A098615 -id:A098615 - cp's OEIS frontend

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

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?
A. Nkwanta and A. Tefera, Curious Relations and Identities Involving the Catalan Generating Function and Numbers, Journal of Integer Sequences, 16 (2013), #13.9.5.
Karim Ritter von Merkl, Computing colored Khovanov homology, arXiv:2505.03916 [math.QA], 2025. See p. 2.
Frank Ruskey and Mark Weston, Spherical Venn Diagrams with Involutory Isometries, Electronic Journal of Combinatorics, 18 (2011), #P191.
L. W. Shapiro, S. Getu, Wen-Jin Woan and L. C. Woodson, The Riordan Group, Discrete Appl. Maths. 34 (1991) 229-239.
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).
Mark C. Wilson, Diagonal asymptotics for products of combinatorial classes.
W.-J. Woan, Area of Catalan Paths, Discrete Math., 226 (2001), 439-444.
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

A054335 A convolution triangle of numbers based on A000984 (central binomial coefficients of even order).

Original entry on oeis.org

1, 2, 1, 6, 4, 1, 20, 16, 6, 1, 70, 64, 30, 8, 1, 252, 256, 140, 48, 10, 1, 924, 1024, 630, 256, 70, 12, 1, 3432, 4096, 2772, 1280, 420, 96, 14, 1, 12870, 16384, 12012, 6144, 2310, 640, 126, 16, 1, 48620, 65536, 51480, 28672, 12012, 3840, 924, 160, 18, 1

Offset: 0

Wolfdieter Lang, Mar 13 2000

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group. The g.f. for the row polynomials p(n,x) (increasing powers of x) is 1/(sqrt(1-4*z)-x*z).
The column sequences are for m=0..20: A000984, A000302 (powers of 4), A002457, A002697, A002802, A038845, A020918, A038846, A020920, A040075, A020922, A045543, A020924, A054337, A020926, A054338, A020928, A054339, A020930, A054340, A020932.
Riordan array (1/sqrt(1-4*x),x/sqrt(1-4*x)). - Paul Barry, May 06 2009
The matrix inverse is apparently given by deleting the leftmost column from A206022. - R. J. Mathar, Mar 12 2013

			Triangle begins:
    1;
    2,    1;
    6,    4,   1;
   20,   16,   6,   1;
   70,   64,  30,   8,  1;
  252,  256, 140,  48, 10,  1;
  924, 1024, 630, 256, 70, 12, 1; ...
Fourth row polynomial (n=3): p(3,x) = 20 + 16*x + 6*x^2 + x^3.
From _Paul Barry_, May 06 2009: (Start)
Production matrix begins
    2,   1;
    2,   2,  1;
    0,   2,  2,  1;
   -2,   0,  2,  2,  1;
    0,  -2,  0,  2,  2,  1;
    4,   0, -2,  0,  2,  2, 1;
    0,   4,  0, -2,  0,  2, 2, 1;
  -10,   0,  4,  0, -2,  0, 2, 2, 1;
    0, -10,  0,  4,  0, -2, 0, 2, 2, 1; (End)

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Paul Barry, Embedding structures associated with Riordan arrays and moment matrices, arXiv preprint arXiv:1312.0583 [math.CO], 2013.
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.

Cf. A000984, A035324, A054336.

Row sums: A026671.

T:= function(n, k)
    if k mod 2=0 then return Binomial(2*n-k, n-Int(k/2))*Binomial(n-Int(k/2),Int(k/2))/Binomial(k,Int(k/2));
    else return 4^(n-k)*Binomial(n-Int((k-1)/2)-1, Int((k-1)/2));
    fi;
  end;
Flat(List([0..10], n-> List([0..n], k-> T(n, k) ))); # G. C. Greubel, Jul 20 2019

T:= func< n, k | (k mod 2) eq 0 select Binomial(2*n-k, n-Floor(k/2))* Binomial(n-Floor(k/2),Floor(k/2))/Binomial(k,Floor(k/2)) else 4^(n-k)*Binomial(n-Floor((k-1)/2)-1, Floor((k-1)/2)) >;
[[T(n,k): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Jul 20 2019

A054335 := proc(n,k)
    if k <0 or k > n then
        0 ;
    elif type(k,odd) then
        kprime := floor(k/2) ;
        binomial(n-kprime-1,kprime)*4^(n-k) ;
    else
        kprime := k/2 ;
        binomial(2*n-k,n-kprime)*binomial(n-kprime,kprime)/binomial(k,kprime) ;
    end if;
end proc: # R. J. Mathar, Mar 12 2013
# Uses function PMatrix from A357368. Adds column 1,0,0,0,... to the left.
PMatrix(10, n -> binomial(2*(n-1), n-1)); # Peter Luschny, Oct 19 2022

Flatten[ CoefficientList[#1, x] & /@ CoefficientList[ Series[1/(Sqrt[1 - 4*z] - x*z), {z, 0, 9}], z]] (* or *)
a[n_, k_?OddQ] := 4^(n-k)*Binomial[(2*n-k-1)/2, (k-1)/2]; a[n_, k_?EvenQ] := (Binomial[n-k/2, k/2]*Binomial[2*n-k, n-k/2])/Binomial[k, k/2]; Table[a[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 08 2011, updated Jan 16 2014 *)

T(n, k) = if(k%2==0, binomial(2*n-k, n-k/2)*binomial(n-k/2,k/2)/binomial(k,k/2), 4^(n-k)*binomial(n-(k-1)/2-1, (k-1)/2));
for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jul 20 2019

def T(n, k):
    if (mod(k,2)==0): return binomial(2*n-k, n-k/2)*binomial(n-k/2,k/2)/binomial(k,k/2)
    else: return 4^(n-k)*binomial(n-(k-1)/2-1, (k-1)/2)
[[T(n,k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Jul 20 2019

a(n, 2*k+1) = binomial(n-k-1, k)*4^(n-2*k-1), a(n, 2*k) = binomial(2*(n-k), n-k)*binomial(n-k, k)/binomial(2*k, k), k >= 0, n >= m >= 0; a(n, m) := 0 if n

Column recursion: a(n, m)=2*(2*n-m-1)*a(n-1, m)/(n-m), n>m >= 0, a(m, m) := 1.

G.f. for column m: cbie(x)*(x*cbie(x))^m, with cbie(x) := 1/sqrt(1-4*x).

G.f.: 1/(1-x*y-2*x/(1-x/(1-x/(1-x/(1-x/(1-... (continued fraction). - Paul Barry, May 06 2009

Sum_{k>=0} T(n,2*k)*(-1)^k*A000108(k) = A000108(n+1). - Philippe Deléham, Jan 30 2012

Sum_{k=0..floor(n/2)} T(n-k,n-2*k) = A098615(n). - Philippe Deléham, Feb 01 2012

T(n,k) = 4*T(n-1,k) + T(n-2,k-2) for k>=1. - Philippe Deléham, Feb 02 2012

Vertical recurrence: T(n,k) = 1*T(n-1,k-1) + 2*T(n-2,k-1) + 6*T(n-3,k-1) + 20*T(n-4,k-1) + ... for k >= 1 (the coefficients 1, 2, 6, 20, ... are the central binomial coefficients A000984). - Peter Bala, Oct 17 2015

A098614 Product of Fibonacci and Catalan numbers: a(n) = A000045(n+1)*A000108(n).

Original entry on oeis.org

1, 1, 4, 15, 70, 336, 1716, 9009, 48620, 267410, 1494844, 8465184, 48466796, 280073300, 1631408400, 9568812015, 56466198990, 335002137360, 1997007404700, 11955535480350, 71850862117320, 433322055191220, 2621615826231480, 15906988165723200, 96775058652983100

Offset: 0

Paul D. Hanna, Oct 09 2004

nonn

Radius of convergence: r = (sqrt(5)-1)/8; A(r) = sqrt(2+2/sqrt(5)). More generally, given {S} such that: S(n) = b*S(n-1) + c*S(n-2), |b|>0, |c|>0, then Sum_{n>=0} S(n)*Catalan(n)*x^n = sqrt( (1-2*b*x - sqrt(1-4*b*x-16*c*x^2))/(2*b^2+8*c) )/x.

a(n) is also the number of nonnesting permutations of {1,1,2,2,...,n,n} that avoid the patterns 1223, 1332, 2113, or the patterns 1123, 1132, 2133. - Amya Luo, Dec 11 2024

			Sequence has the factored form: {1*1, 1*1, 2*2, 3*5, 5*14, 8*42, 13*132, 21*429, ...}.

Paul D. Hanna, Table of n, a(n) for n = 0..1000
Paul Barry, On the duals of the Fibonacci and Catalan-Fibonacci polynomials and Motzkin paths, arXiv:2101.10218 [math.CO], 2021.
Paul Barry and Arnauld Mesinga Mwafise, Classical and Semi-Classical Orthogonal Polynomials Defined by Riordan Arrays, and Their Moment Sequences, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.5.
Sergi Elizalde and Amya Luo, Pattern avoidance in nonnesting permutations, arXiv:2412.00336 [math.CO], 2024.

Cf. A000045, A000108, A098615, A098616, A098618, A249925.

[Fibonacci(n+1)*Catalan(n): n in [0..40]]; // G. C. Greubel, Jul 31 2024

With[{nn=30},Times@@@Thread[{Fibonacci[Range[nn]],CatalanNumber[ Range[ 0,nn-1]]}]] (* Harvey P. Dale, Nov 14 2011 *)

{a(n)=local(X=x+O(x^(n+3)), A); A = sqrt( (1-2*x - sqrt(1-4*X-16*x^2)) / (10*x^2)); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

{a(n)=binomial(2*n,n)/(n+1)*round(((1+sqrt(5))^(n+1)-(1-sqrt(5))^(n+1))/(2^(n+1)*sqrt(5)))}

[fibonacci(n+1)*catalan_number(n) for n in range(41)] # G. C. Greubel, Jul 31 2024

G.f.: A(x) = sqrt( (1-2*x - sqrt(1-4*x-16*x^2))/10 )/x.
G.f. satisfies: A(x) = sqrt( 1 + 2*x*A(x)^2 + 5*x^2*A(x)^4 ).
a(n) == 1 (mod 2) iff n = 2^k - 1 for k>=0.
n*(n+1)*a(n) -2*n*(2*n-1)*a(n-1) -4*(2*n-1)*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 17 2018
Sum_{n>=0} a(n)/8^n = 2*sqrt(2/5). - Amiram Eldar, May 06 2023

A102898 A Catalan-related transform of 3^n.

Original entry on oeis.org

1, 3, 9, 30, 99, 330, 1098, 3660, 12195, 40650, 135486, 451620, 1505358, 5017860, 16726068, 55753560, 185844771, 619482570, 2064940470, 6883134900, 22943778138, 76479260460, 254930851404, 849769504680, 2832564956814

Offset: 0

Paul Barry, Jan 17 2005

Transform of 1/(1-3*x) under the mapping g(x) -> g(x*c(x^2)), where c(x) is the g.f. of the Catalan numbers A000108. The inverse transform is h(x) -> h(x/(1+x^2)).

Maria Paola Bonacina and Nachum Dershowitz, Canonical Inference for Implicational Systems, in Automated Reasoning, Lecture Notes in Computer Science, Volume 5195/2008, Springer-Verlag.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
S. B. Ekhad and M. Yang, Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences, (2017).

Cf. A000108, A098615, A100087.

R:=PowerSeriesRing(Rationals(), 40); Coefficients(R!( 2*x/(3*Sqrt(1-4*x^2)+2*x-3) )); // G. C. Greubel, Jul 08 2022

CoefficientList[Series[2*x/(3*Sqrt[1-4*x^2]+2*x-3), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 01 2014 *)

[1]+[2*sum(k*binomial(n-1, (n-k)//2)*((n-k+1)%2)*3^k/(n+k) for k in (0..n)) for n in (1..40)] # G. C. Greubel, Jul 08 2022

G.f.: 2*x/(3*sqrt(1-4*x^2) + 2*x - 3).
a(n) = Sum_{k=0..n} k*binomial(n-1, (n-k)/2)*(1 + (-1)^(n-k))*3^k/(n+k), n > 0, with a(0) = 1.
3*n*a(n) - 10*n*a(n-1) - 12*(n-3)*a(n-2) + 40*(n-3)*a(n-3) = 0. - R. J. Mathar, Sep 21 2012
a(n) ~ 2^(n+2) * 5^(n-1) / 3^n. - Vaclav Kotesovec, Feb 01 2014

A098617 G.f. A(x) satisfies: A(xG(x)) = G(x), where G(x) is the g.f. for A098616(n) = Pell(n+1)Catalan(n).

Original entry on oeis.org

1, 2, 6, 16, 46, 128, 364, 1024, 2902, 8192, 23188, 65536, 185420, 524288, 1483096, 4194304, 11863910, 33554432, 94908420, 268435456, 759257636, 2147483648, 6074027496, 17179869184, 48592102396, 137438953472, 388736403144

Offset: 0

Paul D. Hanna, Oct 14 2004

nonn

G.f. satisfies: A(x) = x/Series_Reversion(x*G(x)), where G(x) is the g.f. for A098616 = {1*1, 2*1, 5*2, 12*5, 29*14, 70*42, 169*132, ...}.

Hankel transform is 2^n. - Paul Barry Jan 19 2011

			G.f. = 1 + 2*x + 6*x^2 + 16*x^3 + 46*x^4 + 128*x^5 + 364*x^6 + 1024*x^7 + ...

Fung Lam, Table of n, a(n) for n = 0..1000

Cf. A098616, A098615, A000129, A000108.

CoefficientList[Series[(Sqrt[1-4*x^2] + 2*x)/(1-8*x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 31 2014 *)

a(n):=2^n*sum(binomial((n-1)/2, j),j,0,n/2); /* Vladimir Kruchinin, May 18 2011 */

a(n)=polcoeff((sqrt(1-4*x^2+x^2*O(x^n))+2*x)/(1-8*x^2),n)

G.f.: (sqrt(1-4*x^2) + 2*x)/(1-8*x^2).
a(2*n+1) = 2*8^n.
a(n) = sum{k=0..floor((n+1)/2), (C(n,k)-C(n,k-1))*A000129(n-2k+1)}. - Paul Barry Jan 19 2011
a(n) = 2^n*sum(j=0..n/2, binomial((n-1)/2,j)). - Vladimir Kruchinin, May 18 2011
a(n) = Sum_{k, 0<=k<=n} A201093(n,k)*2^k. - Philippe Deléham, Nov 27 2011
G.f.: 1/(1-2x/(1-x/(1+x/(1+x/(1-x/(1-x/(1+x/(1+x/(1-x/(1-... (continued fraction). - Philippe Deléham, Nov 27 2011
Recurrence: (n+6)*a(n)=256*(n+1)*a(n-6)-128*(n+3)*a(n-4)+4*(5*n+23)*a(n-2), for even n. - Fung Lam, Mar 31 2014
Recurrence: n*a(n) = 12*(n-1)*a(n-2) - 32*(n-3)*a(n-4). - Vaclav Kotesovec, Mar 31 2014
Asymptotic approximation: a(n) ~ (4/sqrt(2))^n/sqrt(2)+2^(n+1)/sqrt(2*Pi*n^3), for even n. - Fung Lam, Mar 31 2014
0 = a(n) * (+64*a(n+1) - 8*a(n+3)) + a(n+2) * (-8*a(n+1) + a(n+3)) if n>=0. - Michael Somos, Apr 07 2014

A371458 Expansion of 1/(1 - x/(1 - 9*x^3)^(1/3)).

Original entry on oeis.org

1, 1, 1, 1, 4, 7, 10, 31, 61, 100, 274, 565, 1000, 2551, 5380, 10000, 24376, 52018, 100000, 236389, 507706, 1000000, 2313346, 4986178, 10000000, 22773334, 49180165, 100000000, 225092416, 486575935, 1000000000, 2231117230, 4824998773, 10000000000

Offset: 0

Seiichi Manyama, Jun 07 2024

nonn

Cf. A362206, A371456.

Cf. A066872, A098615, A373510, A373511.

A371458 := proc(n)
    add(9^k*binomial(n/3-1,k),k=0..floor(n/3)) ;
end proc:
seq(A371458(n),n=0..70) ; # R. J. Mathar, Jun 07 2024

a(n) = sum(k=0, n\3, 9^k*binomial(n/3-1, k));

a(3*n) = 10^(n-1) for n > 0.
a(n) = Sum_{k=0..floor(n/3)} 9^k * binomial(n/3-1,k).
D-finite with recurrence (n-1)*(n-2)*a(n) +4*(-7*n^2+48*n-86)*a(n-3) +9*(29*n-141)*(n-6)*a(n-6) -810*(n-6)*(n-9)*a(n-9)=0. - R. J. Mathar, Jun 07 2024
a(n) == 1 (mod 3). - Seiichi Manyama, Jun 11 2024

A373509 Expansion of 1/(1 - x/(1 - 8*x^4)^(1/4)).

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 5, 7, 9, 21, 37, 57, 81, 169, 301, 485, 729, 1431, 2549, 4211, 6561, 12411, 22045, 36975, 59049, 109047, 193029, 326923, 531441, 965511, 1703469, 2903851, 4782969, 8590149, 15111573, 25875081, 43046721, 76670441, 134539837, 231087525

Offset: 0

Seiichi Manyama, Jun 07 2024

nonn

Cf. A098615, A373512.

a(n) = sum(k=0, n\4, 8^k*binomial(n/4-1, k));

a(4*n) = 9^(n-1) for n > 0.

a(n) = Sum_{k=0..floor(n/4)} 8^k * binomial(n/4-1,k).

A111959 Renewal array for aerated central binomial coefficients.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 0, 4, 0, 1, 6, 0, 6, 0, 1, 0, 16, 0, 8, 0, 1, 20, 0, 30, 0, 10, 0, 1, 0, 64, 0, 48, 0, 12, 0, 1, 70, 0, 140, 0, 70, 0, 14, 0, 1, 0, 256, 0, 256, 0, 96, 0, 16, 0, 1, 252, 0, 630, 0, 420, 0, 126, 0, 18, 0, 1, 0, 1024, 0, 1280, 0, 640, 0, 160, 0, 20, 0, 1, 924, 0, 2772, 0

Offset: 0

Paul Barry, Aug 23 2005

Row sums are A098615.
Binomial transform (product with C(n,k)) is A111960.
Diagonal sums are A026671 (with interpolated zeros).
Inverse is (1/sqrt(1+4x^2),x/sqrt(1+4x^2)), or (sqrt(-1))^(n-k)*T(n,k). [corrected by Peter Bala, Aug 13 2021]
The Riordan array (1,x/sqrt(1-4*x^2)) is the same array with an additional column of zeros (besides the top element 1) added to the left. - Vladimir Kruchinin, Feb 17 2011

			From _Peter Bala_, Aug 13 2021: (Start)
Triangle begins
  1;
  0,  1;
  2,  0, 1;
  0,  4, 0, 1;
  6,  0, 6, 0, 1;
  0, 16, 0, 8, 0, 1;
Infinitesimal generator begins
  0;
  0, 0;
  2, 0, 0;
  0, 4, 0, 0;
  0, 0, 6, 0, 0;
  0, 0, 0, 8, 0, 0; (End)

Paul Barry, On the duals of the Fibonacci and Catalan-Fibonacci polynomials and Motzkin paths, arXiv:2101.10218 [math.CO], 2021.
Vladimir Kruchinin, Composition of ordinary generating functions, arXiv:1009.2565 [math.CO], 2010.

Cf. A054335, A038231, A046521.

Riordan array (1/sqrt(1-4x^2), x/sqrt(1-4x^2)); number triangle T(n, k)=(1+(-1)^(n-k))*binomial((n-1)/2, (n-k)/2)*2^(n-k)/2.
G.f.: 1/(1-xy-2x^2/(1-x^2/(1-x^2/(1-x^2/(1-.... (continued fraction). - Paul Barry, Jan 28 2009
From Peter Bala, Aug 13 2021: (Start)
T(2*n,2*k) = A046521(n,k); T(2*n+1,2*k+1) = A038231(n,k).
The row entries, read from right to left, are the coefficients in the n-th order Taylor polynomial of (sqrt(1 + 4*x^2))^((n-1)/2) at x = 0.
The infinitesimal generator of this array has the sequence [2, 4, 6, 8, 10, ...] on the second subdiagonal below the main diagonal and zeros elsewhere.
The m-th power of the array is the Riordan array (1/sqrt(1 - 4*m*x^2), x/sqrt(1 - 4*m*x^2)) with entries given by sqrt(m)^(n-k)*T(n,k). (End)

A373510 Expansion of 1/(1 - x/(1 - 25*x^5)^(1/5)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 6, 11, 16, 21, 26, 106, 211, 341, 496, 676, 2256, 4611, 7866, 12146, 17576, 51781, 106761, 188266, 302671, 456976, 1236306, 2552661, 4602416, 7620071, 11881376, 30218956, 62278561, 114056566, 193134346, 308915776, 749942856, 1540351961

Offset: 0

Seiichi Manyama, Jun 07 2024

nonn

Cf. A066872, A098615, A371458, A373511.

a(n) = sum(k=0, n\5, 25^k*binomial(n/5-1, k));

a(5*n) = 26^(n-1) for n > 0.
a(n) = Sum_{k=0..floor(n/5)} 25^k * binomial(n/5-1,k).
a(n) == 1 (mod 5).

A098619 G.f. A(x) satisfies: A(xG098618(x)) = G098618(x), where G098618 is the g.f. for A098618(n) = A007482(n)Catalan(n).

Original entry on oeis.org

1, 3, 13, 51, 213, 867, 3589, 14739, 60853, 250563, 1033605, 4259571, 17565909, 72412707, 298586661, 1231016019, 5075753589, 20927272323, 86286346693, 355763629491, 1466857936405, 6047981701347, 24936516122469, 102815688922899, 423920292507061, 1747866711689283, 7206641564551429

Offset: 0

Paul D. Hanna, Oct 14 2004

nonn

G.f. satisfies: A(x) = x/(series reversion of x*G098618(x)), where G098618 is the g.f. for A098618 = {1*1,3*1,11*2,39*5,139*14,495*42,1763*132,...}.

Vincenzo Librandi, Table of n, a(n) for n = 0..300

Cf. A098618, A098615, A098617, A007482, A000108.

Flatten[{1,3,13,51,Table[17^(n/2)*(1/2+1/2*(-1)^n + 3/34*Sqrt[17]*(1-(-1)^n) + Sum[(-1)^j*(4/17 + Sum[Binomial[2*k-1,k-1]*2^(k+3)/ ((k+1)*17^(k+1)), {k,1,Floor[(j-1)/2]}]),{j,3,n-1}]),{n,4,20}]}] (* Vaclav Kotesovec, Oct 29 2012 *)

a(n)=polcoeff((sqrt(1-8*x^2+x^2*O(x^n))+3*x)/(1-17*x^2),n);

x='x+O('x^66); Vec((sqrt(1-8*x^2) + 3*x)/(1-17*x^2)) \\ Joerg Arndt, May 12 2013

G.f.: (sqrt(1-8*x^2) + 3*x)/(1-17*x^2).
a(2*n+1) = 3*17^n.
Recurrence: n*a(n) = (25*n-24)*a(n-2) - 136*(n-3)*a(n-4). - Vaclav Kotesovec, Oct 29 2012

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cp's OEIS Frontend

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

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A054335 A convolution triangle of numbers based on A000984 (central binomial coefficients of even order).

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A098614 Product of Fibonacci and Catalan numbers: a(n) = A000045(n+1)*A000108(n).

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A102898 A Catalan-related transform of 3^n.

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A098617 G.f. A(x) satisfies: A(x*G(x)) = G(x), where G(x) is the g.f. for A098616(n) = Pell(n+1)*Catalan(n).

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A371458 Expansion of 1/(1 - x/(1 - 9*x^3)^(1/3)).

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A098617 G.f. A(x) satisfies: A(xG(x)) = G(x), where G(x) is the g.f. for A098616(n) = Pell(n+1)Catalan(n).

A098619 G.f. A(x) satisfies: A(xG098618(x)) = G098618(x), where G098618 is the g.f. for A098618(n) = A007482(n)Catalan(n).