A092392 -id:A092392 - cp's OEIS frontend

A176287 Diagonal sums of number triangle A092392.

1, 2, 7, 23, 81, 291, 1066, 3955, 14818, 55937, 212428, 810664, 3106167, 11942261, 46047897, 178000950, 689580319, 2676598447, 10406929687, 40525045518, 158022343991, 616950024334, 2411395005316, 9434753907065, 36948692202031

Offset: 0

Paul Barry, Apr 14 2010

Hankel transform is A176288.

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

Cf. A000108, A000984, A092392, A132364.

List([0..25], n-> Sum([0..Int(n/2)], k-> Binomial(2*n-3*k, n-k) )); # G. C. Greubel, Nov 25 2019

[ &+[Binomial(2*n-3*k, n-k): k in [0..Floor(n/2)]] : n in [0..25]]; // G. C. Greubel, Nov 25 2019

seq( add(binomial(2*n-3*k, n-k), k=0..floor(n/2)) , n=0..25); # G. C. Greubel, Nov 25 2019

CoefficientList[Series[2/(Sqrt[1-4*x]*(2-x+x*Sqrt[1-4*x])), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 04 2014 *)
a[n_]:= Sum[Binomial[2*n-3*k, n-k], {k, 0, Floor[n/2]}]; Table[a[n], {n,0,25}] (* G. C. Greubel, Oct 19 2016 *)

a(n) = sum(k=0, n\2, binomial(2*n-3*k, n-k)); \\ Michel Marcus, Oct 20 2016

[sum(binomial(2*n-3*k, n-k) for k in (0..floor(n/2))) for n in (0..25)] # G. C. Greubel, Nov 25 2019

G.f.: 1/(sqrt(1-4*x)*(1-x^2*c(x))) = 2/(sqrt(1-4*x)*(2-x+x*sqrt(1-4*x))), c(x) the g.f. of A000108.
a(n) = Sum_{k=0..floor(n/2)} C(2n-3k,n-k).
a(n) = Sum_{k=0..n} A000984(k)*A132364(n-k).
D-finite with recurrence: 2*n*a(n) +(6-11*n)*a(n-1) +(13*n-16)*a(n-2) +2*(5-n)*a(n-3) +3*(2-3*n)*a(n-4) +2*(2*n-5)*a(n-5)=0. - R. J. Mathar, Nov 15 2012 [Verified with Maple's FindRE and MinimalRecurrence functions, Georg Fischer, Nov 03 2022]
a(n) ~ 2^(2*n+3) / (7*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 04 2014

A158793 Triangle read by rows: product of A130595 and A092392 considered as infinite lower triangular arrays.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 7, 4, 1, 1, 19, 9, 5, 1, 1, 51, 26, 11, 6, 1, 1, 141, 70, 34, 13, 7, 1, 1, 393, 197, 92, 43, 15, 8, 1, 1, 1107, 553, 265, 117, 53, 17, 9, 1, 1, 3139, 1570, 751, 346, 145, 64, 19, 10, 1, 1, 8953, 4476, 2156, 991, 441, 176, 76, 21, 11, 1, 1

Offset: 0

Gary W. Adamson & Roger L. Bagula, Mar 26 2009

Riordan array (f(x), x*g(x)) where f(x) is the g.f. of A002426 and where g(x) is the g.f. of A005043. - Philippe Deléham, Dec 05 2009

Matrix product P * Q * P^(-1), where P denotes Pascal's triangle A007318 and Q denotes A061554 (formed from P by sorting the rows into descending order). Cf. A158815 and A171243. - Peter Bala, Jul 13 2021

			First rows of the triangle:
     1;
     1,    1;
     3,    1,    1;
     7,    4,    1,   1;
    19,    9,    5,   1,   1;
    51,   26,   11,   6,   1,   1;
   141,   70,   34,  13,   7,   1,  1;
   393,  197,   92,  43,  15,   8,  1,  1;
  1107,  553,  265, 117,  53,  17,  9,  1,  1;
  3139, 1570,  751, 346, 145,  64, 19, 10,  1, 1;
  8953, 4476, 2156, 991, 441, 176, 76, 21, 11, 1, 1;

T(n, 0) = A002426(n), A005773 (row sums).

Cf. A046899, A007318, A158815, A171243.

A158793 := proc (n, k)
  add((-1)^(n+j)*binomial(n, j)*binomial(2*j-k, j-k), j = k..n);
end proc:
seq(seq(A158793(n, k), k = 0..n), n = 0..10); # Peter Bala, Jul 13 2021

T[n_, k_] := (-1)^(k + n) Binomial[n, k] HypergeometricPFQ[{k/2 + 1/2, k/2 + 1, k - n}, {k + 1, k + 1}, 4];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Peter Luschny, Jul 17 2021 *)

T(n, m) = Sum_{k=m..n-1} A130595(n,k) * A092392(k+1,m+1), with the triangular interpretation of A092392.
Conjecture: T(n,1) = A113682(n-1). - R. J. Mathar, Oct 06 2009
Sum_{k=0..n} T(n,k)*x^k = A002426(n), A005773(n+1), A000244(n), A126932(n) for x = 0,1,2,3 respectively. - Philippe Deléham, Dec 03 2009
T(n, k) = (-1)^(k + n) binomial(n, k) hypergeom([k/2 + 1/2, k/2 + 1, k - n], [k + 1, k + 1], 4). - Peter Luschny, Jul 17 2021

Simplified definition from R. J. Mathar, Oct 06 2009

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

A046899 Triangle in which n-th row is {binomial(n+k,k), k=0..n}, n >= 0.

Original entry on oeis.org

1, 1, 2, 1, 3, 6, 1, 4, 10, 20, 1, 5, 15, 35, 70, 1, 6, 21, 56, 126, 252, 1, 7, 28, 84, 210, 462, 924, 1, 8, 36, 120, 330, 792, 1716, 3432, 1, 9, 45, 165, 495, 1287, 3003, 6435, 12870, 1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, 1, 11, 66, 286, 1001

Offset: 0

N. J. A. Sloane

C(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0) and (0,1). - Joerg Arndt, Jul 01 2011
Row sums are A001700.
T(n, k) is also the number of order-preserving full transformations (of an n-chain) of waist k (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Oct 02 2008
If T(r,c), r=0,1,2,..., c=1,2,...,(r+1), are the triangle elements, then for r > 0, T(r,c) = binomial(r+c-1,c-1) = M(r,c) is the number of monotonic mappings from an ordered set of r elements into an ordered set of c elements. For example, there are 15 monotonic mappings from an ordered set of 4 elements into an ordered set of 3 elements. For c > r+1, use the identity M(r,c) = M(c-1,r+1) = T(c-1,r+1). For example, there are 210 monotonic mappings from an ordered set of 4 elements into an ordered set of 7 elements, because M(4,7) = T(6,5) = 210. Number of monotonic endomorphisms in a set of r elements, M(r,r), therefore appear on the second diagonal of the triangle which coincides with A001700. - Stanislav Sykora, May 26 2012
Start at the origin. Flip a fair coin to determine steps of (1,0) or (0,1). Stop when you are a (perpendicular) distance of n steps from the x axis or the y axis. For k = 0,1,...,n-1, C(n-1,k)/2^(n+k) is the probability that you will stop on the point (n,k). This is equal to the probability that you will stop on the point (k,n). Hence, Sum_{k=0..n} C(n,k)/2^(n+k) = 1. - Geoffrey Critzer, May 13 2017

			The triangle is the lower triangular part of the square array:
  1|  1,  1,   1,   1,    1,    1,     1,     1,     1, ...
  1,  2|  3,   4,   5,    6,    7,     8,     9,    10, ...
  1,  3,  6|  10,  15,   21,   28,    36,    45,    55, ...
  1,  4, 10,  20|  35,   56,   84,   120,   165,   220, ...
  1,  5, 15,  35,  70|  126,  210,   330,   495,   715, ...
  1,  6, 21,  56, 126,  252|  462,   792,  1287,  2002, ...
  1,  7, 28,  84, 210,  462,  924|  1716,  3003,  5005, ...
  1,  8, 36, 120, 330,  792, 1716,  3432|  6435, 11440, ...
  1,  9, 45, 165, 495, 1287, 3003,  6435, 12870| 24310, ...
  1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620| ...
The array read by antidiagonals gives the binomial triangle.
From _Reinhard Zumkeller_, Jul 27 2012: (Start)
Take the first n elements of the n-th diagonal (NW to SE) of left half of Pascal's triangle and write it as n-th row on the triangle on the right side, see above
  0:                 1                    1
  1:               1   _                  1  2
  2:             1   2  __                1  3  6
  3:           1   3  __  __              1  4 10 20
  4:         1   4   6  __  __            1  5 15 35 70
  5:       1   5  10  __  __  __          1  6 21 56 .. ..
  6:     1   6  15  20  __  __  __        1  7 28 .. .. .. ..
  7:   1   7  21  35  __  __  __  __      1  8 .. .. .. .. .. ..
  8: 1   8  28  56  70  __  __  __  __    1 .. .. .. .. .. .. .. .. (End)

H. W. Gould, A class of binomial sums and a series transform, Utilitas Math., 45 (1994), 71-83.

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Karl Dilcher and Maciej Ulas, Arithmetic properties of polynomial solutions of the Diophantine equation P(x)x^(n+1)+Q(x)(x+1)^(n+1)=1, arXiv:1909.11222 [math.NT], 2019. See Qn(x) Table 1 p. 2.
H. W. Gould, A class of binomial sums and a series transform, Utilitas Math., 45 (1994), 71-83. (Annotated scanned copy)
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving partial transformations, Journal of Algebra 278, (2004), 342-359.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving full transformations, Semigroup Forum 72 (2006), 51-62.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A000108, A046900, A001700, A007318, A034868, A092392, A239103, A178300.

import Data.List (transpose)
a046899 n k = a046899_tabl !! n !! k
a046899_row n = a046899_tabl !! n
a046899_tabl = zipWith take [1..] $ transpose a007318_tabl
-- Reinhard Zumkeller, Jul 27 2012

/* As triangle */ [[Binomial(n+k, n): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Aug 18 2015

for n from 0 to 10 do seq( binomial(n+m,n), m = 0 .. n) od; # Zerinvary Lajos, Dec 09 2007

t[n_, k_] := Binomial[n + k, n]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Aug 12 2013 *)

/* same as in A092566 but use */
steps=[[1, 0], [1, 0] ];
/* Joerg Arndt, Jul 01 2011 */

for n in (0..9):
    print([multinomial(n, k) for k in (0..n)]) # Peter Luschny, Dec 24 2020

T(n,k) = A092392(n,n-k), k = 0..n. - Reinhard Zumkeller, Jul 27 2012
T(n,k) = A178300(n,k), n>0, k = 1..n. - L. Edson Jeffery, Jul 23 2014
T(n,k) = (n + 1)*hypergeom([-n, 1 - k], [2], 1). - Peter Luschny, Jan 09 2022
T(n,k) = hypergeom([-n, -k], [1], 1). - Peter Luschny, Mar 21 2024
G.f.: 1/((1-2x*y*C(x*y))*(1-x*C(x*y))), where C(x) is the g.f. for A000108, the Catalan numbers. - Michael D. Weiner, Jul 31 2024

More terms from James Sellers

A100100 Triangle T(n,k) = binomial(2*n-k-1, n-k) read by rows.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 10, 6, 3, 1, 35, 20, 10, 4, 1, 126, 70, 35, 15, 5, 1, 462, 252, 126, 56, 21, 6, 1, 1716, 924, 462, 210, 84, 28, 7, 1, 6435, 3432, 1716, 792, 330, 120, 36, 8, 1, 24310, 12870, 6435, 3003, 1287, 495, 165, 45, 9, 1, 92378, 48620, 24310, 11440, 5005, 2002

Offset: 0

Paul Barry, Nov 08 2004

Sometimes called a Catalan triangle, although there are many other triangles that carry that name - see A009766, A008315, A028364, A033184, A053121, A059365, A062103.
Number of nodes of outdegree k in all ordered trees with n edges. Equivalently, number of ascents of length k in all Dyck paths of semilength n. Example: T(3,2) = 3 because the Dyck paths of semilength 3 are UDUDUD, UD(UU)DD, (UU)DDUD, (UU)DUDD and UUUDDD, where U = (1,1), D = (1,-1), the ascents of length 2 being shown between parentheses. - Emeric Deutsch, Nov 19 2006
Riordan array (f(x), x*g(x)) where f(x) is the g.f. of A088218 and g(x) is the g.f. of A000108. - Philippe Deléham, Jan 23 2010
T(n,k) is the number of nonnegative paths of upsteps U = (1,1) and downsteps D = (1,-1) of length 2*n with k returns to  ground level, the horizontal line through the initial vertex. Example: T(2,1) = 2 counts UDUU, UUDD. Also, T(n,k) = number of these paths whose last descent has length k, that is, k downsteps follow the last upstep. Example: T(2,1) = 2 counts UUUD, UDUD. - David Callan, Nov 21 2011
Belongs to the hitting-time subgroup of the Riordan group. Multiplying this triangle by the square Pascal matrix gives A092392 read as a square array. See the example below. - Peter Bala, Nov 03 2015

			From _Paul Barry_, Mar 15 2010: (Start)
Triangle begins in row n=0 with columns 0<=k<=n as:
    1;
    1,   1;
    3,   2,   1;
   10,   6,   3,  1;
   35,  20,  10,  4,  1;
  126,  70,  35, 15,  5, 1;
  462, 252, 126, 56, 21, 6, 1;
Production matrix begins
  1, 1;
  2, 1, 1;
  3, 1, 1, 1;
  4, 1, 1, 1, 1;
  5, 1, 1, 1, 1, 1;
  6, 1, 1, 1, 1, 1, 1;
  7, 1, 1, 1, 1, 1, 1, 1;
(End)
A092392 as a square array = A100100 * square Pascal matrix:
/1   1  1  1 ...\   / 1          \/1 1  1  1 ...\
|2   3  4  5 ...|   | 1 1        ||1 2  3  4 ...|
|6  10 15 21 ...| = | 3 2 1      ||1 3  6 10 ...|
|20 35 56 84 ...|   |10 6 3 1    ||1 4 10 20 ...|
|70 ...         |   |35 ...      ||1 ...        |
- _Peter Bala_, Nov 03 2015

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
Gi-Sang Cheon, Hana Kim, and Louis W. Shapiro, An algebraic structure for Faber polynomials, Lin. Alg. Applic. 433 (2010) 1170-1179.
Tian-Xiao He and Renzo Sprugnoli, Sequence characterization of Riordan arrays, Discrete Math. 309 (2009), no. 12, 3962-3974.

Row sums are A000984. Equivalent to A092392, to which A088218 has been added as a first column. Columns include A088218, A000984, A001700, A001791, A002054, A002694. Diagonal sums are A100217. Matrix inverse is A100218.

Cf. A059481 (mirrored). Cf. A033184, A094527, A113955.

a100100 n k = a100100_tabl !! n !! n
a100100_row n = a100100_tabl !! n
a100100_tabl = [1] : f a092392_tabl where
   f (us : wss'@(vs : wss)) = (vs !! 1 : us) : f wss'
-- Reinhard Zumkeller, Jan 15 2014

/* As triangle */ [[Binomial(2*n - k - 1, n - k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Nov 21 2018

A100100 := proc(n,k)
    binomial(2*n-k-1,n-1) ;
end proc:
seq(seq(A100100(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Feb 06 2015

Flatten[Table[Binomial[2 n - k - 1, n - k], {n, 0, 11}, {k, 0, n}]] (* Vincenzo Librandi, Nov 21 2018 *)

T(n,k)=binomial(2*n-k-1,n-k) \\ Charles R Greathouse IV, Jan 16 2012

From Peter Bala, Sep 06 2015: (Start)
Matrix product A094527 * P^(-1) = A113955 * P^(-2), where P denotes Pascal's triangle A007318.
Essentially, the logarithmic derivative of A033184. (End)
Let column(k) = [T(n, k), n >= k], then the generating function for column(k) is (2/(sqrt(1-4*x)+1))^(k-1)/sqrt(1-4*x). - Peter Luschny, Mar 19 2021
O.g.f. row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k, i.e. o.g.f. of the triangle, G(z,x) = 1/((2 - c(z))*(1 - x*z*c(z))), with c the o.g.f. of A000108 (Catalan). See the Riordan coomment by Philippe Deléham above. - Wolfdieter Lang, Apr 06 2021

A100257 Triangle of expansions of 2^(k-1)*x^k in terms of T(n,x), in descending degrees n of T, with T the Chebyshev polynomials.

Original entry on oeis.org

1, 1, 0, 1, 0, 1, 1, 0, 3, 0, 1, 0, 4, 0, 3, 1, 0, 5, 0, 10, 0, 1, 0, 6, 0, 15, 0, 10, 1, 0, 7, 0, 21, 0, 35, 0, 1, 0, 8, 0, 28, 0, 56, 0, 35, 1, 0, 9, 0, 36, 0, 84, 0, 126, 0, 1, 0, 10, 0, 45, 0, 120, 0, 210, 0, 126, 1, 0, 11, 0, 55, 0, 165, 0, 330, 0, 462, 0, 1, 0, 12, 0, 66, 0, 220, 0

Offset: 0

Ralf Stephan, Nov 13 2004

			x^0 = T(0,x)
x^1 = T(1,x) + 0T(0,x)
2x^2 = T(2,x) + 0T(1,x) + 1T(0,x)
4x^3 = T(3,x) + 0T(2,x) + 3T(1,x) + 0T(0,x)
8x^4 = T(4,x) + 0T(3,x) + 4T(2,x) + 0T(1,x) + 3T(0,x)
16x^5 = T(5,x) + 0T(4,x) + 5T(3,x) + 0T(2,x) + 10T(1,x) + 0T(0,x)

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.

Vincenzo Librandi, Table of n, a(n) for n = 0..6104
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].
H. J. Brothers, Pascal's Prism: Supplementary Material.
Daniel J. Greenhoe, Frames and Bases: Structure and Design, Version 0.20, Signal Processing ABCs series (2019) Vol. 4, see page 175.
Daniel J. Greenhoe, A Book Concerning Transforms, Version 0.10, Signal Processing ABCs series (2019) Vol. 5, see page 97.
Index entries for sequences related to Chebyshev polynomials.

Without zeros: A008311. Row sums are A011782. Cf. A092392.

Diagonals are (with interleaved zeros) twice A001700, A001791, A002054, A002694, A003516, A002696, A030053, A004310, A030054, A004311, A030055, A004312, A030056, A004313.

a[k_, n_] := If[k == 1, 1, If[EvenQ[n] || k < 0 || n > k, 0, If[n >= k - 1, Binomial[2*Floor[k/2], Floor[k/2]]/2, Binomial[k - 1, Floor[n/2]]]]];
Table[a[k, n], {k, 1, 13}, {n, 1, k}] // Flatten (* Jean-François Alcover, May 04 2017, translated from PARI *)

a(k,n)=if(k==1,1,if(n%2==0||k<0||n>k,0,if(n>=k-1,binomial(2*floor(k/2),floor(k/2))/2,binomial(k-1,floor(n/2)))))

A104684 Triangle read by rows: T(n,k) is the number of lattice paths from (0,0) to (n,n) using steps E=(1,0), N=(0,1) and D=(1,1) (i.e., bilateral Schroeder paths), having k D=(1,1) steps.

Original entry on oeis.org

1, 2, 1, 6, 6, 1, 20, 30, 12, 1, 70, 140, 90, 20, 1, 252, 630, 560, 210, 30, 1, 924, 2772, 3150, 1680, 420, 42, 1, 3432, 12012, 16632, 11550, 4200, 756, 56, 1, 12870, 51480, 84084, 72072, 34650, 9240, 1260, 72, 1, 48620, 218790, 411840, 420420, 252252

Offset: 0

Emeric Deutsch, Apr 24 2005

Row sums are the central Delannoy numbers (A001850). T(n,0)=A000984(n) (the central binomial numbers). Alternating row sums = 1 See the Bataille link.
  Row reversed version of A063007.
Another version of [0, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] = 1; 0, 1; 0, 2, 1; 0, 6, 6, 1; 0, 20, 30, 12, 1; 0, 70, 140, 90, 20, 1; ..., where DELTA is the operator defined in A084938. - Philippe Deléham, Apr 25 2005
Terms in row n are the coefficients of the Legendre polynomial P(n,2x+1) with decreasing powers of x.
Coefficient array of x^n*Legendre_P(n,2/x+1). - Paul Barry, Apr 19 2009

			T(2,1)=6 because we have NED, NDE, EDN, END, DEN and DNE.
The triangle T(n, k) begins:
n\k    0     1     2     3     4    5    6  7 8 ...
0:     1
1:     2     1
2:     6     6     1
3:    20    30    12     1
4:    70   140    90    20    1
5:   252   630   560   210   30     1
6:   924  2772  3150  1680  420    42    1
7:  3432 12012 16632 11550 4200   756   56  1
8: 12870 51480 84084 72072 34650 9240 1260 72 1
...
row n=9: 48620 218790 411840 420420 252252 90090 18480 1980 90 1,
row n=10: 184756 923780 1969110 2333760 1681680 756756 210210 34320 2970 110 1.
... reformatted by _Wolfdieter Lang_, Sep 11 2016

T. D. Noe, Rows n=0..100 of triangle, flattened
Michel Bataille, Quickie Q.944, Maths. Magazine, 77, No. 4, p. 321, Answer A.944, Maths. Magazine, 77, No. 4, p. 327.
H. J. Brothers, Pascal's Prism: Supplementary Material.
Ömür Deveci and Anthony G. Shannon, Some aspects of Neyman triangles and Delannoy arrays, Mathematica Montisnigri, Volume L, 2021.
Robert A. Sulanke, Objects Counted by the Central Delannoy Numbers, Journal of Integer Sequences, Volume 6, 2003, Article 03.1.5.

Cf. A000984, A001850, A063007.

a104684 n k = a104684_tabl !! n !! k
a104684_row n = a104684_tabl !! n
a104684_tabl = map (map abs) $
               zipWith (zipWith (*)) a130595_tabl a092392_tabl
-- Reinhard Zumkeller, Dec 20 2013

T:=(n,k)->binomial(n,k)*binomial(2*n-k,n): for n from 0 to 9 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

T[n_, k_] := Binomial[n, k] Binomial[2n-k, n];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 19 2018 *)

T(n, k) = binomial(n, k)*binomial(2n-k, n) (0 <= k <= n).
G.f.: G(t, z) = 1/sqrt((1-tz)^2 - 4z).
T(n,k) = binomial(2(n-k),n-k)*binomial(2n-k,k). - Paul Barry, Mar 14 2006
T(2n,n) = C(2n,n)*C(3n,n) = C(n,n)*C(2n,n)*C(3n,n) = A006480(n). - Paul Barry, Mar 14 2006
G.f.: 1/(1-xy-2x/(1-xy-x/(1-xy-x/(1-xy-x/(1-xy-x... (continued fraction). - Paul Barry, Jan 06 2009
T(n,k) = Sum_{j=0..n} C(n,j)^2*C(j,k). - Paul Barry, May 28 2009
T(n,k) = [x^k]F(-n,-n;1;1+x). - Paul Barry, Oct 05 2010
T(n,k) = (n-k+1)*A060693(n,k). - Peter Luschny, May 17 2011
T(n,k) = A054142(n,k)*A000984(n-k). - Philippe Deléham, Nov 19 2011.
T(n,k) = abs(A130595(n,k)*A092392(n,k)). - Reinhard Zumkeller, Dec 20 2013

A264772 Triangle T(n,k) = binomial(3n - 2k, 2*n - k), 0 <= k <= n.

Original entry on oeis.org

1, 3, 1, 15, 4, 1, 84, 21, 5, 1, 495, 120, 28, 6, 1, 3003, 715, 165, 36, 7, 1, 18564, 4368, 1001, 220, 45, 8, 1, 116280, 27132, 6188, 1365, 286, 55, 9, 1, 735471, 170544, 38760, 8568, 1820, 364, 66, 10, 1, 4686825, 1081575, 245157, 54264, 11628, 2380, 455, 78, 11, 1

Offset: 0

Peter Bala, Nov 24 2015

Riordan array (f(x), x*g(x)), where g(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + ... is the o.g.f. for A001764 and f(x) = g(x)/(3 - 2*g(x)) = 1 + 3*x + 15*x^2 + 84*x^3 + 495*x^4 + ... is the o.g.f. for A005809.
The even numbered columns give the Riordan array A119301, the odd numbered columns give the Riordan array A144484. A159841 is the array formed from columns 1,4,7,10,....
More generally, if R = (R(n,k))n,k>=0 is a proper Riordan array, m is a nonnegative integer and a > b are integers then the array with (n,k)-th element R((m + 1)*n - a*k, m*n - b*k) is also a Riordan array (not necessarily proper). Here we take R as Pascal's triangle and m = a = 2, b = 1. See A092392, A264773, A264774 and A113139 for further examples.

			Triangle begins
.n\k.|......0.....1....2....3...4..5...6..7...
----------------------------------------------
..0..|      1
..1..|      3     1
..2..|     15     4    1
..3..|     84    21    5    1
..4..|    495   120   28    6   1
..5..|   3003   715  165   36   7  1
..6..|  18564  4368 1001  220  45  8  1
..7..| 116280 27132 6188 1365 286 55  9  1
...

Michael De Vlieger, Table of n, a(n) for n = 0..11475
Peter Bala, A 4-parameter family of embedded Riordan arrays
Paul Barry, On the halves of a Riordan array and their antecedents, arXiv:1906.06373 [math.CO], 2019.
E. Lebensztayn, On the asymptotic enumeration of accessible automata, Discrete Mathematics and Theoretical Computer Science, Vol. 12, No.3, 2010, 75-80, Section 2.
R. Sprugnoli, An Introduction to Mathematical Methods in Combinatorics, CreateSpace Independent Publishing Platform 2006, Section 5.6, ISBN-13: 978-1502925244.

Cf. A005809 (column 0), A045721 (column 1), A025174 (column 2), A004319 (column 3), A236194 (column 4), A013698 (column 5). Cf. A001764, A007318, A092392, A119301 (C(3n-k,2n)), A144484 (C(3n+1-k,2n+1)), A159841 (C(3n+1,2n+k+1)), A264773, A264774.

/* As triangle */ [[Binomial(3*n-2*k, n-k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Dec 02 2015

A264772:= proc(n,k) binomial(3*n - 2*k, 2*n - k); end proc:
seq(seq(A264772(n,k), k = 0..n), n = 0..10);

Table[Binomial[3 n - 2 k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 01 2015 *)

T(n,k) = binomial(3*n - 2*k, n - k).

O.g.f.: f(x)/(1 - t*x*g(x)), where f(x) = Sum_{n >= 0} binomial(3*n,n)*x^n and g(x) = Sum_{n >= 0} 1/(2*n + 1)*binomial(3*n,n)*x^n.

A119301 Triangle read by rows: T(n,k) = binomial(3*n-k,n-k).

Original entry on oeis.org

1, 3, 1, 15, 5, 1, 84, 28, 7, 1, 495, 165, 45, 9, 1, 3003, 1001, 286, 66, 11, 1, 18564, 6188, 1820, 455, 91, 13, 1, 116280, 38760, 11628, 3060, 680, 120, 15, 1, 735471, 245157, 74613, 20349, 4845, 969, 153, 17, 1, 4686825, 1562275, 480700, 134596, 33649, 7315

Offset: 0

Paul Barry, May 13 2006

First column is A005809. Second column is A025174.

Row sums are A045721. Inverse is Riordan array (1-3x,x(1-x)^2), A119302.

			Triangle begins
  1,
  3, 1,
  15, 5, 1,
  84, 28, 7, 1,
  495, 165, 45, 9, 1,
  3003, 1001, 286, 66, 11, 1,
  18564, 6188, 1820, 455, 91, 13, 1,
  116280, 38760, 11628, 3060, 680, 120, 15, 1
  ...
Horizontal recurrence: T(4,1) = 1*84 + 2*28 + 3*7 + 4*1 = 165. - _Peter Bala_, Dec 29 2014

Cf. A092392, A119304.

Cf. A005809, A025174, A045721, A119302.

T := proc(n,k) option remember;
`if`(n = 0, 1, add(i*T(n-1,k-2+i),i=1..n+1-k)) end:
for n from 0 to 9 do print(seq(T(n,k),k=0..n)) od; # Peter Luschny, Dec 30 2014

Flatten[Table[Binomial[3n-k,n-k],{n,0,10},{k,0,n}]] (* Harvey P. Dale, Jul 28 2012 *)

G.f. g(x) = 2*sin(arcsin(3*sqrt(3*x)/2)/3)/sqrt(3*x) satisfies g(x) = 1/(1-x*g(x)^2).
Riordan array (1/(1-3*x*g(x)^2),x*g(x)^2) where g(x)=1+x*g(x)^3.
'Horizontal' recurrence equation: T(n,0) = binomial(3*n,n) and for k >= 1, T(n,k) = Sum_{i = 1..n+1-k} i*T(n-1,k-2+i). - Peter Bala, Dec 28 2014
T(n, k) = Sum_{j = 0..n} binomial(n+j-1, j)*binomial(2*n-k-j, n). - Peter Bala, Jun 04 2024

A178300 Triangle T(n,k) = binomial(n+k-1,n) read by rows, 1 <= k <= n.

Original entry on oeis.org

1, 1, 3, 1, 4, 10, 1, 5, 15, 35, 1, 6, 21, 56, 126, 1, 7, 28, 84, 210, 462, 1, 8, 36, 120, 330, 792, 1716, 1, 9, 45, 165, 495, 1287, 3003, 6435, 1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 1, 11, 66, 286, 1001, 3003, 8008, 19448, 43758, 92378, 1, 12, 78, 364, 1365, 4368, 12376, 31824, 75582, 167960, 352716, 1, 13, 91, 455, 1820, 6188, 18564, 50388, 125970, 293930, 646646, 1352078

Offset: 1

Alford Arnold, May 24 2010

Obtained from A176992 by reversing entries in each row, from A092392 by removing the left column and reversing entries in each row, or from A100100 by removing the first two columns and reversing entries in each row.
Also T(n,k) = count of degree k monomials in the Monomial symmetric polynomials m(mu,k) summed over all partitions mu of n.
T(n,k) is the number of ways to put n indistinguishable balls into k distinguishable boxes. - Dennis P. Walsh, Apr 11 2012
T(n,k) is the number of compositions of n into k parts if zeros are allowed as parts. - L. Edson Jeffery, Jul 23 2014
T(n,k) is the number of compositions (ordered partitions) of n+k into exactly k parts. - Juergen Will, Jan 23 2016
T(n,k) is the number of binary strings with exactly n zeros and k-1 ones. - Dennis P. Walsh, Apr 09 2016
T(n,k) is the number of functions f:[k-1]->[n+1] that are nondecreasing. There is a unique correspondence between such a function and a binary string with exactly n zeros and k-1 ones. Given a string, let the corresponding function f be defined by f(i)=1 + (the number of zeros in the string that precede the i-th one in the string) for i=1,..,k-1. - Dennis P. Walsh, Apr 09 2016

			Triangle begins
  1;
  1,    3;
  1,    4,   10;
  1,    5,   15,   35;
  1,    6,   21,   56,  126;
  1,    7,   28,   84,  210,  462;
  1,    8,   36,  120,  330,  792, 1716;
T(3,3)=10 since there are 10 ways to put 3 identical balls into 3 distinguishable boxes, namely, (OOO)()(), ()(OOO)(), ()()(OOO), (OO)(O)(), (OO)()(O), (O)(OO)(), ()(OO)(O), (O)()(OO), ()(O)(OO), and (O)(O)(O). - _Dennis P. Walsh_, Apr 11 2012
For example, T(3,3)=10 since there are ten functions f:[2]->[4] that are nondecreasing, namely, <f(1),f(2)> = <1,1> or <1,2> or <1,3> or <1,4> or <2,2> or <2,3> or <2,4> or <3,3> or <3,4> or <4,4>. - _Dennis P. Walsh_, Apr 09 2016

P. A. MacMahon, Memoir on the Theory of the Compositions of Numbers, Phil. Trans. Royal Soc. London A, 184 (1893), 835-901.
Ch. Stover and E. W. Weisstein, Composition. From MathWorld - A Wolfram Web Resource.
Wikipedia, Symmetric Polynomials

Cf. A000142, A000312, A007318, A001791 (row sums), A209664-A209673.
Cf. A000217, A000292, A000332, A001700, A008292, A038675, A046899, A088218.
Cf. A176992, A092392, A100100.

// As triangle
[[Binomial(n+k-1,n): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Jan 24 2016

seq(seq(binomial(n+k-1,n),k=1..n),n=1..15); # Dennis P. Walsh, Apr 11 2012

m[par_?PartitionQ, v_] := Block[{le = Length[par], it }, If[le > v, Return[0]]; it = Permutations[PadRight[par, v]]; Tr[ Apply[Times, Table[Subscript[x, j], {j, v}]^# & /@ it, {1}]]];
Table[Tr[(m[#, k] & /@ Partitions[l]) /. Subscript[x, ] -> 1], {l, 11}, {k, l}](* _Wouter Meeussen, Mar 11 2012 *)
Quiet[Needs["Combinatorica`"], All]; Grid[Table[Length[Combinatorica`Compositions[n, k]], {n, 10}, {k, n}]] (* L. Edson Jeffery, Jul 24 2014 *)
t[n_, k_] := Binomial[n + k - 1, n]; Table[ t[n, k], {n, 10}, {k, n}] // Flatten (* Robert G. Wilson v, Jul 24 2014 *)

T(n,k) = A046899(n,k-1) = A038675(n,k)/A008292(n,k).
T(n,1) = 1.
T(n,2) = n+1.
T(n,3) = A000217(n+1).
T(n,4) = A000292(n+1).
T(n,5) = A000332(n+4).
T(n,n) = A001700(n-1) = A088218(n). - Dennis P. Walsh, Apr 10 2012

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A264772 Triangle T(n,k) = binomial(3n - 2k, 2*n - k), 0 <= k <= n.