A001700 -id:A001700 - cp's OEIS frontend

A092392 Triangle read by rows: T(n,k) = C(2*n - k,n), 0 <= k <= n.

1, 2, 1, 6, 3, 1, 20, 10, 4, 1, 70, 35, 15, 5, 1, 252, 126, 56, 21, 6, 1, 924, 462, 210, 84, 28, 7, 1, 3432, 1716, 792, 330, 120, 36, 8, 1, 12870, 6435, 3003, 1287, 495, 165, 45, 9, 1, 48620, 24310, 11440, 5005, 2002, 715, 220, 55, 10, 1, 184756, 92378, 43758, 19448, 8008, 3003, 1001, 286, 66, 11, 1

Offset: 0

Ralf Stephan, Mar 21 2004

First column is C(2*n,n) or A000984. Central coefficients are C(3*n,n) or A005809. - Paul Barry, Oct 14 2009
T(n,k) = A046899(n,n-k), k = 0..n-1. - Reinhard Zumkeller, Jul 27 2012
From  Peter Bala, Nov 03 2015: (Start)
Viewed as the square array [binomial (2*n + k, n + k)]n,k>=0  this is the generalized Riordan array ( 1/sqrt(1 - 4*x),c(x) ) in the sense of the Bala link, where c(x) is the o.g.f. for A000108.
The square array factorizes as ( 1/(2 - c(x)),x*c(x) ) * ( 1/(1 - x),1/(1 - x) ), which equals the matrix product of A100100 with the square Pascal matrix [binomial (n + k,k)]n,k>=0. See the example below. (End)

			From _Paul Barry_, Oct 14 2009: (Start)
Triangle begins
  1,
  2, 1,
  6, 3, 1,
  20, 10, 4, 1,
  70, 35, 15, 5, 1,
  252, 126, 56, 21, 6, 1,
  924, 462, 210, 84, 28, 7, 1,
  3432, 1716, 792, 330, 120, 36, 8, 1
Production array is
  2, 1,
  2, 1, 1,
  2, 1, 1, 1,
  2, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1, 1, 1, 1, 1 (End)
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..149 of triangle, flattened
P. Bala, Notes on generalized Riordan arrays
P. Barry, On the Central Coefficients of Riordan Matrices, Journal of Integer Sequences, 16 (2013), #13.5.1.
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
Ik-Pyo Kim, Michael J. Tsatsomeros, Inverse Relations in Shapiro's Open Questions, arXiv:1707.06590 [math.CO], 2017. See p. 7.

Rows 0-14 are A000984, A001700, A001791, A002054, A002694, A003516, A002696, A030053, A004310, A030054, A004311, A030055, A004312, A030056, A004313.
Columns are A000217, A000292, A000332, A000389, A000579.
Diagonals are A005809, A045721, A025174, A004319, A013698, A003408.
Cf. A100100.

a092392 n k = a092392_tabl !! (n-1) !! (k-1)
a092392_row n = a092392_tabl !! (n-1)
a092392_tabl = map reverse a046899_tabl
-- Reinhard Zumkeller, Jul 27 2012

/* As a triangle */ [[Binomial(2*n-k, n): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Nov 22 2017

A092392 := proc(n,k)
    binomial(2*n-k,n-k) ;
end proc:
seq(seq(A092392(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Feb 06 2015

Table[Binomial[2 n - k, n], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Mar 19 2016 *)

C(x):=(1-sqrt(1-4*x))/2;
A(x,y):=(1/sqrt(1-4*x))/(1-y*C(x));
taylor(A(x,y),y,0,10,x,0,10); /* Vladimir Kruchinin, Mar 19 2016 */

for(n=0,10, for(k=0,n, print1(binomial(2*n - k,n), ", "))) \\ G. C. Greubel, Nov 22 2017

As a number triangle, this is T(n, k) = if(k <= n, C(2*n - k, n), 0). Its row sums are C(2*n + 1, n + 1) = A001700. Its diagonal sums are A176287. - Paul Barry, Apr 23 2005
G.f. of column k: 2^k/[sqrt(1 - 4*x)*(1 + sqrt(1 - 4*x))^k].
As a number triangle, this is the Riordan array (1/sqrt(1 - 4*x), x*c(x)), c(x) the g.f. of A000108. - Paul Barry, Jun 24 2005
G.f.: A(x,y)=1/sqrt(1 - 4*x)/(1-y*x*C(x)), where C(x) is g.f. of Catalan numbers. - Vladimir Kruchinin, Mar 19 2016

Diagonal sums comment corrected by Paul Barry, Apr 14 2010

Offset corrected by R. J. Mathar, Feb 08 2013

A165817 Number of compositions (= ordered integer partitions) of n into 2n parts.

Original entry on oeis.org

1, 2, 10, 56, 330, 2002, 12376, 77520, 490314, 3124550, 20030010, 129024480, 834451800, 5414950296, 35240152720, 229911617056, 1503232609098, 9847379391150, 64617565719070, 424655979547800, 2794563003870330, 18412956934908690, 121455445321173600

Offset: 0

Thomas Wieder, Sep 29 2009

Number of ways to put n indistinguishable balls into 2*n distinguishable boxes.

Number of rankings of n unlabeled elements for 2*n levels.

			Let [1,1,1], [1,2] and [3] be the integer partitions of n=3.
Then [0,0,0,1,1,1], [0,0,0,0,1,2] and [0,0,0,0,0,3] are the corresponding partitions occupying 2*n = 6 positions.
We have to take into account the multiplicities of the parts including the multiplicities of the zeros.
Then
  [0,0,0,1,1,1] --> 6!/(3!*3!) = 20
  [0,0,0,0,1,2] --> 6!/(4!*1!*1!) = 30
  [0,0,0,0,0,3] --> 6!/(5!*1!) = 6
and thus a(3) = 20+30+6=56.
a(2)=10, since we have 10 ordered partitions of n=2 where the parts are distributed over 2*n=4 boxes:
  [0, 0, 0, 2]
  [0, 0, 1, 1]
  [0, 0, 2, 0]
  [0, 1, 0, 1]
  [0, 1, 1, 0]
  [0, 2, 0, 0]
  [1, 0, 0, 1]
  [1, 0, 1, 0]
  [1, 1, 0, 0]
  [2, 0, 0, 0].

Alois P. Heinz, Table of n, a(n) for n = 0..400
W. Mlotkowski and K. A. Penson, Probability distributions with binomial moments, arXiv preprint arXiv:1309.0595 [math.PR], 2013.

Cf. A000079, A001700, A059481, A081204, A001764, A004319, A006013, A005809, A013698, A025174, A045721, A117671, A236194.

[Binomial(3*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014

for n from 0 to 16 do
a[n] := 9*sqrt(3)*GAMMA(n+5/3)*GAMMA(n+4/3)*27^n/(Pi*GAMMA(2*n+3))
end do;

Table[Binomial[3 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)

vector(30, n, n--; binomial(3*n-1, n)) \\ Altug Alkan, Nov 04 2015

from math import comb
def A165817(n): return comb(3*n-1,n) if n else 1 # Chai Wah Wu, Oct 11 2023

def A165817(n):
    return rising_factorial(2*n,n)/falling_factorial(n,n)
[A165817(n) for n in (0..22)]   # Peter Luschny, Nov 21 2012

a(n) = 9*sqrt(3)*GAMMA(n+5/3)*GAMMA(n+4/3)*27^n/(Pi*GAMMA(2*n+3)).
a(n) = binomial(3*n-1, n);
Let denote P(n) = the number of integer partitions of n,
p(i) = the number of parts of the i-th partition of n,
d(i) = the number of different parts of the i-th partition of n,
m(i,j) = multiplicity of the j-th part of the i-th partition of n.
Then one has:
a(n) = Sum_{i=1..P(n)} (2*n)!/((2*n-p(i))!*(Prod_{j=1..d(i)} m(i,j)!)).
a(n) = rf(2*n,n)/ff(n,n), where rf is the rising factorial and ff the falling factorial. - Peter Luschny, Nov 21 2012
G.f.: A(x) = x*B'(x)/B(x), where B(x) satisfies B(x)^3-2*B(x)^2+B(x)=x, B(x)=A006013(x). - Vladimir Kruchinin, Feb 06 2013
G.f.: A(x) = sqrt(3*x)*cot(asin((3^(3/2)*sqrt(x))/2)/3)/(sqrt(4-27*x)). - Vladimir Kruchinin, Mar 18 2015
a(n) = Sum_{k=0..n} binomial(n-1,n-k)*binomial(2*n,k). - Vladimir Kruchinin, Oct 06 2015
From Peter Bala, Nov 04 2015: (Start)
The o.g.f. equals f(x)/g(x), where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A045721 (k = 1), A025174 (k = 2), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5) and A117671 (k = -2). (End)
a(n) = [x^n] 1/(1 - x)^(2*n). - Ilya Gutkovskiy, Oct 03 2017
a(n) = A059481(2n,n). - Alois P. Heinz, Oct 17 2022
From Peter Bala, Feb 14 2024: (Start)
a(n) = (-1)^n * binomial(-2*n, n).
a(n) = hypergeom([1 - 2*n, -n], [1], 1).
The g.f. A(x) satisfies A(x/(1 + x)^3) = 1/(1 - 2*x).
Sum_{n >= 0} a(n)/9^n = (1 + 4*cos(Pi/9))/3.
Sum_{n >= 0} a(n)/27^n = (3 + 4*sqrt(3)*cos(Pi/18))/9.
Sum_{n >= 0} a(n)*(2/27)^n = (2 + sqrt(3))/3. (End)
From Peter Bala, Sep 16 2024: (Start)
a(n) = Sum_{k = 0..n} binomial(n+k-1, k)*binomial(2*n-k-1, n-k).
More generally, a(n) = Sum_{k = 0..n} (-1)^k*binomial(x*n, k)*binomial((x+3)*n-k-1, n-k) for arbitrary x.
a(n) = (2/3) * Sum_{k = 0..n} (-1)^k*binomial(x*n+k-1, k)*binomial((x+3)*n, n-k) for n >= 1 and arbitrary x. (End)
G.f.: 1/(3-2*g) where g = 1+x*g^3 is the g.f. of A001764. - Seiichi Manyama, Aug 17 2025

a(0) prepended and more terms from Alois P. Heinz, Apr 04 2012

A060540 Square array read by antidiagonals downwards: T(n,k) = (nk)!/(k!^nn!), (n>=1, k>=1), the number of ways of dividing nk labeled items into n unlabeled boxes with k items in each box.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 10, 15, 1, 1, 35, 280, 105, 1, 1, 126, 5775, 15400, 945, 1, 1, 462, 126126, 2627625, 1401400, 10395, 1, 1, 1716, 2858856, 488864376, 2546168625, 190590400, 135135, 1, 1, 6435, 66512160, 96197645544, 5194672859376, 4509264634875, 36212176000, 2027025, 1

Offset: 1

Henry Bottomley, Apr 02 2001

The Copeland link gives the associations of this entry with the operator calculus of Appell Sheffer polynomials, the combinatorics of simple set partitions encoded in the Faa di Bruno formula for composition of analytic functions (formal Taylor series), the Pascal matrix, and the geometry of the n-dimensional simplices (hypertriangles, or hypertetrahedra). These, in turn, are related to simple instances of the application of the exponential formula / principle / schema giving the number of not-necessarily-connected objects composed from an ensemble of connected objects. - Tom Copeland, Jun 09 2021

			Array begins:
  1,   1,       1,          1,             1,                 1, ...
  1,   3,      10,         35,           126,               462, ...
  1,  15,     280,       5775,        126126,           2858856, ...
  1, 105,   15400,    2627625,     488864376,       96197645544, ...
  1, 945, 1401400, 2546168625, 5194672859376, 11423951396577720, ...
  ...

Seiichi Manyama, Antidiagonals n = 1..50, flattened (first 20 antidiagonals from Harry J. Smith)
Tom Copeland, Calculus, Combinatorics, and Geometry Underlying OEIS A060540, and the Exponential Formula, 2021.
Nattawut Phetmak and Jittat Fakcharoenphol, Uniformly Generating Derangements with Fixed Number of Cycles in Polynomial Time, Thai J. Math. (2023) Vol. 21, No. 4, 899-915. See pp. 901, 914.
Elena L. Wang and Guoce Xin, On Ward Numbers and Increasing Schröder Trees, arXiv:2507.15654 [math.CO], 2025. See p. 13.

Rows include A000012, A001700, A060542, A082368, A322252.
Columns k=1..10 give A000012, A001147, A025035, A025036, A025037, A025038, A025040, A025041, A025042.
Main diagonal is A057599.
Related to A057599, see also A096126 and A246048.
Cf. A060358, A361948 (includes row/col 0).
Cf. A000217, A000292, A000332, A000389, A000579, A000580, A007318, A036040, A099174, A133314, A132440, A135278 (associations in Copeland link).

T[n_, k_] := (n*k)!/(k!^n*n!);
Table[T[n-k+1, k], {n, 1, 10}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Jun 29 2018 *)

{ i=0; for (m=1, 20, for (n=1, m, k=m - n + 1; write("b060540.txt", i++, " ", (n*k)!/(k!^n*n!))); ) } \\ Harry J. Smith, Jul 06 2009

T(n,k) = (n*k)!/(k!^n*n!) = T(n-1,k)*A060543(n,k) = A060538(n,k)/k!.

T(n,k) = Product_{j=2..n} binomial(j*k-1,k-1). - M. F. Hasler, Aug 22 2014

Definition reworded by M. F. Hasler, Aug 23 2014

A349053 Number of non-weakly alternating integer compositions of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 4, 12, 37, 95, 232, 533, 1198, 2613, 5619, 11915, 25011, 52064, 107694, 221558, 453850, 926309, 1884942, 3825968, 7749312, 15667596, 31628516, 63766109, 128415848, 258365323, 519392582, 1043405306, 2094829709, 4203577778, 8431313237, 16904555958

Offset: 0

Gus Wiseman, Dec 16 2021

nonn

We define a sequence to be weakly alternating if it is alternately weakly increasing and weakly decreasing, starting with either. Then a sequence is (strongly) alternating iff it is a weakly alternating anti-run.

			The a(6) = 12 compositions:
  (1,1,2,2,1)  (1,1,2,3)  (1,2,4)
  (1,2,1,1,2)  (1,2,3,1)  (4,2,1)
  (1,2,2,1,1)  (1,3,2,1)
  (2,1,1,2,1)  (2,1,1,3)
               (3,1,1,2)
               (3,2,1,1)

Andrew Howroyd, Table of n, a(n) for n = 0..1000 (terms 0..55 from Martin Ehrenstein)
Wikipedia, Alternating permutation

Complementary directed versions are A129852/A129853, strong A025048/A025049.
The strong version is A345192.
The complement is counted by A349052.
These compositions are ranked by A349057, strong A345168.
The complementary version for patterns is A349058, strong A345194.
The complementary multiplicative version is A349059, strong A348610.
An unordered version (partitions) is A349061, complement A349060.
The version for ordered prime factorizations is A349797, complement A349056.
The version for patterns is A350138, strong A350252.
The version for ordered factorizations is A350139.
A001250 counts alternating permutations, complement A348615.
A001700 counts compositions of 2n with alternating sum 0.
A003242 counts Carlitz (anti-run) compositions.
A011782 counts compositions, unordered A000041.
A025047 counts alternating compositions, ranked by A345167.
A106356 counts compositions by number of maximal anti-runs.
A344604 counts alternating compositions with twins.
A345164 counts alternating ordered prime factorizations.
A349054 counts strict alternating compositions.
Cf. A102726, A114901, A128761, A261983, A333213, A333755, A344614, A344615, A345165, A345170, A345195, A349799, A349800, A350251, A350252.

wwkQ[y_]:=And@@Table[If[EvenQ[m],y[[m]]<=y[[m+1]],y[[m]]>=y[[m+1]]],{m,1,Length[y]-1}]||And@@Table[If[EvenQ[m],y[[m]]>=y[[m+1]],y[[m]]<=y[[m+1]]],{m,1,Length[y]-1}];
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],!wwkQ[#]&]],{n,0,10}]

a(n) = A011782(n) - A349052(n).

a(21)-a(35) from Martin Ehrenstein, Jan 08 2022

A357641 Number of integer compositions of 2n whose half-alternating sum is 0.

Original entry on oeis.org

1, 0, 2, 8, 28, 104, 396, 1504, 5720, 21872, 83980, 323344, 1248072, 4828784, 18721080, 72711552, 282861360, 1101980000, 4298748300, 16789002736, 65641204200, 256895795312, 1006308200040, 3945185586368, 15478849767888, 60774329914144, 238775589937976

Offset: 0

Gus Wiseman, Oct 12 2022

nonn

We define the half-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A + B - C - D + E + F - G - ...

			The a(0) = 1 through a(3) = 8 compositions:
  ()  .  (112)   (123)
         (1111)  (213)
                 (1212)
                 (1221)
                 (2112)
                 (2121)
                 (11121)
                 (11211)

Alois P. Heinz, Table of n, a(n) for n = 0..1664

The skew-alternating version appears to be A001700.
The version for partitions is A035363.
The skew-alternating form is A088218 (also for full alternating sum).
These compositions are ranked by A357625, reverse A357626.
For reversed partitions we have A357639, ranked by A357631.
A124754 gives alternating sum of standard compositions, reverse A344618.
A357621 = half-alternating sum of standard compositions, reverse A357622.
A357637 counts partitions by half-alternating sum, skew A357638.
Cf. A000583, A001511, A053251, A344619, A357136, A357182, A357627, A357628, A357629, A357633, A357642.

a:= proc(n) option remember; `if`(n<3, [1, 0, 2][n+1],
      (8*(n-3)*(5*n-7)*(2*n-5)*a(n-3) -4*(5*n-12)*(n-2)^2*a(n-2)
       +2*(2*n-5)*(5*n-7)*n*a(n-1))/((5*n-12)*(n+1)*(n-2)))
    end:
seq(a(n), n=0..40);  # Alois P. Heinz, Oct 19 2022

halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
Table[Length[Select[Join@@Permutations/@IntegerPartitions[2n],halfats[#]==0&]],{n,0,7}]

a(11)-a(26) from Alois P. Heinz, Oct 19 2022

A059481 Triangle read by rows. T(n, k) = binomial(n+k-1, k) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 126, 1, 6, 21, 56, 126, 252, 462, 1, 7, 28, 84, 210, 462, 924, 1716, 1, 8, 36, 120, 330, 792, 1716, 3432, 6435, 1, 9, 45, 165, 495, 1287, 3003, 6435, 12870, 24310, 1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, 92378

Offset: 0

Fabian Rothelius, Feb 04 2001

T(n,k) is the number of ways to distribute k identical objects in n distinct containers; containers may be left empty.
T(n,k) is the number of nondecreasing functions f from {1,...,k} to {1,...,n}. - Dennis P. Walsh, Apr 07 2011
Coefficients of Faber polynomials for function x^2/(x-1). - Michael Somos, Sep 09 2003
Consider k-fold Cartesian products CP(n,k) of the vector A(n)=[1,2,3,...,n].
An element of CP(n,k) is a n-tuple T_t of the form T_t=[i_1,i_2,i_3,...,i_k] with t=1,...,n^k.
We count members T of CP(n,k) which satisfy some condition delta(T_t), so delta(.) is an indicator function which attains values of 1 or 0 depending on whether T_t is to be counted or not; the summation sum_{CP(n,k)} delta(T_t) over all elements T_t of CP produces the count.
For the triangle here we have delta(T_t) = 0 if for any two i_j, i_(j+1) in T_t one has i_j > i_(j+1), T(n,k) = Sum_{CP(n,k)} delta(T_t) = Sum_{CP(n,k)} delta(i_j > i_(j+1)).
The indicator function which tests on i_j = i_(j+1) generates A158497, which contains further examples of this type of counting.
Triangle of the numbers of combinations of k elements with repetitions from n elements {1,2,...,n} (when every element i, i=1,...,n, appears in a k-combination either 0, or 1, or 2, ..., or k times). - Vladimir Shevelev, Jun 19 2012
G.f. for Faber polynomials is -log(-t*x-(1-sqrt(1-4*t))/2+1)=sum(n>0, T(n,k)*t^k/n). - Vladimir Kruchinin, Jul 04 2013
Values of complete homogeneous symmetric polynomials with all arguments equal to 1, or, equivalently, the number of monomials of degree k in n variables. - Tom Copeland, Apr 07 2014
Row k >= 0 of the infinite square array A[k,n] = C(n,k), n >= 0, would start with k zeros in front of the first nonzero element C(k,k) = 1; this here is the triangle obtained by taking the first k+1 nonzero terms C(k .. 2k, k) of rows k = 0, 1, 2, ... of that array. - M. F. Hasler, Mar 05 2017

			The triangle T(n,k), n >= 0, 0 <= k <= n, begins
  1      A000217
  1 1   /     A000292
  1 2  3    /    A000332
  1 3  6  10    /    A000389
  1 4 10  20  35    /     A000579
  1 5 15  35  70 126     /
  1 6 21  56 126 252  462
  1 7 28  84 210 462  924 1716
  1 8 36 120 330 792 1716 3432 6435
.
T(3,2)=6 considers the CP with the 3^2=9 elements (1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3), and does not count the 3 of them which are (2,1),(3,1) and (3,2).
T(3,3) = 10 because the ways to distribute the 3 objects into the three containers are: (3,0,0) (0,3,0) (0,0,3) (2,1,0) (1,2,0) (2,0,1) (1,0,2) (0,1,2) (0,2,1) (1,1,1), for a total of 10 possibilities.
T(3,3)=10 since (x^2/(x-1))^3 = (x+1+1/x+O(1/x^2))^3 = x^3+3x^2+6x+10+O(x).
T(4,2)=10 since there are 10 nondecreasing functions f from {1,2} to {1,2,3,4}. Using <f(1),f(2)> to denote such a function, the ten functions are <1,1>, <1,2>, <1,3>, <1,4>, <2,2>, <2,3>, <2,4>, <3,3>, <3,4>, and <4,4>. - _Dennis P. Walsh_, Apr 07 2011
T(4,0) + T(4,1) + T(4,2) + T(4,3) = 1 + 4 + 10 + 20 = 35 = T(4,4). - _Jonathan Sondow_, Jun 28 2014
From _Paul Curtz_, Jun 18 2018: (Start)
Consider the array
2,    1,    1,    1,    1,    1,     ... = A054977(n)
1,    1/2,  1/3,  1/4,  1/5,  1/6,   ... = 1/(n+1) = 1/A000027(n)
1/3,  1/6,  1/10, 1/15, 1/21, 1/28,  ... = 2/((n+2)*(n+3)) = 1/A000217(n+2)
1/10, 1/20, 1/35, 1/56, 1/84, 1/120, ... = 6/((n+3)*(n+4)*(n+5)) =1/A000292(n+2) (see the triangle T(n,k)).
Every row is an autosequence of the second kind. (See OEIS Wiki, Autosequence.)
By decreasing antidiagonals the denominator of the array is a(n).
Successive vertical denominators: A088218(n), A000984(n), A001700(n), A001791(n+1), A002054(n), A002694(n).
Successive diagonal denominators: A165817(n), A005809(n), A045721(n), A025174(n+1), A004319(n). (End)
Without the first row (2, 1, 1, 1, ... ), the array leads to A165257(n) instead of a(n). - _Paul Curtz_, Jun 19 2018

R. Grimaldi, Discrete and Combinatorial Mathematics, Addison-Wesley, 4th edition, chapter 1.4.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
J. Abate and W. Whitt, Brownian Motion and the Generalized Catalan Numbers, J. Int. Seq. 14 (2011) # 11.2.6, (referring to A158498 before recycling)
M. A. A. Obaid, S. K. Nauman, W. M. Fakieh, and C. M. Ringel, The numbers of support-tilting modules for a Dynkin algebra, 2014 and J. Int. Seq. 18 (2015) 15.10.6.
C. M. Ringel, The Catalan combinatorics of the hereditary artin algebras, arXiv:1502.06553 [math.RT], 2015.
Dennis Walsh, Notes on finite monotonic and non-monotonic functions

Columns: T(n,1) = A000027(n), n >= 1. T(n,2) = A000217(n) = A161680(n+1), n >= 2. T(n,3) = A000292(n), n >= 3. T(n,4) = A000332(n+3), n >= 4. T(n,5) = A000389(n+4), n >= 5. T(n,6) = A000579(n+5), n >=6. T(n,k) = A001405(n+k-1) for k <= n <= k+2. [Corrected and extended by M. F. Hasler, Mar 05 2017]
Rows: T(5,k) = A000332(k+4). T(6,k) = A000389(k+5). T(7,k) = A000579(k+6).
Diagonals: T(n,n) = A001700(n-1). T(n,n-1) = A000984(n-1).
T(n,k) = A046899(n-1,k). - R. J. Mathar, Mar 26 2009
Take Pascal's triangle A007318, delete entries to the right of a vertical line just right of center, then scan the diagonals.
For a signed version of this triangle see A027555.
Row sums give A000984.
Cf. A007318, A158497, A100100 (mirrored), A009766.
A000984, A001700, A001791, A002054, A002694, A004319, A005809, A025174, A045721, A088218, A165817, A054977, A165257, A000027, A000217, A006134 (trace of the symmetric Pascal matrix).

Flat(List([0..10], n->List([0..n], k->Binomial(n+k-1, k)))); # Stefano Spezia, Oct 30 2018

a059481 n k = a059481_tabl !! n !! n
a059481_row n = a059481_tabl !! n
a059481_tabl = map reverse a100100_tabl
-- Reinhard Zumkeller, Jan 15 2014

&cat [[&*[ Binomial(n+k-1,k)]: k in [0..n]]: n in [0..30] ]; // Vincenzo Librandi, Apr 08 2011

for n from 0 to 10 do for k from 0 to n do print(binomial(n+k-1,k)) ; od: od: # R. J. Mathar, Mar 31 2009

t[n_, k_] := Binomial[n+k-1, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 09 2013 *)
(* The combinatorial objects defined in the first comment can, for n >= 1, be generated by: *) r[n_, k_] := FrobeniusSolve[ConstantArray[1,n],k]; (* Peter Luschny, Jan 24 2019 *)

sjoin(v, j) := apply(sconcat, rest(join(makelist(j, length(v)), v)))$ display_triangle(n) := for i from 0 thru n do disp(sjoin(makelist(binomial(i+j-1, j), j, 0, i), " ")); display_triangle(10); /* triangle output */ /* Stefano Spezia, Oct 30 2018 */

{T(n, k) = binomial( n+k-1, k)}; \\ Michael Somos, Sep 09 2003, edited by M. F. Hasler, Mar 05 2017

{T(n, k) = if( n<0, 0, polcoeff( Pol(((1 / (x - x^2) + x * O(x^n))^n + O(x)) * x^n), k))}; /* Michael Somos, Sep 09 2003 */

[[binomial(n+k-1,k) for k in range(n+1)] for n in range(11)] # G. C. Greubel, Nov 21 2018

T(n,0) + T(n,1) + . . . + T(n,n-1) = T(n,n). - Jonathan Sondow, Jun 28 2014
From Peter Bala, Jul 21 2015: (Start)
T(n, k) = Sum_{j = k..n} (-1)^(k+j)*binomial(2*n,n+j)*binomial(n+j-1,j)* binomial(j,k) (gives the correct value T(n,k) = 0 for k > n).
O.g.f.: 1/2*( x*(2*x - 1)/(sqrt(1 - 4*t*x)*(1 - x - t)) + (1 + 2*x)/sqrt(1 - 4*t*x) + (1 - t)/(1 - x - t) ) = 1 + (1 + t)*x + (1 + 2*t + 3*t^2)*x^2 + (1 + 3*t + 6*t^2 + 10*t^3)*x^3 + ....
n-th row polynomial R(n,t) = [x^n] ( (1 + x)^2/(1 + x(1 - t)) )^n.
exp( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (1 + t)*x + (1 + 2*t + 2*t^2)*x^2 + (1 + 3*t + 5*t^2 + 5*t^3)*x^3 + ... is the o.g.f for A009766. (End)
a(n) = abs(A027555(n)). - M. F. Hasler, Mar 05 2017
For n >= k > 0, T(n, k) = Sum_{j=1..n} binomial(k + j - 2, k - 1) = Sum_{j=1..n} A007318(k + j - 2, k - 1). - Stefano Spezia, Oct 30 2018
T(n, k) = RisingFactorial(n, k) / k!. - Peter Luschny, Nov 24 2023

Offset changed from 1 to 0 by R. J. Mathar, Jan 15 2013

Edited by M. F. Hasler, Mar 05 2017

A349052 Number of weakly alternating compositions of n.

Original entry on oeis.org

1, 1, 2, 4, 8, 16, 28, 52, 91, 161, 280, 491, 850, 1483, 2573, 4469, 7757, 13472, 23378, 40586, 70438, 122267, 212210, 368336, 639296, 1109620, 1925916, 3342755, 5801880, 10070133, 17478330, 30336518, 52653939, 91389518, 158621355, 275313226, 477850887, 829388075

Offset: 0

Gus Wiseman, Nov 29 2021

nonn

We define a sequence to be weakly alternating if it is alternately weakly increasing and weakly decreasing, starting with either. A sequence is alternating iff it is a weakly alternating anti-run.

			The a(5) = 16 compositions:
  (1,1,1,1,1)  (1,1,1,2)  (1,1,3)  (1,4)  (5)
               (1,1,2,1)  (1,2,2)  (2,3)
               (1,2,1,1)  (1,3,1)  (3,2)
               (2,1,1,1)  (2,1,2)  (4,1)
                          (2,2,1)
                          (3,1,1)
The a(6) = 28 compositions:
  (111111)  (11112)  (1113)  (114)  (15)  (6)
            (11121)  (1122)  (132)  (24)
            (11211)  (1131)  (141)  (33)
            (12111)  (1212)  (213)  (42)
            (21111)  (1311)  (222)  (51)
                     (2121)  (231)
                     (2211)  (312)
                     (3111)  (411)

Andrew Howroyd, Table of n, a(n) for n = 0..1000 (Terms 0..55 from Martin Ehrenstein)

The strong case is A025047, ranked by A345167.
The directed versions are A129852 and A129853, strong A025048 and A025049.
The complement is counted by A349053, strong A345192.
The version for permutations of prime indices is A349056, strong A345164.
The complement is ranked by A349057, strong A345168.
The version for patterns is A349058, strong A345194.
The multiplicative version is A349059, strong A348610.
An unordered version (partitions) is A349060, complement A349061.
The non-alternating case is A349800, ranked by A349799.
A001250 counts alternating permutations, complement A348615.
A001700 counts compositions of 2n with alternating sum 0.
A003242 counts Carlitz (anti-run) compositions.
A011782 counts compositions.
A106356 counts compositions by number of maximal anti-runs.
A344604 counts alternating compositions with twins.
A345170 counts partitions w/ an alternating permutation, ranked by A345172.
A349054 counts strict alternating compositions.
Cf. A000041, A008965, A102726, A114901, A128761, A261983, A333213, A333755, A344614, A344615, A345165, A345195.

whkQ[y_]:=And@@Table[If[EvenQ[m],y[[m]]<=y[[m+1]],y[[m]]>=y[[m+1]]],{m,1,Length[y]-1}];
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],whkQ[#]||whkQ[-#]&]],{n,0,10}]

C(n,f)={my(M=matrix(n,n,j,k,k>=j), s=M[,n]); for(b=1, n, f=!f; M=matrix(n,n,j,k, if(k1,M[j-k,k-1]) ))); for(k=2, n, M[,k]+=M[,k-1]); s+=M[,n]); s~}
seq(n) = concat([1], C(n,0) + C(n,1) - vector(n,j,numdiv(j))) \\ Andrew Howroyd, Jan 31 2024

a(21)-a(37) from Martin Ehrenstein, Jan 08 2022

A028364 Triangle T(n,m) = Sum_{k=0..m} Catalan(n-k)*Catalan(k).

Original entry on oeis.org

1, 1, 2, 2, 3, 5, 5, 7, 9, 14, 14, 19, 23, 28, 42, 42, 56, 66, 76, 90, 132, 132, 174, 202, 227, 255, 297, 429, 429, 561, 645, 715, 785, 869, 1001, 1430, 1430, 1859, 2123, 2333, 2529, 2739, 3003, 3432, 4862, 4862, 6292, 7150, 7810, 8398, 8986, 9646, 10504, 11934, 16796

Offset: 0

Wouter Meeussen

There are several versions of a Catalan triangle: see A009766, A008315, A028364.
The subtriangle [1], [2, 3], [5, 7, 9], ..., namely T(N,M-1), for N >= 1, M=1..N, appears as one-point function in the totally asymmetric exclusion process for the parameters alpha=1=beta. See the Derrida et al. and Liggett references given under A067323, where these triangle entries are called T_{N,N+M-1} for the given alpha and beta values. See the row reversed triangle A067323.
Consider a Dyck path as a path with steps N=(0,1) and E=(1,0) from (0,0) to (n,n) that stays weakly above y=x. T(n,m) is the number of Dyck paths of semilength n+1 where the (m+1)st north step is followed by an east step. - Lara Pudwell, Apr 12 2023

			Triangle begins
   1;
   1,  2;
   2,  3,  5;
   5,  7,  9, 14;
  14, 19, 23, 28, 42;

Alois P. Heinz, Rows n = 0..140, flattened
Ayomikun Adeniran and Lara Pudwell, Pattern avoidance in parking functions, Enumer. Comb. Appl. 3:3 (2023), Article S2R17.
G. Chatel and V. Pilaud, Cambrian Hopf Algebras, arXiv:1411.3704 [math.CO], 2014-2015.
A. Sapounakis et al., Ordered trees and the inorder transversal, Disc. Math., 306 (2006), 1732-1741.

Cf. A000108 (column 0 and main diagonal), A001700 (row sums), A065097 (T(2*n-1, n-1)), A201205 (central terms).

Cf. A067323, A009766, A039598, A039599, A028377, A028378, A028376.

b:= proc(n, i) option remember; `if`(n=0, 1, add(
      expand(b(n-1, j)*`if`(i>n, x, 1)), j=1..i))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b((n+1)$2)):
seq(T(n), n=0..10);  # Alois P. Heinz, Nov 28 2015

t[n_, k_] = Sum[CatalanNumber[n-j]*CatalanNumber[j], {j, 0, k}]; Flatten[Table[t[n, k], {n, 0, 8}, {k, 0, n}]] (* Jean-François Alcover, Jul 22 2011 *)

T(n,k) = Sum_{j>=0} A039598(k,j)*A039599(n-k,j). - Philippe Deléham, Feb 18 2004
Sum_{k>=0} T(n,k) = A001700(n). T(n,k) = A067323(n,n-k), n >= k >= 0, otherwise 0. - Philippe Deléham, May 26 2005
G.f. for column sequences m >= 0: (-(c(m,x)-1)/x+c(m,x)*c(x))/x^m with the g.f. c(x) of A000108 (Catalan) and c(m,x):=sum(C(k)*x^k,k=0..m) with C(n):=A000108(n). - Wolfdieter Lang, Mar 24 2006
G.f. for column sequences m >= 0 (without leading zeros): c(x)*Sum_{k=0..m} C(m,k)*c(x)^k with the g.f. c(x) of A000108 (Catalan) and C(n,m) is the Catalan triangle A033184(n,m). - Wolfdieter Lang, Mar 24 2006
T(n,n) = T(n,k) + T(n,n-1-k) = A000108(n+1), n > 0, k = 0..floor((n+1)/2). - Yuchun Ji, Jan 09 2019
G.f. for triangle: Sum_{n>=0, m>=0} T(n, m)*x^n*y^m = (c(x)-c(xy))/(x(1-y)c(x)) with the g.f. c(x) of A000108 (Catalan). - Lara Pudwell, Apr 12 2023

A347456 Number of factorizations of n with alternating product >= 1.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 2, 2, 1, 2, 1, 4, 1, 1, 1, 6, 1, 1, 1, 3, 1, 2, 1, 2, 2, 1, 1, 6, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 5, 1, 1, 2, 8, 1, 2, 1, 2, 1, 2, 1, 8, 1, 1, 2, 2, 1, 2, 1, 6, 4, 1, 1, 5, 1, 1, 1

Offset: 1

Gus Wiseman, Oct 09 2021

nonn

We define the alternating product of a sequence (y_1,...,y_k) to be Product_i y_i^((-1)^(i-1)).
A factorization of n is a weakly increasing sequence of positive integers > 1 with product n.
Also the number of factorizations of n with alternating sum >= 0.

			The a(n) factorizations for n = 4, 16, 24, 36, 60, 64, 96:
  4     16        24      36        60       64            96
  2*2   4*4       2*2*6   6*6       2*5*6    8*8           2*6*8
        2*2*4     2*3*4   2*2*9     3*4*5    2*4*8         3*4*8
        2*2*2*2           2*3*6     2*2*15   4*4*4         4*4*6
                          3*3*4     2*3*10   2*2*16        2*2*24
                          2*2*3*3            2*2*4*4       2*3*16
                                             2*2*2*2*4     2*4*12
                                             2*2*2*2*2*2   2*2*2*2*6
                                                           2*2*2*3*4

The case of partitions is A000041, reverse A344607.
The reverse version is A001055, strict A347705.
Positions of 3's appear to be A065036.
Positions of 1's are 1 and A167171.
The opposite version (<= instead of >=) is A339846.
The strict version (> instead of >=) is A339890, also the odd-length case.
Allowing any integer alternating product gives A347437.
The case of alternating product 1 is A347438, also the even-length case.
Allowing any integer reciprocal alternating product gives A347439.
The complement (< instead of >=) is A347440.
Allowing any integer reverse-alternating product gives A347442.
A038548 counts factorizations with a wiggly permutation.
A045778 counts strict factorizations.
A074206 counts ordered factorizations.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A119620 counts partitions with alternating product 1.
A347447 counts strict factorizations with alternating product > 1.
Cf. A001700, A028983, A316523, A347441, A347443, A347446, A347448, A347450, A347454, A347463, A347708.

facs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
altprod[q_]:=Product[q[[i]]^(-1)^(i-1),{i,Length[q]}];
Table[Length[Select[facs[n],altprod[#]>=1&]],{n,100}]

a(n) = A347438(n) + A347440(n).

A060693 Triangle (0 <= k <= n) read by rows: T(n, k) is the number of Schröder paths from (0,0) to (2n,0) having k peaks.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 10, 6, 1, 14, 35, 30, 10, 1, 42, 126, 140, 70, 15, 1, 132, 462, 630, 420, 140, 21, 1, 429, 1716, 2772, 2310, 1050, 252, 28, 1, 1430, 6435, 12012, 12012, 6930, 2310, 420, 36, 1, 4862, 24310, 51480, 60060, 42042, 18018, 4620, 660, 45, 1, 16796

Offset: 0

F. Chapoton, Apr 20 2001

The rows sum to A006318 (Schroeder numbers), the left column is A000108 (Catalan numbers); the next-to-left column is A001700, the alternating sum in each row but the first is 0.
T(n,k) is the number of Schroeder paths (i.e., consisting of steps U=(1,1), D=(1,-1), H=(2,0) and never going below the x-axis) from (0,0) to (2n,0), having k peaks. Example: T(2,1)=3 because we have UU*DD, U*DH and HU*D, the peaks being shown by *. E.g., T(n,k) = binomial(n,k)*binomial(2n-k,n-1)/n for n>0. - Emeric Deutsch, Dec 06 2003
A090181*A007318 as infinite lower triangular matrices. - Philippe Deléham, Oct 14 2008
T(n,k) is also the number of rooted plane trees with maximal degree 3 and k vertices of degree 2 (a node may have at most 2 children, and there are exactly k nodes with 1 child). Equivalently, T(n,k) is the number of syntactically different expressions that can be formed that use a unary operation k times, a binary operation n-k times, and nothing else (sequence of operands is fixed). - Lars Hellstrom (Lars.Hellstrom(AT)residenset.net), Dec 08 2009

			Triangle begins:
00: [    1]
01: [    1,     1]
02: [    2,     3,      1]
03: [    5,    10,      6,      1]
04: [   14,    35,     30,     10,      1]
05: [   42,   126,    140,     70,     15,      1]
06: [  132,   462,    630,    420,    140,     21,     1]
07: [  429,  1716,   2772,   2310,   1050,    252,    28,    1]
08: [ 1430,  6435,  12012,  12012,   6930,   2310,   420,   36,   1]
09: [ 4862, 24310,  51480,  60060,  42042,  18018,  4620,  660,  45,  1]
10: [16796, 92378, 218790, 291720, 240240, 126126, 42042, 8580, 990, 55, 1]
...

Vincenzo Librandi, Rows n = 0..100, flattened
J. Agapito, A. Mestre, P. Petrullo, and M. Torres, Counting noncrossing partitions via Catalan triangles, CEAFEL Seminar, June 30, 2015.
Jean-Christophe Aval and François Bergeron, Rectangular Schröder Parking Functions Combinatorics, arXiv:1603.09487 [math.CO], 2016.
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, J. Integer Sequ., Vol. 9 (2006), Article 06.2.4.
Paul Barry, Three Études on a sequence transformation pipeline, arXiv:1803.06408 [math.CO], 2018.
Paul Barry, Riordan Pseudo-Involutions, Continued Fractions and Somos 4 Sequences, arXiv:1807.05794 [math.CO], 2018.
Paul Barry, The Central Coefficients of a Family of Pascal-like Triangles and Colored Lattice Paths, J. Int. Seq., Vol. 22 (2019), Article 19.1.3.
Paul Barry, Generalized Catalan Numbers Associated with a Family of Pascal-like Triangles, J. Int. Seq., Vol. 22 (2019), Article 19.5.8.
Paul Barry, On the inversion of Riordan arrays, arXiv:2101.06713 [math.CO], 2021.
Paul Barry, On Motzkin-Schröder Paths, Riordan Arrays, and Somos-4 Sequences, J. Int. Seq. (2023) Vol. 26, Art. 23.4.7.
Paul Barry, Notes on Riordan arrays and lattice paths, arXiv:2504.09719 [math.CO], 2025. See p. 25.
David Callan and Toufik Mansour, Five subsets of permutations enumerated as weak sorting permutations, arXiv:1602.05182 [math.CO], 2016.
G. E. Cossali, A Common Generating Function of Catalan Numbers and Other Integer Sequences, J. Int. Seqs. 6 (2003).
Dan Drake, Bijections from Weighted Dyck Paths to Schröder Paths, J. Int. Seq. 13 (2010) # 10.9.2
Samuele Giraudo, Tree series and pattern avoidance in syntax trees, arXiv:1903.00677 [math.CO], 2019.
Nate Kube and Frank Ruskey, Sequences That Satisfy a(n-a(n))=0, Journal of Integer Sequences, Vol. 8 (2005), Article 05.5.5.
Krishna Menon and Anurag Singh, Grassmannian permutations avoiding identity, arXiv:2212.13794 [math.CO], 2022.
Jean-Christophe Novelli and Jean-Yves Thibon, Duplicial algebras and Lagrange inversion, arXiv preprint arXiv:1209.5959 [math.CO], 2012.

Cf. A006318, A000108, A001700.
Triangle in A088617 transposed.
Diagonals give A000108, A001700, A002457, A002802, A002803; A000012, A000217, A034827, A000910, A088625, A088626.
T(2n,n) gives A007004.
Cf. A001263, A033282, A086810, A088617, A097610, A101282.

A060693 := (n,k) -> binomial(n,k)*binomial(2*n-k,n)/(n-k+1); # Peter Luschny, May 17 2011

t[n_, k_] := Binomial[n, k]*Binomial[2 n - k, n]/(n - k + 1); Flatten[Table[t[n, k], {n, 0, 9}, {k, 0, n}]] (* Robert G. Wilson v, May 30 2011 *)

T(n, k) = binomial(n, k)*binomial(2*n - k, n)/(n - k + 1);
for(n=0, 10, for(k=0, n, print1(T(n, k),", ")); print); \\ Indranil Ghosh, Jul 28 2017

from sympy import binomial
def T(n, k): return binomial(n, k) * binomial(2 * n - k, n) / (n - k + 1)
for n in range(11): print([T(n, k) for k in range(n + 1)])  # Indranil Ghosh, Jul 28 2017

Triangle T(n, k) (0 <= k <= n) read by rows; given by [1, 1, 1, 1, 1, ...] DELTA [1, 0, 1, 0, 1, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 12 2003
If C_n(x) is the g.f. of row n of the Narayana numbers (A001263), C_n(x) = Sum_{k=1..n} binomial(n,k-1)*(binomial(n-1,k-1)/k) * x^k and T_n(x) is the g.f. of row n of T(n,k), then T_n(x) = C_n(x+1), or T(n,k) = [x^n]Sum_{k=1..n}(A001263(n,k)*(x+1)^k). - Mitch Harris, Jan 16 2007, Jan 31 2007
G.f.: (1 - t*y - sqrt((1-y*t)^2 - 4*y)) / 2.
T(n, k) = binomial(2n-k, n)*binomial(n, k)/(n-k+1). - Philippe Deléham, Dec 07 2003
A060693(n, k) = binomial(2*n-k, k)*A000108(n-k); A000108: Catalan numbers. - Philippe Deléham, Dec 30 2003
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000108(n), A006318(n), A047891(n+1), A082298(n), A082301(n), A082302(n), A082305(n), A082366(n), A082367(n), for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, respectively. - Philippe Deléham, Apr 01 2007
T(n,k) = Sum_{j>=0} A090181(n,j)*binomial(j,k). - Philippe Deléham, May 04 2007
Sum_{k=0..n} T(n,k)*x^(n-k) = (-1)^n*A107841(n), A080243(n), A000007(n), A000012(n), A006318(n), A103210(n), A103211(n), A133305(n), A133306(n), A133307(n), A133308(n), A133309(n) for x = -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, respectively. - Philippe Deléham, Oct 18 2007
From Paul Barry, Jan 29 2009: (Start)
G.f.: 1/(1-xy-x/(1-xy-x/(1-xy-x/(1-xy-x/(1-xy-x/(1-.... (continued fraction);
G.f.: 1/(1-(x+xy)/(1-x/(1-(x+xy)/(1-x/(1-(x+xy)/(1-.... (continued fraction). (End)
T(n,k) = [k<=n]*(Sum_{j=0..n} binomial(n,j)^2*binomial(j,k))/(n-k+1). - Paul Barry, May 28 2009
T(n,k) = A104684(n,k)/(n-k+1). - Peter Luschny, May 17 2011
From Tom Copeland, Sep 21 2011: (Start)
With F(x,t) = (1-(2+t)*x-sqrt(1-2*(2+t)*x+(t*x)^2))/(2*x) an o.g.f. (nulling the n=0 term) in x for the A060693 polynomials in t,
  G(x,t) = x/(1+t+(2+t)*x+x^2) is the compositional inverse in x.
Consequently, with H(x,t) = 1/(dG(x,t)/dx) = (1+t+(2+t)*x+x^2)^2 / (1+t-x^2), the n-th A060693 polynomial in t is given by (1/n!)*((H(x,t)*d/dx)^n) x evaluated at x=0, i.e., F(x,t) = exp(x*H(u,t)*d/d) u, evaluated at u = 0.
  Also, dF(x,t)/dx = H(F(x,t),t). (End)
See my 2008 formulas in A033282 to relate this entry to A088617, A001263, A086810, and other matrices. - Tom Copeland, Jan 22 2016
Rows of this entry are non-vanishing antidiagonals of A097610. See p. 14 of Agapito et al. for a bivariate generating function and its inverse. - Tom Copeland, Feb 03 2016
From Werner Schulte, Jan 09 2017: (Start)
T(n,k) = A126216(n,k-1) + A126216(n,k) for 0 < k < n;
Sum_{k=0..n} (-1)^k*(1+x*(n-k))*T(n,k) = x + (1-x)*A000007(n).
(End)
Conjecture: Sum_{k=0..n} (-1)^k*T(n,k)*(n+1-k)^2 = 1+n+n^2. - Werner Schulte, Jan 11 2017

More terms from Vladeta Jovovic, Apr 21 2001
New description from Philippe Deléham, Aug 12 2003
New name using a comment by Emeric Deutsch from Peter Luschny, Jul 26 2017

cp's OEIS Frontend

A092392 Triangle read by rows: T(n,k) = C(2*n - k,n), 0 <= k <= n.

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A165817 Number of compositions (= ordered integer partitions) of n into 2n parts.

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A060540 Square array read by antidiagonals downwards: T(n,k) = (n*k)!/(k!^n*n!), (n>=1, k>=1), the number of ways of dividing nk labeled items into n unlabeled boxes with k items in each box.

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A349053 Number of non-weakly alternating integer compositions of n.

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A357641 Number of integer compositions of 2n whose half-alternating sum is 0.

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A059481 Triangle read by rows. T(n, k) = binomial(n+k-1, k) for 0 <= k <= n.

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A060540 Square array read by antidiagonals downwards: T(n,k) = (nk)!/(k!^nn!), (n>=1, k>=1), the number of ways of dividing nk labeled items into n unlabeled boxes with k items in each box.