n), A064613(n), A005572(n), A005573(n) for x = -2, -1, 0, 1 respectively. - Philippe Deléham, Nov 28 2009
As an infinite lower triangular matrix = the binomial transform of A091965 and 4th binomial transform of A053121. - Gary W. Adamson, Aug 03 2011
G.f.: 2/(1 - 4*x - 2*x*y + sqrt(1 - 8*x + 12*x^2)). - Daniel Checa, Aug 17 2022
G.f. for the m-th column: x^m*(A(x))^(m+1), where A(x) is the g.f. of the sequence counting the walks on the cubic lattice starting and finishing on the xy plane and never going below it (A005572). Explicitly, the g.f. is x^m*((1 - 4*x - sqrt(1 - 8*x + 12*x^2))/(2*x^2))^(m+1). - Daniel Checa, Aug 28 2022

A089942 Inverse binomial matrix applied to A039599.

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 1, 3, 2, 1, 3, 6, 6, 3, 1, 6, 15, 15, 10, 4, 1, 15, 36, 40, 29, 15, 5, 1, 36, 91, 105, 84, 49, 21, 6, 1, 91, 232, 280, 238, 154, 76, 28, 7, 1, 232, 603, 750, 672, 468, 258, 111, 36, 8, 1, 603, 1585, 2025, 1890, 1398, 837, 405, 155, 45, 9, 1, 1585, 4213, 5500

Offset: 0

Paul Barry, Nov 16 2003

Reverse of A071947 - related to lattice paths. First column is A005043.
Triangle T(n,k), 0 <= k <= n, defined by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = T(n-1,1), T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-1,k+1) for k >= 1. - Philippe Deléham, Feb 27 2007
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 from 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. - Philippe Deléham, Sep 25 2007
Riordan array (f(x),x*g(x)), where f(x)is the o.g.f. of A005043 and g(x)is the o.g.f. of A001006. - Philippe Deléham, Nov 22 2009
Riordan array ((1+x-sqrt(1-2x-3x^2))/(2x(1+x)), (1-x-sqrt(1-2x-3x^2))/(2x)). Inverse of Riordan array ((1+x)/(1+x+x^2),x/(1+x+x^2)). E.g.f. of column k is exp(x)*(Bessel_I(k,2x)-Bessel_I(k+1,2x)).
Diagonal sums are A187306.
Simultaneous equations using the first n rows solve for diagonal lengths of odd N = (2n+1) regular polygons, with constants c^0, c^1, c^2, ...; where c = 1 + 2*cos( 2*Pi/N) = sin(3*Pi/N)/sin(Pi/N) = the third longest diagonal of N>5.  By way of example, take the first 4 rows relating to the 9-gon (nonagon), N=(2*4 + 1), with c = 1 + 2*cos(2*Pi/9) = 2.5320888.... The simultaneous equations are (1,0,0,0) = 1; (0,1,0,0) = c; (1,1,1,0) = c^2, (1,3,2,1) = c^3. The answers are 1, 2.532..., 2.879..., and 1.879...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. - Gary W. Adamson, Sep 07 2011
Number of linearly independent irreducible representations of a given weight j in a tensor of given rank n. - Mikkel N. Schmidt, Aug 20 2025

			Triangle begins
   1,
   0,   1,
   1,   1,   1,
   1,   3,   2,   1,
   3,   6,   6,   3,   1,
   6,  15,  15,  10,   4,  1,
  15,  36,  40,  29,  15,  5,  1,
  36,  91, 105,  84,  49, 21,  6, 1,
  91, 232, 280, 238, 154, 76, 28, 7, 1
Production matrix is
  0, 1,
  1, 1, 1,
  0, 1, 1, 1,
  0, 0, 1, 1, 1,
  0, 0, 0, 1, 1, 1,
  0, 0, 0, 0, 1, 1, 1,
  0, 0, 0, 0, 0, 1, 1, 1,
  0, 0, 0, 0, 0, 0, 1, 1, 1,
  0, 0, 0, 0, 0, 0, 0, 1, 1, 1

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Paul Barry and Aoife Hennessy, Four-term Recurrences, Orthogonal Polynomials and Riordan Arrays, Journal of Integer Sequences, 2012, article 12.4.2. - From _N. J. A. Sloane_, Sep 21 2012
J. A. R. Coope, R. F. Snider, and F. R. McCourt, Irreducible Cartesian Tensors, The Journal of Chemical Physics, vol 43, no 7, 1965.
Emeric Deutsch, Luca Ferrari, and Simone Rinaldi, Production Matrices, Advances in Applied Mathematics, 34 (2005) pp. 101-122.
Toufik Mansour and Mark Shattuck, Enumeration of Catalan and smooth words according to capacity, Integers (2025) Vol. 25, Art. No. A5. See p. 29.
Donatella Merlini, Douglas G. Rogers, Renzo Sprugnoli, and M. Cecilia Verri, On some alternative characterizations of Riordan arrays, Canad J. Math., 49 (1997), 301-320.
Yidong Sun and Luping Ma, Minors of a class of Riordan arrays related to weighted partial Motzkin paths, Eur. J. Comb. (2014) Vol. 39, 157-169. See Table 2.2.

Row sums give A002426 (central trinomial coefficients).

T:= (n,k) -> simplify(GegenbauerC(n-k,-n+1,-1/2)-GegenbauerC(n-k-1,-n+1,-1/2)): for n from 1 to 9 do seq(T(n,k), k=1..n) od; # Peter Luschny, May 12 2016
# Or by recurrence:
T := proc(n, k) option remember;
if n = k then 1 elif k < 0 or n < 0 or k > n then 0
elif k = 0 then T(n-1, 1) else T(n-1, k-1) + T(n-1, k) + T(n-1, k+1) fi end:
for n from 0 to 9 do seq(T(n, k), k = 0..n) od; # Peter Luschny, May 25 2021

T[n_, k_] := GegenbauerC[n - k, -n + 1, -1/2] - GegenbauerC[n - k - 1, -n + 1, -1/2]; Table[T[n, k], {n,1,10}, {k,1,n}] // Flatten (* G. C. Greubel, Feb 28 2017 *)

G.f.: (1+z-q)/[(1+z)(2z-t+tz+tq)], where q = sqrt(1-2z-3z^2).
Sum_{k>=0} T(m,k)*T(n,k) = T(m+n,0) = A005043(m+n). - Philippe Deléham, Mar 22 2007
Sum_{k=0..n} T(n,k)*(2k+1) = 3^n. - Philippe Deléham, Mar 22 2007
Sum_{k=0..n} T(n,k)*2^k = A112657(n). - Philippe Deléham, Apr 01 2007
T(n,2k) + T(n,2k+1) = A109195(n,k). - Philippe Deléham, Nov 11 2008
T(n,k) = GegenbauerC(n-k,-n+1,-1/2) - GegenbauerC(n-k-1,-n+1,-1/2) for 1 <= k <= n. - Peter Luschny, May 12 2016

Edited by Emeric Deutsch, Mar 04 2004

A091965 Triangle read by rows: T(n,k) = number of lattice paths from (0,0) to (n,k) that do not go below the line y=0 and consist of steps U=(1,1), D=(1,-1) and three types of steps H=(1,0) (left factors of 3-Motzkin steps).

Original entry on oeis.org

1, 3, 1, 10, 6, 1, 36, 29, 9, 1, 137, 132, 57, 12, 1, 543, 590, 315, 94, 15, 1, 2219, 2628, 1629, 612, 140, 18, 1, 9285, 11732, 8127, 3605, 1050, 195, 21, 1, 39587, 52608, 39718, 19992, 6950, 1656, 259, 24, 1, 171369, 237129, 191754, 106644, 42498, 12177, 2457

Offset: 0

Emeric Deutsch, Mar 13 2004

T(n,0) = A002212(n+1), T(n,1) = A045445(n+1); row sums give A026378.
The inverse is A207815. - Gary W. Adamson, Dec 17 2006 [corrected by Philippe Deléham, Feb 22 2012]
Reversal of A084536. - Philippe Deléham, Mar 23 2007
Triangle T(n,k), 0 <= k <= n, read by rows given by T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = 3*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + 3*T(n-1,k) + T(n-1,k+1) for k >= 1. - Philippe Deléham, Mar 27 2007
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. - Philippe Deléham, Sep 25 2007
5^n = (n-th row terms) dot (first n+1 terms in (1,2,3,...)). Example for row 4: 5^4 = 625 = (137, 132, 57, 12, 1) dot (1, 2, 3, 4, 5) = (137 + 264 + 171 + 48 + 5) = 625. - Gary W. Adamson, Jun 15 2011
Riordan array ((1-3*x-sqrt(1-6*x+5*x^2))/(2*x^2), (1-3*x-sqrt(1-6*x+5*x^2))/(2*x)). - Philippe Deléham, Feb 19 2012

			Triangle begins:
     1;
     3,    1;
    10,    6,    1;
    36,   29,    9,    1;
   137,  132,   57,   12,    1;
   543,  590,  315,   94,   15,    1;
  2219, 2628, 1629,  612,  140,   18,    1;
T(3,1)=29 because we have UDU, UUD, 9 HHU paths, 9 HUH paths and 9 UHH paths.
Production matrix begins
  3, 1;
  1, 3, 1;
  0, 1, 3, 1;
  0, 0, 1, 3, 1;
  0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 0, 0, 0, 0, 1, 3, 1;
- _Philippe Deléham_, Nov 07 2011

A. Nkwanta, Lattice paths and RNA secondary structures, DIMACS Series in Discrete Math. and Theoretical Computer Science, 34, 1997, 137-147.

Vincenzo Librandi, Rows n = 0..100, flattened
Shu-Chiuan Chang and Robert Shrock, Structure of the Partition Function and Transfer Matrices for the Potts Model in a Magnetic Field on Lattice Strips, J. Stat. Physics 137 (2009) 667, table 5.
Helmut Prodinger, The amplitude of Motzkin paths, arXiv:2104.07596 [math.CO], 2021. Mentions this sequence.
Helmut Prodinger, Multi-edge trees and 3-coloured Motzkin paths: bijective issues, arXiv:2105.03350 [math.CO], 2021.
Helmut Prodinger, Weighted unary-binary trees, Hex-trees, marked ordered trees, and related structures, arXiv:2106.14782 [math.CO], 2021.

Cf. A002212, A045445, A026378.

Cf. A123965.

nmax = 9; t[n_, k_] := ((k+1)*n!*Hypergeometric2F1[k+3/2, k-n, 2k+3, -4]) / ((k+1)!*(n-k)!); Flatten[ Table[ t[n, k], {n, 0, nmax}, {k, 0, n}]] (* Jean-François Alcover, Nov 14 2011, after Vladimir Kruchinin *)
T[0, 0, x_, y_] := 1; T[n_, 0, x_, y_] := x*T[n - 1, 0, x, y] + T[n - 1, 1, x, y]; T[n_, k_, x_, y_] := T[n, k, x, y] = If[k < 0 || k > n, 0,
T[n - 1, k - 1, x, y] + y*T[n - 1, k, x, y] + T[n - 1, k + 1, x, y]];
Table[T[n, k, 3, 3], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, May 22 2017 *)

T(n,k):=(k+1)*sum((binomial(2*(m+1),m-k)*binomial(n,m))/(m+1),m,k,n); /* Vladimir Kruchinin, Oct 08 2011 */

@CachedFunction
def A091965(n,k):
    if n==0 and k==0: return 1
    if k<0 or k>n: return 0
    if k==0: return 3*A091965(n-1,0)+A091965(n-1,1)
    return A091965(n-1,k-1)+3*A091965(n-1,k)+A091965(n-1,k+1)
for n in (0..7):
    [A091965(n,k) for k in (0..n)] # Peter Luschny, Nov 05 2012

G.f.: G = 2/(1 - 3*z - 2*t*z + sqrt(1-6*z+5*z^2)). Alternatively, G = M/(1 - t*z*M), where M = 1 + 3*z*M + z^2*M^2.
Sum_{k>=0} T(m, k)*T(n, k) = T(m+n, 0) = A002212(m+n+1). - Philippe Deléham, Sep 14 2005
The triangle may also be generated from M^n * [1,0,0,0,...], where M = an infinite tridiagonal matrix with 1's in the super and subdiagonals and [3,3,3,...] in the main diagonal. - Gary W. Adamson, Dec 17 2006
Sum_{k=0..n} T(n,k)*(k+1) = 5^n. - Philippe Deléham, Mar 27 2007
Sum_{k=0..n} T(n,k)*x^k = A117641(n), A033321(n), A007317(n), A002212(n+1), A026378(n+1) for x = -3, -2, -1, 0, 1 respectively. - Philippe Deléham, Nov 28 2009
T(n,k) = (k+1)*Sum_{m=k..n} binomial(2*(m+1),m-k)*binomial(n,m)/(m+1). - Vladimir Kruchinin, Oct 08 2011
The n-th row polynomial R(n,x) equals the n-th degree Taylor polynomial of the function (1 - x^2)*(1 + 3*x + x^2)^n expanded about the point x = 0. - Peter Bala, Sep 06 2022

A038622 Triangular array that counts rooted polyominoes.

Original entry on oeis.org

1, 2, 1, 5, 3, 1, 13, 9, 4, 1, 35, 26, 14, 5, 1, 96, 75, 45, 20, 6, 1, 267, 216, 140, 71, 27, 7, 1, 750, 623, 427, 238, 105, 35, 8, 1, 2123, 1800, 1288, 770, 378, 148, 44, 9, 1, 6046, 5211, 3858, 2436, 1296, 570, 201, 54, 10, 1, 17303, 15115, 11505, 7590, 4302, 2067, 825, 265

Offset: 0

N. J. A. Sloane, torsten.sillke(AT)lhsystems.com

The PARI program gives any row k and any n-th term for this triangular array in square or right triangle array format. - Randall L Rathbun, Jan 20 2002
Triangle T(n,k), 0 <= k <= n, read by rows given by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = 2*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-1,k+1) for k >= 1. - Philippe Deléham, Mar 27 2007
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. - Philippe Deléham, Sep 25 2007
Triangle read by rows = partial sums of A064189 terms starting from the right. - Gary W. Adamson, Oct 25 2008
Column k has e.g.f. exp(x)*(Bessel_I(k,2x)+Bessel_I(k+1,2x)). - Paul Barry, Mar 08 2011

			From _Paul Barry_, Mar 08 2011: (Start)
Triangle begins
     1;
     2,    1;
     5,    3,    1;
    13,    9,    4,   1;
    35,   26,   14,   5,   1;
    96,   75,   45,  20,   6,   1;
   267,  216,  140,  71,  27,   7,  1;
   750,  623,  427, 238, 105,  35,  8, 1;
  2123, 1800, 1288, 770, 378, 148, 44, 9, 1;
Production matrix is
  2, 1,
  1, 1, 1,
  0, 1, 1, 1,
  0, 0, 1, 1, 1,
  0, 0, 0, 1, 1, 1,
  0, 0, 0, 0, 1, 1, 1,
  0, 0, 0, 0, 0, 1, 1, 1,
  0, 0, 0, 0, 0, 0, 1, 1, 1,
  0, 0, 0, 0, 0, 0, 0, 1, 1, 1
(End)

Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened
D. Gouyou-Beauchamps and G. Viennot, Equivalence of the two-dimensional directed animal problem to a one-dimensional path problem, Adv. in Appl. Math. 9 (1988), no. 3, 334-357. See Table 1 on page 340.

Cf. A005773 (1st column), A005774 (2nd column), A005775, A066822, A000244 (row sums).

Cf. A064189, A007318, A061554, A092392.

import Data.List (transpose)
a038622 n k = a038622_tabl !! n !! k
a038622_row n = a038622_tabl !! n
a038622_tabl = iterate (\row -> map sum $
   transpose [tail row ++ [0,0], row ++ [0], [head row] ++ row]) [1]
-- Reinhard Zumkeller, Feb 26 2013

T := (n,k) -> simplify(GegenbauerC(n-k,-n+1,-1/2)+GegenbauerC(n-k-1,-n+1,-1/2)):
for n from 1 to 9 do seq(T(n,k),k=1..n) od; # Peter Luschny, May 12 2016

nmax = 10; t[n_ /; n > 0, k_ /; k >= 1] := t[n, k] = t[n-1, k-1] + t[n-1, k] + t[n-1, k+1]; t[0, 0] = 1; t[0, ] = 0; t[?Negative, ?Negative] = 0; t[n, 0] := 2 t[n-1, 0] + t[n-1, 1]; Flatten[ Table[ t[n, k], {n, 0, nmax}, {k, 0, n}]](* Jean-François Alcover, Nov 09 2011 *)

s=[0,1]; {A038622(n,k)=if(n==0,1,t=(2*(n+k)*(n+k-1)*s[2]+3*(n+k-1)*(n+k-2)*s[1])/((n+2*k-1)*n); s[1]=s[2]; s[2]=t; t)}

a(n, k) = a(n-1, k-1) + a(n-1, k) + a(n-1, k+1) for k>0, a(n, k) = 2*a(n-1, k) + a(n-1, k+1) for k=0.
Riordan array ((sqrt(1-2x-3x^2)+3x-1)/(2x(1-3x)),(1-x-sqrt(1-2x-3x^2))/(2x)). Inverse of Riordan array ((1-x)/(1+x+x^2),x/(1+x+x^2)). First column is A005773(n+1). Row sums are 3^n (A000244). If L=A038622, then L*L' is the Hankel matrix for A005773(n+1), where L' is the transpose of L. - Paul Barry, Sep 18 2006
T(n,k) = GegenbauerC(n-k,-n+1,-1/2) + GegenbauerC(n-k-1,-n+1,-1/2). In this form also the missing first column of the triangle 1,1,1,3,7,19,... (cf. A002426) can be computed. - Peter Luschny, May 12 2016
From Peter Bala, Jul 12 2021: (Start)
T(n,k) = Sum_{j = k..n} binomial(n,j)*binomial(j,floor((j-k)/2)).
Matrix product of Riordan arrays ( 1/(1 - x), x/(1 - x) ) * ( (1 - x*c(x^2))/(1 - 2*x), x*c(x^2) ) = A007318 * A061554 (triangle version), where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108.
Triangle equals A007318^(-1) * A092392 * A007318. (End)
The n-th row polynomial R(n,x) equals the n-th degree Taylor polynomial of the function (1 + x)*(1 + x + x^2)^n expanded about the point x = 0. - Peter Bala, Sep 06 2022

More terms from David W. Wilson

A111418 Right-hand side of odd-numbered rows of Pascal's triangle.

Original entry on oeis.org

1, 3, 1, 10, 5, 1, 35, 21, 7, 1, 126, 84, 36, 9, 1, 462, 330, 165, 55, 11, 1, 1716, 1287, 715, 286, 78, 13, 1, 6435, 5005, 3003, 1365, 455, 105, 15, 1, 24310, 19448, 12376, 6188, 2380, 680, 136, 17, 1, 92378, 75582, 50388

Offset: 0

Philippe Deléham, Nov 13 2005

Riordan array (c(x)/sqrt(1-4*x),x*c(x)^2) where c(x) is g.f. of A000108. Unsigned version of A113187. Diagonal sums are A014301(n+1).
Triangle T(n,k),0<=k<=n, read by rows defined by :T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=3*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+2*T(n-1,k)+T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 22 2007
Reversal of A122366. - Philippe Deléham, Mar 22 2007
Column k has e.g.f. exp(2x)(Bessel_I(k,2x)+Bessel_I(k+1,2x)). - Paul Barry, Jun 06 2007
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. - Philippe Deléham, Sep 25 2007
Diagonal sums are A014301(n+1). - Paul Barry, Mar 08 2011
This triangle T(n,k) appears in the expansion of odd powers of Fibonacci numbers F=A000045 in terms of F-numbers with multiples of odd numbers as indices. See the Ozeki reference, p. 108, Lemma 2. The formula is: F_l^(2*n+1) = sum(T(n,k)*(-1)^((n-k)*(l+1))* F_{(2*k+1)*l}, k=0..n)/5^n, n >= 0, l >= 0. - Wolfdieter Lang, Aug 24 2012
Central terms give A052203. - Reinhard Zumkeller, Mar 14 2014
This triangle appears in the expansion of (4*x)^n in terms of the polynomials Todd(n, x):= T(2*n+1, sqrt(x))/sqrt(x) = sum(A084930(n,m)*x^m), n >= 0. This follows from the inversion of the lower triangular Riordan matrix built from A084930 and comparing the g.f. of the row polynomials. - Wolfdieter Lang, Aug 05 2014
From Wolfdieter Lang, Aug 15 2014: (Start)
This triangle is the inverse of the signed Riordan triangle (-1)^(n-m)*A111125(n,m).
This triangle T(n,k) appears in the expansion of x^n in terms of the polynomials todd(k, x):= T(2*k+1, sqrt(x)/2)/(sqrt(x)/2) = S(k, x-2) - S(k-1, x-2) with the row polynomials T and S for the triangles A053120 and A049310, respectively: x^n = sum(T(n,k)*todd(k, x), k=0..n). Compare this with the preceding comment.
The A- and Z-sequences for this Riordan triangle are [1, 2, 1, repeated 0] and [3, 1, repeated 0]. For A- and Z-sequences for Riordan triangles see the W. Lang link under A006232. This corresponds to the recurrences given in the Philippe Deléham, Mar 22 2007 comment above. (End)

			From _Wolfdieter Lang_, Aug 05 2014: (Start)
The triangle T(n,k) begins:
n\k      0      1      2      3     4     5    6    7   8  9  10 ...
0:       1
1:       3      1
2:      10      5      1
3:      35     21      7      1
4:     126     84     36      9     1
5:     462    330    165     55    11     1
6:    1716   1287    715    286    78    13    1
7:    6435   5005   3003   1365   455   105   15    1
8:   24310  19448  12376   6188  2380   680  136   17   1
9:   92378  75582  50388  27132 11628  3876  969  171  19  1
10: 352716 293930 203490 116280 54264 20349 5985 1330 210 21   1
...
Expansion examples (for the Todd polynomials see A084930 and a comment above):
(4*x)^2 = 10*Todd(n,  0) + 5*Todd(n, 1) + 1*Todd(n, 2) = 10*1 + 5*(-3 + 4*x) + 1*(5 - 20*x + 16*x^2).
(4*x)^3 =  35*1 + 21*(-3 + 4*x) + 7*(5 - 20*x + 16*x^2) + (-7 + 56*x - 112*x^2 +64*x^3)*1. (End)
---------------------------------------------------------------------
Production matrix is
3, 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,
0, 0, 0, 0, 0, 0, 1, 2, 1,
0, 0, 0, 0, 0, 0, 0, 1, 2, 1
- _Paul Barry_, Mar 08 2011
Application to odd powers of Fibonacci numbers F, row n=2:
F_l^5 = (10*(-1)^(2*(l+1))*F_l + 5*(-1)^(1*(l+1))*F_{3*l} + 1*F_{5*l})/5^2, l >= 0. - _Wolfdieter Lang_, Aug 24 2012

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Paul Barry, On the Hurwitz Transform of Sequences, Journal of Integer Sequences, Vol. 15 (2012), #12.8.7.
E. Deutsch, L. Ferrari and S. Rinaldi, Production Matrices, Advances in Applied Mathematics, 34 (2005) pp. 101-122.
Asamoah Nkwanta and Earl R. Barnes, Two Catalan-type Riordan Arrays and their Connections to the Chebyshev Polynomials of the First Kind, Journal of Integer Sequences, Article 12.3.3, 2012.
A. Nkwanta, A. Tefera, Curious Relations and Identities Involving the Catalan Generating Function and Numbers, Journal of Integer Sequences, 16 (2013), #13.9.5.
K. Ozeki, On Melham's sum, The Fibonacci Quart. 46/47 (2008/2009), no. 2, 107-110.
Sun, Yidong; Ma, Luping Minors of a class of Riordan arrays related to weighted partial Motzkin paths. Eur. J. Comb. 39, 157-169 (2014), Table 2.2.

Cf. A000108, A113187.

Columns are : A001700, A002054, A003516, A030053, A030054, A030055, A030056.

a111418 n k = a111418_tabl !! n !! k
a111418_row n = a111418_tabl !! n
a111418_tabl = map reverse a122366_tabl
-- Reinhard Zumkeller, Mar 14 2014

Table[Binomial[2*n+1, n-k], {n,0,10}, {k,0,n}] (* G. C. Greubel, May 22 2017 *)
T[0, 0, x_, y_] := 1; T[n_, 0, x_, y_] := x*T[n - 1, 0, x, y] + T[n - 1, 1, x, y]; T[n_, k_, x_, y_] := T[n, k, x, y] = If[k < 0 || k > n, 0,
T[n - 1, k - 1, x, y] + y*T[n - 1, k, x, y] + T[n - 1, k + 1, x, y]];
Table[T[n, k, 3, 2], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, May 22 2017 *)

T(n, k) = C(2*n+1, n-k).
Sum_{k=0..n} T(n, k) = 4^n.
Sum_{k, 0<=k<=n}(-1)^k *T(n,k) = binomial(2*n,n) = A000984(n). - Philippe Deléham, Mar 22 2007
T(n,k) = sum{j=k..n, C(n,j)*2^(n-j)*C(j,floor((j-k)/2))}. - Paul Barry, Jun 06 2007
Sum_{k, k>=0} T(m,k)*T(n,k) = T(m+n,0)= A001700(m+n). - Philippe Deléham, Nov 22 2009
G.f. row polynomials: ((1+x) - (1-x)/sqrt(1-4*z))/(2*(x - (1+x)^2*z))
(see the Riordan property mentioned in a comment above). - Wolfdieter Lang, Aug 05 2014

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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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A053121 Catalan triangle (with 0's) read by rows.

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A039598 Triangle formed from odd-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x). Sometimes called Catalan's triangle.

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A064189 Triangle T(n,k), 0 <= k <= n, read by rows, defined by: T(0,0)=1, T(n,k)=0 if n < k, T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-1,k+1).

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A061554 Square table read by antidiagonals: a(n,k) = binomial(n+k, floor(k/2)).

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