cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A001263 Triangle of Narayana numbers T(n,k) = C(n-1,k-1)*C(n,k-1)/k with 1 <= k <= n, read by rows. Also called the Catalan triangle.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 6, 1, 1, 10, 20, 10, 1, 1, 15, 50, 50, 15, 1, 1, 21, 105, 175, 105, 21, 1, 1, 28, 196, 490, 490, 196, 28, 1, 1, 36, 336, 1176, 1764, 1176, 336, 36, 1, 1, 45, 540, 2520, 5292, 5292, 2520, 540, 45, 1, 1, 55, 825, 4950, 13860, 19404, 13860, 4950, 825
Offset: 1

Views

Author

Keywords

Comments

Number of antichains (or order ideals) in the poset 2*(k-1)*(n-k) or plane partitions with rows <= k-1, columns <= n-k and entries <= 2. - Mitch Harris, Jul 15 2000
T(n,k) is the number of Dyck n-paths with exactly k peaks. a(n,k) = number of pairs (P,Q) of lattice paths from (0,0) to (k,n+1-k), each consisting of unit steps East or North, such that P lies strictly above Q except at the endpoints. - David Callan, Mar 23 2004
Number of permutations of [n] which avoid-132 and have k-1 descents. - Mike Zabrocki, Aug 26 2004
T(n,k) is the number of paths through n panes of glass, entering and leaving from one side, of length 2n with k reflections (where traversing one pane of glass is the unit length). - Mitch Harris, Jul 06 2006
Antidiagonal sums given by A004148 (without first term).
T(n,k) is the number of full binary trees with n internal nodes and k-1 jumps. In the preorder traversal of a full binary tree, any transition from a node at a deeper level to a node on a strictly higher level is called a jump. - Emeric Deutsch, Jan 18 2007
From Gary W. Adamson, Oct 22 2007: (Start)
The n-th row can be generated by the following operation using an ascending row of (n-1) triangular terms, (A) and a descending row, (B); e.g., row 6:
A: 1....3....6....10....15
B: 15...10....6.....3.....1
C: 1...15...50....50....15....1 = row 6.
Leftmost column of A,B -> first two terms of C; then followed by the operation B*C/A of current column = next term of row C, (e.g., 10*15/3 = 50). Continuing with the operation, we get row 6: (1, 15, 50, 50, 15, 1). (End)
The previous comment can be upgraded to: The ConvOffsStoT transform of the triangular series; and by rows, row 6 is the ConvOffs transform of (1, 3, 6, 10, 15). Refer to triangle A117401 as another example of the ConvOffsStoT transform, and OEIS under Maple Transforms. - Gary W. Adamson, Jul 09 2012
For a connection to Lagrange inversion, see A134264. - Tom Copeland, Aug 15 2008
T(n,k) is also the number of order-decreasing and order-preserving mappings (of an n-element set) of height k (height of a mapping is the cardinal of its image set). - Abdullahi Umar, Aug 21 2008
Row n of this triangle is the h-vector of the simplicial complex dual to an associahedron of type A_n [Fomin & Reading, p.60]. See A033282 for the corresponding array of f-vectors for associahedra of type A_n. See A008459 and A145903 for the h-vectors for associahedra of type B and type D respectively. The Hilbert transform of this triangle (see A145905 for the definition of this transform) is A145904. - Peter Bala, Oct 27 2008
T(n,k) is also the number of noncrossing set partitions of [n] into k blocks. Given a partition P of the set {1,2,...,n}, a crossing in P are four integers [a, b, c, d] with 1 <= a < b < c < d <= n for which a, c are together in a block, and b, d are together in a different block. A noncrossing partition is a partition with no crossings. - Peter Luschny, Apr 29 2011
Noncrossing set partitions are also called genus 0 partitions. In terms of genus-dependent Stirling numbers of the second kind S2(n,k,g) that count partitions of genus g of an n-set into k nonempty subsets, one has T(n,k) = S2(n,k,0). - Robert Coquereaux, Feb 15 2024
Diagonals of A089732 are rows of A001263. - Tom Copeland, May 14 2012
From Peter Bala, Aug 07 2013: (Start)
Let E(y) = Sum_{n >= 0} y^n/(n!*(n+1)!) = 1/sqrt(y)*BesselI(1,2*sqrt(y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence n!*(n+1)! as defined in Wang and Wang.
Generating function E(y)*E(x*y) = 1 + (1 + x)*y/(1!*2!) + (1 + 3*x + x^2)*y^2/(2!*3!) + (1 + 6*x + 6*x^2 + x^3)*y^3/(3!*4!) + .... Cf. A105278 with a generating function exp(y)*E(x*y).
The n-th power of this array has a generating function E(y)^n*E(x*y). In particular, the matrix inverse A103364 has a generating function E(x*y)/E(y). (End)
T(n,k) is the number of nonintersecting n arches above the x axis, starting and ending on vertices 1 to 2n, with k being the number of arches starting on an odd vertice and ending on a higher even vertice. Example: T(3,2)=3 [16,25,34] [14,23,56] [12,36,45]. - Roger Ford, Jun 14 2014
Fomin and Reading on p. 31 state that the rows of the Narayana matrix are the h-vectors of the associahedra as well as its dual. - Tom Copeland, Jun 27 2017
The row polynomials P(n, x) = Sum_{k=1..n} T(n, k)*x^(k-1), together with P(0, x) = 1, multiplied by (n+1) are the numerator polynomials of the o.g.f.s of the diagonal sequences of the triangle A103371: G(n, x) = (n+1)*P(n, x)/(1 - x)^{2*n+1}, for n >= 0. This is proved with Lagrange's theorem applied to the Riordan triangle A135278 = (1/(1 - x)^2, x/(1 - x)). See an example below. - Wolfdieter Lang, Jul 31 2017
T(n,k) is the number of Dyck paths of semilength n with k-1 uu-blocks (pairs of consecutive up-steps). - Alexander Burstein, Jun 22 2020
In case you were searching for Narayama numbers, the correct spelling is Narayana. - N. J. A. Sloane, Nov 11 2020
Named after the Canadian mathematician Tadepalli Venkata Narayana (1930-1987). They were also called "Runyon numbers" after John P. Runyon (1922-2013) of Bell Telephone Laboratories, who used them in a study of a telephone traffic system. - Amiram Eldar, Apr 15 2021 The Narayana numbers were first studied by Percy Alexander MacMahon (see reference, Article 495) as pointed out by Bóna and Sagan (see link). - Peter Luschny, Apr 28 2022
From Andrea Arlette España, Nov 14 2022: (Start)
T(n,k) is the degree distribution of the paths towards synchronization in the transition diagram associated with the Laplacian system over the complete graph K_n, corresponding to ordered initial conditions x_1 < x_2 < ... < x_n.
T(n,k) for n=2N+1 and k=N+1 is the number of states in the transition diagram associated with the Laplacian system over the complete bipartite graph K_{N,N}, corresponding to ordered (x_1 < x_2 < ... < x_N and x_{N+1} < x_{N+2} < ... < x_{2N}) and balanced (Sum_{i=1..N} x_i/N = Sum_{i=N+1..2N} x_i/N) initial conditions. (End)
From Gus Wiseman, Jan 23 2023: (Start)
Also the number of unlabeled ordered rooted trees with n nodes and k leaves. See the link by Marko Riedel. For example, row n = 5 counts the following trees:
((((o)))) (((o))o) ((o)oo) (oooo)
(((o)o)) ((oo)o)
(((oo))) ((ooo))
((o)(o)) (o(o)o)
((o(o))) (o(oo))
(o((o))) (oo(o))
The unordered version is A055277. Leaves in standard ordered trees are counted by A358371. (End)

Examples

			The initial rows of the triangle are:
  [1] 1
  [2] 1,  1
  [3] 1,  3,   1
  [4] 1,  6,   6,    1
  [5] 1, 10,  20,   10,    1
  [6] 1, 15,  50,   50,   15,    1
  [7] 1, 21, 105,  175,  105,   21,   1
  [8] 1, 28, 196,  490,  490,  196,  28,  1
  [9] 1, 36, 336, 1176, 1764, 1176, 336, 36, 1;
  ...
For all n, 12...n (1 block) and 1|2|3|...|n (n blocks) are noncrossing set partitions.
Example of umbral representation:
  A007318(5,k)=[1,5/1,5*4/(2*1),...,1]=(1,5,10,10,5,1),
  so A001263(5,k)={1,b(5)/b(1),b(5)*b(4)/[b(2)*b(1)],...,1}
  = [1,30/2,30*20/(6*2),...,1]=(1,15,50,50,15,1).
  First = last term = b.(5!)/[b.(0!)*b.(5!)]= 1. - _Tom Copeland_, Sep 21 2011
Row polynomials and diagonal sequences of A103371: n = 4,  P(4, x) = 1 + 6*x + 6*x^2 + x^3, and the o.g.f. of fifth diagonal is G(4, x) = 5* P(4, x)/(1 - x)^9, namely [5, 75, 525, ...]. See a comment above. - _Wolfdieter Lang_, Jul 31 2017
		

References

  • Berman and Koehler, Cardinalities of finite distributive lattices, Mitteilungen aus dem Mathematischen Seminar Giessen, 121 (1976), pp. 103-124.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 196.
  • P. A. MacMahon, Combinatory Analysis, Vols. 1 and 2, Cambridge University Press, 1915, 1916; reprinted by Chelsea, 1960, Sect. 495.
  • T. V. Narayana, Lattice Path Combinatorics with Statistical Applications. Univ. Toronto Press, 1979, pp. 100-101.
  • A. Nkwanta, Lattice paths and RNA secondary structures, in African Americans in Mathematics, ed. N. Dean, Amer. Math. Soc., 1997, pp. 137-147.
  • T. K. Petersen, Eulerian Numbers, Birkhäuser, 2015, Chapter 2.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 17.
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 6.36(a) and (b).

Crossrefs

Other versions are in A090181 and A131198. - Philippe Deléham, Nov 18 2007
Cf. variants: A181143, A181144. - Paul D. Hanna, Oct 13 2010
Row sums give A000108 (Catalan numbers), n>0.
A008459 (h-vectors type B associahedra), A033282 (f-vectors type A associahedra), A145903 (h-vectors type D associahedra), A145904 (Hilbert transform). - Peter Bala, Oct 27 2008
Cf. A016098 and A189232 for numbers of crossing set partitions.
Cf. A243752.
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • GAP
    Flat(List([1..11],n->List([1..n],k->Binomial(n-1,k-1)*Binomial(n,k-1)/k))); # Muniru A Asiru, Jul 12 2018
  • Haskell
    a001263 n k = a001263_tabl !! (n-1) !! (k-1)
    a001263_row n = a001263_tabl !! (n-1)
    a001263_tabl = zipWith dt a007318_tabl (tail a007318_tabl) where
       dt us vs = zipWith (-) (zipWith (*) us (tail vs))
                              (zipWith (*) (tail us ++ [0]) (init vs))
    -- Reinhard Zumkeller, Oct 10 2013
    
  • Magma
    /* triangle */ [[Binomial(n-1,k-1)*Binomial(n,k-1)/k : k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 19 2014
    
  • Maple
    A001263 := (n,k)->binomial(n-1,k-1)*binomial(n,k-1)/k;
    a:=proc(n,k) option remember; local i; if k=1 or k=n then 1 else add(binomial(n+i-1, 2*k-2)*a(k-1,i),i=1..k-1); fi; end:
    # Alternatively, as a (0,0)-based triangle:
    R := n -> simplify(hypergeom([-n, -n-1], [2], x)): Trow := n -> seq(coeff(R(n,x),x,j), j=0..n): seq(Trow(n), n=0..9); # Peter Luschny, Mar 19 2018
  • Mathematica
    T[n_, k_] := If[k==0, 0, Binomial[n-1, k-1] Binomial[n, k-1] / k];
    Flatten[Table[Binomial[n-1,k-1] Binomial[n,k-1]/k,{n,15},{k,n}]] (* Harvey P. Dale, Feb 29 2012 *)
    TRow[n_] := CoefficientList[Hypergeometric2F1[1 - n, -n, 2, x], x];
    Table[TRow[n], {n, 1, 11}] // Flatten (* Peter Luschny, Mar 19 2018 *)
    aot[n_]:=If[n==1,{{}},Join@@Table[Tuples[aot/@c],{c,Join@@Permutations/@IntegerPartitions[n-1]}]];
    Table[Length[Select[aot[n],Length[Position[#,{}]]==k&]],{n,2,9},{k,1,n-1}] (* Gus Wiseman, Jan 23 2023 *)
    T[1, 1] := 1; T[n_, k_]/;1<=k<=n := T[n, k] = (2n/k-1) T[n-1,k-1] + T[n-1, k]; T[n_, k_] := 0; Flatten@Table[T[n, k], {n, 1, 11}, {k, 1, n}] (* Oliver Seipel, Dec 31 2024 *)
  • PARI
    {a(n, k) = if(k==0, 0, binomial(n-1, k-1) * binomial(n, k-1) / k)};
    
  • PARI
    {T(n,k)=polcoeff(polcoeff(exp(sum(m=1,n,sum(j=0,m,binomial(m,j)^2*y^j)*x^m/m) +O(x^(n+1))),n,x),k,y)} \\ Paul D. Hanna, Oct 13 2010
    
  • Sage
    @CachedFunction
    def T(n, k):
        if k == n or k == 1: return 1
        if k <= 0 or k > n: return 0
        return binomial(n, 2) * (T(n-1, k)/((n-k)*(n-k+1)) + T(n-1, k-1)/(k*(k-1)))
    for n in (1..9): print([T(n, k) for k in (1..n)])  # Peter Luschny, Oct 28 2014
    

Formula

a(n, k) = C(n-1, k-1)*C(n, k-1)/k for k!=0; a(n, 0)=0.
Triangle equals [0, 1, 0, 1, 0, 1, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...] where DELTA is Deléham's operator defined in A084938.
0Mike Zabrocki, Aug 26 2004
T(n, k) = C(n, k)*C(n-1, k-1) - C(n, k-1)*C(n-1, k) (determinant of a 2 X 2 subarray of Pascal's triangle A007318). - Gerald McGarvey, Feb 24 2005
T(n, k) = binomial(n-1, k-1)^2 - binomial(n-1, k)*binomial(n-1, k-2). - David Callan, Nov 02 2005
a(n,k) = C(n,2) (a(n-1,k)/((n-k)*(n-k+1)) + a(n-1,k-1)/(k*(k-1))) a(n,k) = C(n,k)*C(n,k-1)/n. - Mitch Harris, Jul 06 2006
Central column = A000891, (2n)!*(2n+1)! / (n!*(n+1)!)^2. - Zerinvary Lajos, Oct 29 2006
G.f.: (1-x*(1+y)-sqrt((1-x*(1+y))^2-4*y*x^2))/(2*x) = Sum_{n>0, k>0} a(n, k)*x^n*y^k.
From Peter Bala, Oct 22 2008: (Start)
Relation with Jacobi polynomials of parameter (1,1):
Row n+1 generating polynomial equals 1/(n+1)*x*(1-x)^n*Jacobi_P(n,1,1,(1+x)/(1-x)). It follows that the zeros of the Narayana polynomials are all real and nonpositive, as noted above. O.g.f for column k+2: 1/(k+1) * y^(k+2)/(1-y)^(k+3) * Jacobi_P(k,1,1,(1+y)/(1-y)). Cf. A008459.
T(n+1,k) is the number of walks of n unit steps on the square lattice (i.e., each step in the direction either up (U), down (D), right (R) or left (L)) starting from the origin and finishing at lattice points on the x axis and which remain in the upper half-plane y >= 0 [Guy]. For example, T(4,3) = 6 counts the six walks RRL, LRR, RLR, UDL, URD and RUD, from the origin to the lattice point (1,0), each of 3 steps. Compare with tables A145596 - A145599.
Define a functional I on formal power series of the form f(x) = 1 + ax + bx^2 + ... by the following iterative process. Define inductively f^(1)(x) = f(x) and f^(n+1)(x) = f(x*f^(n)(x)) for n >= 1. Then set I(f(x)) = lim_{n -> infinity} f^(n)(x) in the x-adic topology on the ring of formal power series; the operator I may also be defined by I(f(x)) := 1/x*series reversion of x/f(x).
The o.g.f. for this array is I(1 + t*x + t*x^2 + t*x^3 + ...) = 1 + t*x + (t + t^2)*x^2 + (t + 3*t^2 + t^3)*x^3 + ... = 1/(1 - x*t/(1 - x/(1 - x*t/(1 - x/(1 - ...))))) (as a continued fraction). Cf. A108767, A132081 and A141618. (End)
G.f.: 1/(1-x-xy-x^2y/(1-x-xy-x^2y/(1-... (continued fraction). - Paul Barry, Sep 28 2010
E.g.f.: exp((1+y)x)*Bessel_I(1,2*sqrt(y)x)/(sqrt(y)*x). - Paul Barry, Sep 28 2010
G.f.: A(x,y) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^2*y^k] * x^n/n ). - Paul D. Hanna, Oct 13 2010
With F(x,t) = (1-(1+t)*x-sqrt(1-2*(1+t)*x+((t-1)*x)^2))/(2*x) an o.g.f. in x for the Narayana polynomials in t, G(x,t) = x/(t+(1+t)*x+x^2) is the compositional inverse in x. Consequently, with H(x,t) = 1/ (dG(x,t)/dx) = (t+(1+t)*x+x^2)^2 / (t-x^2), the n-th Narayana polynomial in t is given by (1/n!)*((H(x,t)*D_x)^n)x evaluated at x=0, i.e., F(x,t) = exp(x*H(u,t)*D_u)u, evaluated at u = 0. Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 04 2011
With offset 0, A001263 = Sum_{j>=0} A132710^j / A010790(j), a normalized Bessel fct. May be represented as the Pascal matrix A007318, n!/[(n-k)!*k!], umbralized with b(n)=A002378(n) for n>0 and b(0)=1: A001263(n,k)= b.(n!)/{b.[(n-k)!]*b.(k!)} where b.(n!) = b(n)*b(n-1)...*b(0), a generalized factorial (see example). - Tom Copeland, Sep 21 2011
With F(x,t) = {1-(1-t)*x-sqrt[1-2*(1+t)*x+[(t-1)*x]^2]}/2 a shifted o.g.f. in x for the Narayana polynomials in t, G(x,t)= x/[t-1+1/(1-x)] is the compositional inverse in x. Therefore, with H(x,t)=1/(dG(x,t)/dx)=[t-1+1/(1-x)]^2/{t-[x/(1-x)]^2}, (see A119900), the (n-1)-th Narayana 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/du) u, evaluated at u = 0. Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 30 2011
T(n,k) = binomial(n-1,k-1)*binomial(n+1,k)-binomial(n,k-1)*binomial(n,k). - Philippe Deléham, Nov 05 2011
A166360(n-k) = T(n,k) mod 2. - Reinhard Zumkeller, Oct 10 2013
Damped sum of a column, in leading order: lim_{d->0} d^(2k-1) Sum_{N>=k} T(N,k)(1-d)^N=Catalan(n). - Joachim Wuttke, Sep 11 2014
Multiplying the n-th column by n! generates the revert of the unsigned Lah numbers, A089231. - Tom Copeland, Jan 07 2016
Row polynomials: (x - 1)^(n+1)*(P(n+1,(1 + x)/(x - 1)) - P(n-1,(1 + x)/(x - 1)))/((4*n + 2)), n = 1,2,... and where P(n,x) denotes the n-th Legendre polynomial. - Peter Bala, Mar 03 2017
The coefficients of the row polynomials R(n, x) = hypergeom([-n,-n-1], [2], x) generate the triangle based in (0,0). - Peter Luschny, Mar 19 2018
Multiplying the n-th diagonal by n!, with the main diagonal n=1, generates the Lah matrix A105278. With G equal to the infinitesimal generator of A132710, the Narayana triangle equals Sum_{n >= 0} G^n/((n+1)!*n!) = (sqrt(G))^(-1) * I_1(2*sqrt(G)), where G^0 is the identity matrix and I_1(x) is the modified Bessel function of the first kind of order 1. (cf. Sep 21 2011 formula also.) - Tom Copeland, Sep 23 2020
T(n,k) = T(n,k-1)*C(n-k+2,2)/C(k,2). - Yuchun Ji, Dec 21 2020
From Sergii Voloshyn, Nov 25 2024: (Start)
G.f.: F(x,y) = (1-x*(1+y)-sqrt((1-x*(1+y))^2-4*y*x^2))/(2*x) is the solution of the differential equation x^3 * d^2(x*F(x,y))/dx^2 = y * d^2(x*F(x,y))/dy^2.
Let E be the operator x*D*D, where D denotes the derivative operator d/dx. Then (1/(n! (1 + n)!)) * E^n(x/(1 - x)) = (row n generating polynomial)/(1 - x)^(2*n+1) = Sum_{k >= 0} C(n-1, k-1)*C(n, k-1)/k*x^k. For example, when n = 4 we have (1/4!/5!)*E^3(x/(1 - x)) = x (1 + 6 x + 6 x^2 + x^3)/(1 - x)^9. (End)

Extensions

Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A033282 Triangle read by rows: T(n, k) is the number of diagonal dissections of a convex n-gon into k+1 regions.

Original entry on oeis.org

1, 1, 2, 1, 5, 5, 1, 9, 21, 14, 1, 14, 56, 84, 42, 1, 20, 120, 300, 330, 132, 1, 27, 225, 825, 1485, 1287, 429, 1, 35, 385, 1925, 5005, 7007, 5005, 1430, 1, 44, 616, 4004, 14014, 28028, 32032, 19448, 4862, 1, 54, 936, 7644, 34398, 91728, 148512, 143208, 75582, 16796
Offset: 3

Views

Author

Keywords

Comments

T(n+3, k) is also the number of compatible k-sets of cluster variables in Fomin and Zelevinsky's cluster algebra of finite type A_n. Take a row of this triangle regarded as a polynomial in x and rewrite as a polynomial in y := x+1. The coefficients of the polynomial in y give a row of the triangle of Narayana numbers A001263. For example, x^2 + 5*x + 5 = y^2 + 3*y + 1. - Paul Boddington, Mar 07 2003
Number of standard Young tableaux of shape (k+1,k+1,1^(n-k-3)), where 1^(n-k-3) denotes a sequence of n-k-3 1's (see the Stanley reference).
Number of k-dimensional 'faces' of the n-dimensional associahedron (see Simion, p. 168). - Mitch Harris, Jan 16 2007
Mirror image of triangle A126216. - Philippe Deléham, Oct 19 2007
For relation to Lagrange inversion or series reversion and the geometry of associahedra or Stasheff polytopes (and other combinatorial objects) see A133437. - Tom Copeland, Sep 29 2008
Row generating polynomials 1/(n+1)*Jacobi_P(n,1,1,2*x+1). Row n of this triangle is the f-vector of the simplicial complex dual to an associahedron of type A_n [Fomin & Reading, p. 60]. See A001263 for the corresponding array of h-vectors for associahedra of type A_n. See A063007 and A080721 for the f-vectors for associahedra of type B and type D respectively. - Peter Bala, Oct 28 2008
f-vectors of secondary polytopes for Grobner bases for optimization and integer programming (see De Loera et al. and Thomas). - Tom Copeland, Oct 11 2011
From Devadoss and O'Rourke's book: The Fulton-MacPherson compactification of the configuration space of n free particles on a line segment with a fixed particle at each end is the n-Dim Stasheff associahedron whose refined f-vector is given in A133437 which reduces to A033282. - Tom Copeland, Nov 29 2011
Diagonals of A132081 are rows of A033282. - Tom Copeland, May 08 2012
The general results on the convolution of the refined partition polynomials of A133437, with u_1 = 1 and u_n = -t otherwise, can be applied here to obtain results of convolutions of these polynomials. - Tom Copeland, Sep 20 2016
The signed triangle t(n, k) =(-1)^k* T(n+2, k-1), n >= 1, k = 1..n, seems to be obtainable from the partition array A111785 (in Abramowitz-Stegun order) by adding the entries corresponding to the partitions of n with the number of parts k. E.g., triangle t, row n=4: -1, (6+3) = 9, -21, 14. - Wolfdieter Lang, Mar 17 2017
The preceding conjecture by Lang is true. It is implicit in Copeland's 2011 comments in A086810 on the relations among a gf and its compositional inverse for that entry and inversion through A133437 (a differently normalized version of A111785), whose integer partitions are the same as those for A134685. (An inversion pair in Copeland's 2008 formulas below can also be used to prove the conjecture.) In addition, it follows from the relation between the inversion formula of A111785/A133437 and the enumeration of distinct faces of associahedra. See the MathOverflow link concernimg Loday and the Aguiar and Ardila reference in A133437 for proofs of the relations between the partition polynomials for inversion and enumeration of the distinct faces of the A_n associahedra, or Stasheff polytopes. - Tom Copeland, Dec 21 2017
The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial (x+1)*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1)/(n!*(n+1)!) in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 07 2022
Chapoton's observation above is correct: the precise expansion is (x+1)*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1)/ (n!*(n+1)!) = Sum_{k = 0..n-1} (-1)^k*T(n+2,n-k-1)*binomial(x+2*n-k,2*n-k), as can be verified using the WZ algorithm. For example, n = 4 gives (x+1)*(x+2)^2*(x+3)^2*(x+4)^2*(x+5)/(4!*5!) = 14*binomial(x+8,8) - 21*binomial(x+7,7) + 9*binomial(x+6,6) - binomial(x+5,5). - Peter Bala, Jun 24 2023

Examples

			The triangle T(n, k) begins:
n\k  0  1   2    3     4     5      6      7     8     9
3:   1
4:   1  2
5:   1  5   5
6:   1  9  21   14
7:   1 14  56   84    42
8:   1 20 120  300   330   132
9:   1 27 225  825  1485  1287    429
10:  1 35 385 1925  5005  7007   5005   1430
11:  1 44 616 4004 14014 28028  32032  19448  4862
12:  1 54 936 7644 34398 91728 148512 143208 75582 16796
... reformatted. - _Wolfdieter Lang_, Mar 17 2017
		

References

  • S. Devadoss and J. O'Rourke, Discrete and Computational Geometry, Princeton Univ. Press, 2011 (See p. 241.)
  • Ronald L. Graham, Donald E. Knuth, Oren Patashnik, Concrete Mathematics, 2nd ed., Addison-Wesley, 1994. Exercise 7.50, pages 379, 573.
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Section 5.8.

Crossrefs

Cf. diagonals: A000012, A000096, A033275, A033276, A033277, A033278, A033279; A000108, A002054, A002055, A002056, A007160, A033280, A033281; row sums: A001003 (Schroeder numbers, first term omitted). See A086810 for another version.
A007160 is a diagonal. Cf. A001263.
With leading zero: A086810.
Cf. A019538 'faces' of the permutohedron.
Cf. A063007 (f-vectors type B associahedra), A080721 (f-vectors type D associahedra), A126216 (mirror image).
Cf. A248727 for a relation to f-polynomials of simplices.
Cf. A111785 (contracted partition array, unsigned; see a comment above).
Antidiagonal sums give A005043. - Jordan Tirrell, Jun 01 2017

Programs

  • Magma
    [[Binomial(n-3, k)*Binomial(n+k-1, k)/(k+1): k in [0..(n-3)]]: n in [3..12]];  // G. C. Greubel, Nov 19 2018
    
  • Maple
    T:=(n,k)->binomial(n-3,k)*binomial(n+k-1,k)/(k+1): seq(seq(T(n,k),k=0..n-3),n=3..12); # Muniru A Asiru, Nov 24 2018
  • Mathematica
    t[n_, k_] = Binomial[n-3, k]*Binomial[n+k-1, k]/(k+1);
    Flatten[Table[t[n, k], {n, 3, 12}, {k, 0, n-3}]][[1 ;; 52]] (* Jean-François Alcover, Jun 16 2011 *)
  • PARI
    Q=(1+z-(1-(4*w+2+O(w^20))*z+z^2+O(z^20))^(1/2))/(2*(1+w)*z);for(n=3,12,for(m=1,n-2,print1(polcoef(polcoef(Q,n-2,z),m,w),", "))) \\ Hugo Pfoertner, Nov 19 2018
    
  • PARI
    for(n=3,12, for(k=0,n-3, print1(binomial(n-3,k)*binomial(n+k-1,k)/(k+1), ", "))) \\ G. C. Greubel, Nov 19 2018
    
  • Sage
    [[ binomial(n-3,k)*binomial(n+k-1,k)/(k+1) for k in (0..(n-3))] for n in (3..12)] # G. C. Greubel, Nov 19 2018

Formula

G.f. G = G(t, z) satisfies (1+t)*G^2 - z*(1-z-2*t*z)*G + t*z^4 = 0.
T(n, k) = binomial(n-3, k)*binomial(n+k-1, k)/(k+1) for n >= 3, 0 <= k <= n-3.
From Tom Copeland, Nov 03 2008: (Start)
Two g.f.s (f1 and f2) for A033282 and their inverses (x1 and x2) can be derived from the Drake and Barry references.
1. a: f1(x,t) = y = {1 - (2t+1) x - sqrt[1 - (2t+1) 2x + x^2]}/[2x (t+1)] = t x + (t + 2 t^2) x^2 + (t + 5 t^2 + 5 t^3) x^3 + ...
b: x1 = y/[t + (2t+1)y + (t+1)y^2] = y {1/[t/(t+1) + y] - 1/(1+y)} = (y/t) - (1+2t)(y/t)^2 + (1+ 3t + 3t^2)(y/t)^3 +...
2. a: f2(x,t) = y = {1 - x - sqrt[(1-x)^2 - 4xt]}/[2(t+1)] = (t/(t+1)) x + t x^2 + (t + 2 t^2) x^3 + (t + 5 t^2 + 5 t^3) x^4 + ...
b: x2 = y(t+1) [1- y(t+1)]/[t + y(t+1)] = (t+1) (y/t) - (t+1)^3 (y/t)^2 + (t+1)^4 (y/t)^3 + ...
c: y/x2(y,t) = [t/(t+1) + y] / [1- y(t+1)] = t/(t+1) + (1+t) y + (1+t)^2 y^2 + (1+t)^3 y^3 + ...
x2(y,t) can be used along with the Lagrange inversion for an o.g.f. (A133437) to generate A033282 and show that A133437 is a refinement of A033282, i.e., a refinement of the f-polynomials of the associahedra, the Stasheff polytopes.
y/x2(y,t) can be used along with the indirect Lagrange inversion (A134264) to generate A033282 and show that A134264 is a refinement of A001263, i.e., a refinement of the h-polynomials of the associahedra.
f1[x,t](t+1) gives a generator for A088617.
f1[xt,1/t](t+1) gives a generator for A060693, with inverse y/[1 + t + (2+t) y + y^2].
f1[x(t-1),1/(t-1)]t gives a generator for A001263, with inverse y/[t + (1+t) y + y^2].
The unsigned coefficients of x1(y t,t) are A074909, reverse rows of A135278. (End)
G.f.: 1/(1-x*y-(x+x*y)/(1-x*y/(1-(x+x*y)/(1-x*y/(1-(x+x*y)/(1-x*y/(1-.... (continued fraction). - Paul Barry, Feb 06 2009
Let h(t) = (1-t)^2/(1+(u-1)*(1-t)^2) = 1/(u + 2*t + 3*t^2 + 4*t^3 + ...), then a signed (n-1)-th row polynomial of A033282 is given by u^(2n-1)*(1/n!)*((h(t)*d/dt)^n) t, evaluated at t=0, with initial n=2. The power series expansion of h(t) is related to A181289 (cf. A086810). - Tom Copeland, Sep 06 2011
With a different offset, the row polynomials equal 1/(1 + x)*Integral_{0..x} R(n,t) dt, where R(n,t) = Sum_{k = 0..n} binomial(n,k)*binomial(n+k,k)*t^k are the row polynomials of A063007. - Peter Bala, Jun 23 2016
n-th row polynomial = ( LegendreP(n-1,2*x + 1) - LegendreP(n-3,2*x + 1) )/((4*n - 6)*x*(x + 1)), n >= 3. - Peter Bala, Feb 22 2017
n*T(n+1, k) = (4n-6)*T(n, k-1) + (2n-3)*T(n, k) - (n-3)*T(n-1, k) for n >= 4. - Fang Lixing, May 07 2019

Extensions

Missing factor of 2 for expansions of f1 and f2 added by Tom Copeland, Apr 12 2009

A086810 Triangle obtained by adding a leading diagonal 1,0,0,0,... to A033282.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 5, 5, 0, 1, 9, 21, 14, 0, 1, 14, 56, 84, 42, 0, 1, 20, 120, 300, 330, 132, 0, 1, 27, 225, 825, 1485, 1287, 429, 0, 1, 35, 385, 1925, 5005, 7007, 5005, 1430, 0, 1, 44, 616, 4004, 14014, 28028, 32032, 19448, 4862, 0, 1, 54, 936, 7644, 34398, 91728
Offset: 0

Views

Author

Philippe Deléham, Aug 05 2003

Keywords

Comments

Mirror image of triangle A133336. - Philippe Deléham, Dec 10 2008
From Tom Copeland, Oct 09 2011: (Start)
With polynomials
P(0,t) = 0
P(1,t) = 1
P(2,t) = t
P(3,t) = t + 2 t^2
P(4,t) = t + 5 t^2 + 5 t^3
P(5,t) = t + 9 t^2 + 21 t^3 + 14 t^4
The o.g.f. A(x,t) = {1+x-sqrt[(1-x)^2-4xt]}/[2(1+t)] (see Drake et al.).
B(x,t)= x-t x^2/(1-x)= x-t(x^2+x^3+x^4+...) is the comp. inverse in x.
Let h(x,t) = 1/(dB/dx) = (1-x)^2/(1+(1+t)*x*(x-2)) = 1/(1-t(2x+3x^2+4x^3+...)), an o.g.f. in x for row polynomials in t of A181289. Then P(n,t) is given by (1/n!)(h(x,t)*d/dx)^n x, evaluated at x=0, A = exp(x*h(y,t)*d/dy) y, eval. at y=0, and dA/dx = h(A(x,t),t). These results are a special case of A133437 with u(x,t) = B(x,t), i.e., u_1=1 and (u_n)=-t for n > 1. See A001003 for t = 1. (End)
Let U(x,t) = [A(x,t)-x]/t, then U(x,0) = -dB(x,t)/dt and U satisfies dU/dt = UdU/dx, the inviscid Burgers' equation (Wikipedia), also called the Hopf equation (see Buchstaber et al.). Also U(x,t) = U(A(x,t),0) = U(x+tU,0) since U(x,0) = [x-B(x,t)]/t. - Tom Copeland, Mar 12 2012
Diagonals of A132081 are essentially rows of this sequence. - Tom Copeland, May 08 2012
T(r, s) is the number of [0,r]-covering hierarchies with s segments (see Kreweras). - Michel Marcus, Nov 22 2014
From Yu Hin Au, Dec 07 2019: (Start)
T(n,k) is the number of small Schröder n-paths (lattice paths from (0,0) to (2n,0) using steps U=(1,1), F=(2,0), D=(1,-1) with no F step on the x-axis) that has exactly k U steps.
T(n,k) is the number of Schröder trees (plane rooted tree where each internal node has at least two children) with exactly n+1 leaves and k internal nodes. (End)

Examples

			Triangle starts:
  1;
  0,  1;
  0,  1,  2;
  0,  1,  5,  5;
  0,  1,  9, 21, 14;
  ...
		

Crossrefs

The diagonals (except for A000007) are also the diagonals of A033282.
Row sums: A001003 (Schroeder numbers).

Programs

  • Mathematica
    Table[Boole[n == 2] + If[# == -1, 0, Binomial[n - 3, #] Binomial[n + # - 1, #]/(# + 1)] &[k - 1], {n, 2, 12}, {k, 0, n - 2}] // Flatten (* after Jean-François Alcover at A033282, or *)
    Table[If[n == 0, 1, Binomial[n, k] Binomial[n + k, k - 1]/n], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Aug 22 2016 *)
  • PARI
    t(n, k) = if (n==0, 1, binomial(n, k)*binomial(n+k, k-1)/n);
    tabl(nn) = {for (n=0, nn, for (k=0, n, print1(t(n,k), ", ");); print(););} \\ Michel Marcus, Nov 22 2014

Formula

Triangle T(n, k) read by rows; given by [0, 1, 0, 1, 0, 1, ...] DELTA [1, 1, 1, 1, 1, 1, 1, 1, 1, ...] where DELTA is Deléham's operator defined in A084938.
For k>0, T(n, k) = binomial(n-1, k-1)*binomial(n+k, k)/(n+1); T(0, 0) = 1 and T(n, 0) = 0 if n > 0. [corrected by Marko Riedel, May 04 2023]
Sum_{k>=0} T(n, k)*2^k = A107841(n). - Philippe Deléham, May 26 2005
Sum_{k>=0} T(n-k, k) = A005043(n). - Philippe Deléham, May 30 2005
T(n, k) = A108263(n+k, k). - Philippe Deléham, May 30 2005
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A001003(n), A107841(n), A131763(n), A131765(n), A131846(n), A131926(n), A131869(n), A131927(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, respectively. - Philippe Deléham, Nov 05 2007
Sum_{k=0..n} T(n,k)*5^k*(-2)^(n-k) = A152601(n). - Philippe Deléham, Dec 10 2008
Sum_{k=0..n} T(n,k)*(-1)^k*3^(n-k) = A154825(n). - Philippe Deléham, Jan 17 2009
Umbrally, P(n,t) = Lah[n-1,-t*a.]/n! = (1/n)*Sum_{k=1..n-1} binomial(n-2,k-1)a_k t^k/k!, where (a.)^k = a_k = (n-1+k)!/(n-1)!, the rising factorial, and Lah(n,t) = n!*Laguerre(n,-1,t) are the Lah polynomials A008297 related to the Laguerre polynomials of order -1. - Tom Copeland, Oct 04 2014
T(n, k) = binomial(n, k)*binomial(n+k, k-1)/n, for k >= 0; T(0, 0) = 1 (see Kreweras, p. 21). - Michel Marcus, Nov 22 2014
P(n,t) = Lah[n-1,-:Dt:]/n! t^(n-1) with (:Dt:)^k = (d/dt)^k t^k = k! Laguerre(k,0,-:tD:) with (:tD:)^j = t^j D^j. The normalized Laguerre polynomials of 0 order are given in A021009. - Tom Copeland, Aug 22 2016

Extensions

Typo in a(60) corrected by Michael De Vlieger, Nov 21 2019

A100754 Triangle read by rows: T(n, k) = number of hill-free Dyck paths (i.e., no peaks at height 1) of semilength n and having k peaks.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 8, 8, 1, 1, 13, 29, 13, 1, 1, 19, 73, 73, 19, 1, 1, 26, 151, 266, 151, 26, 1, 1, 34, 276, 749, 749, 276, 34, 1, 1, 43, 463, 1781, 2762, 1781, 463, 43, 1, 1, 53, 729, 3758, 8321, 8321, 3758, 729, 53, 1, 1, 64, 1093, 7253, 21659, 31004, 21659, 7253, 1093, 64, 1
Offset: 2

Views

Author

Emeric Deutsch, Jan 14 2005

Keywords

Comments

Row n has n - 1 terms. Row sums yield the Fine numbers (A000957).
Related to the number of certain sets of non-crossing partitions for the root system A_n (p. 11, Athanasiadis and Savvidou). - Tom Copeland, Oct 19 2014
T(n,k) is the number of permutations pi of [n-1] with k - 1 descents such that s(pi) avoids the patterns 132, 231, and 312, where s is West's stack-sorting map. - Colin Defant, Sep 16 2018
The absolute values of the polynomials at -1 and j (cube root of 1) seem to be given by A126120 and A005043. - F. Chapoton, Nov 16 2021
Don Knuth observes that this sequence also arrises from the enumeration of restricted max-and-min-closed relations, only there it appears as an array read by antidiagonals: see the Knuth "Notes" link and A372068. Knuth also gives a formula expressing the array A372368 in terms of this array. He also reports that there is strong experimental evidence that the n-th term of row m in this array is a polynomial of degree 2*m-2 in n. - N. J. A. Sloane, May 12 2024

Examples

			T(4, 2) = 4 because we have UU*DDUU*DD, UU*DUU*DDD, UUU*DDU*DD and UUU*DU*DDD, where U = (1, 1), D = (1,-1) and * indicates the peaks.
Triangle starts:
   1;
   1,  1;
   1,  4,   1;
   1,  8,   8,    1;
   1, 13,  29,   13,    1;
   1, 19,  73,   73,   19,    1;
   1, 26, 151,  266,  151,   26,    1;
   1, 34, 276,  749,  749,  276,   34,   1;
   1, 43, 463, 1781, 2762, 1781,  463,  43,  1;
   1, 53, 729, 3758, 8321, 8321, 3758, 729, 53, 1;
   ...
As an array (for which the rows of the preceding triangle are the antidiagonals):
   1,  1,    1,     1,      1,      1,       1,        1,        1, ...
   1,  4,    8,    13,     19,     26,      34,       43,       53, ...
   1,  8,   29,    73,    151,    276,     463,      729,     1093, ...
   1, 13,   73,   266,    749,   1781,    3758,     7253,    13061, ...
   1, 19,  151,   749,   2762,   8321,   21659,    50471,   107833, ...
   1, 26,  276,  1781,   8321,  31004,   97754,   271125,   679355, ...
   1, 34,  463,  3758,  21659,  97754,  367285,  1196665,  3478915, ...
   1, 43,  729,  7253,  50471, 271125, 1196665,  4526470, 15118415, ...
   1, 53, 1093, 13061, 107833, 679355, 3478915, 15118415, 57500480, ...
   ...
		

Crossrefs

Programs

  • Maple
    T := (n, k) -> add((j/(n-j))*binomial(n-j, k-j)*binomial(n-j,k), j=0..min(k,n-k)): for n from 2 to 13 do seq(T(n, k), k = 1..n-1) od; # yields the sequence in triangular form
  • Mathematica
    T[n_, k_] := Sum[(j/(n-j))*Binomial[n-j, k-j]*Binomial[n-j, k], {j, 0, Min[k, n-k]}]; Table[T[n, k], {n, 2, 13}, {k, 1, n-1}] // Flatten (* Jean-François Alcover, Feb 19 2017, translated from Maple *)

Formula

T(n, k) = Sum_{j=0..min(k, n-k)} (j/(n-j)) * C(n-j, k-j) * C(n-j, k), n >= 2.
G.f.: t*z*r/(1 - t*z*r), where r = r(t, z) is the Narayana function defined by r = z*(1 + r)*(1 + t*r).
From Tom Copeland, Oct 19 2014: (Start)
With offset 0 for A108263 and offset 1 for A132081, row polynomials of this entry P(n, x) = Sum_{i} A108263(n, i)*x^i*(1 + x)^(n - 2*i) = Sum_{i} A132081(n - 2, i)*x^i*(1 + x)^(n - 2*i).
E.g., P(4, x) = 1*x*(1 + x)^(4 - 2*1) + 2*x^2*(1 + x)^(4 - 2*2) = x + 4*x^2 + x^3.
Equivalently, let Q(n, x) be the row polynomials of A108263. Then P(n, x) = (1 + x)^n * Q(n, x/(1 + x)^2).
E.g., P(4, x) = (1 + x)^4 * (x/(1 + x)^2 + 2 * (x/(1 + x)^2)^2).
See Athanasiadis and Savvidou (p. 7). (End)

A108263 Triangle read by rows: T(n,k) is the number of short bushes with n edges and k branchnodes (i.e., nodes of outdegree at least two). A short bush is an ordered tree with no nodes of outdegree 1.

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 0, 1, 2, 0, 1, 5, 0, 1, 9, 5, 0, 1, 14, 21, 0, 1, 20, 56, 14, 0, 1, 27, 120, 84, 0, 1, 35, 225, 300, 42, 0, 1, 44, 385, 825, 330, 0, 1, 54, 616, 1925, 1485, 132, 0, 1, 65, 936, 4004, 5005, 1287, 0, 1, 77, 1365, 7644, 14014, 7007, 429, 0, 1, 90, 1925, 13650, 34398
Offset: 0

Views

Author

Emeric Deutsch, May 29 2005

Keywords

Comments

Row n has 1+floor(n/2) terms. Row sums are the Riordan numbers (A005043). Column 3 yields A033275; column 4 yields A033276.
Related to the number of certain non-crossing partitions for the root system A_n. Cf. p. 12, Athanasiadis and Savvidou. Diagonals are A033282/A086810. Also see A132081 and A100754.- Tom Copeland, Oct 19 2014

Examples

			T(6,3)=5 because the only short bushes with 6 edges and 3 branchnodes are the five full binary trees with 6 edges.
Triangle begins:
1;
0;
0,1;
0,1;
0,1,2;
0,1,5;
0,1,9,5
		

Crossrefs

Programs

  • Maple
    G:=(1+z-sqrt((1-z)^2-4*t*z^2))/2/z/(1+t*z): Gser:=simplify(series(G,z=0,18)): P[0]:=1: for n from 1 to 16 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 16 do seq(coeff(t*P[n],t^k),k=1..1+floor(n/2)) od; # yields sequence in triangular form
    A108263 := (n,k) -> binomial(n-k-1,n-2*k)*binomial(n,k)/(n-k+1);
    seq(print(seq(A108263(n,k),k=0..ceil((n-1)/2))),n=0..8); # Peter Luschny, Sep 25 2014
  • Mathematica
    T[n_,k_]:=Binomial[n-k-1,n-2k]*Binomial[n,k]/(n-k+1); Flatten[Table[T[n,k],{n,0,11},{k,0,Ceiling[(n-1)/2]}]] (* Indranil Ghosh, Feb 20 2017 *)

Formula

G.f. G=G(t, z) satisfies z*(1+t*z)*G^2 - (1+z)*G + 1 = 0.
T(n, k) = A086810(n-k, k). - Philippe Deléham, May 30 2005

A108767 Triangle read by rows: T(n,k) is number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(1,1), d=(1,-2) and have k peaks (i.e., ud's).

Original entry on oeis.org

1, 1, 2, 1, 6, 5, 1, 12, 28, 14, 1, 20, 90, 120, 42, 1, 30, 220, 550, 495, 132, 1, 42, 455, 1820, 3003, 2002, 429, 1, 56, 840, 4900, 12740, 15288, 8008, 1430, 1, 72, 1428, 11424, 42840, 79968, 74256, 31824, 4862, 1, 90, 2280, 23940, 122094, 325584, 465120
Offset: 1

Views

Author

Emeric Deutsch, Jun 24 2005

Keywords

Comments

Row sums yield A001764.
From Peter Bala, Sep 16 2012: (Start)
The number of 2-Dyck paths of order n with k peaks (Cigler). A 2-Dyck path of order n is a lattice path from (0,0) to (2*n,n) with steps (0,1) (North) and (1,0) (East) that never goes below the diagonal {2*i,i} 0 <= i <= n. A peak is a consecutive North East pair.
Row reverse of A120986. Described in A173020 as generalized Runyon numbers R_{n,k}^(2).
(End)
From Alexander Burstein, Jun 15 2020: (Start)
T(n,k) is the number of paths from (0,0) to (3n,0) that stay on or above the horizontal axis, with unit steps u=(1,1) and d=(1,-2), that have n+1-k peaks at even height.
T(n,k) is also the number of paths from (0,0) to (3n,0) that stay on or above the horizontal axis, with unit steps u=(1,1) and d=(1,-2), that have n-k peaks at odd height. (End)
An apparent refinement is A338135. - Tom Copeland, Oct 12 2020

Examples

			T(3,2)=6 because we have uuduuuudd, uuuduuudd, uuuuduudd, uuuudduud, uuuuududd and uuuuuddud.
Triangle starts:
  1;
  1,  2;
  1,  6,   5;
  1, 12,  28,   14;
  1, 20,  90,  120,   42;
  1, 30, 220,  550,  495,  132;
  1, 42, 455, 1820, 3003, 2002, 429;
  ...
		

Crossrefs

Runyon numbers R_{n,k}^(m): A010054 (m=0), A001263 (m=1), this sequence (m=2), A173020 (m=3).

Programs

  • Magma
    A108767:= func< n,k,m | Binomial(n,k)*Binomial(m*n,k-1)/n >;
    [A108767(n,k,2): k in [1..n], n in [1..12]]; // G. C. Greubel, Feb 20 2021
  • Maple
    T:=(n,k)->binomial(n,k)*binomial(2*n,k-1)/n: for n from 1 to 10 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form
  • Mathematica
    T[n_, k_] := Binomial[n, k]*Binomial[2*n, k - 1]/n;
    Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Nov 11 2017, from Maple *)
  • PARI
    T(n,k) = binomial(n, k)*binomial(2*n, k-1)/n; \\ Andrew Howroyd, Nov 06 2017
    
  • Python
    from sympy import binomial
    def T(n, k): return binomial(n, k)*binomial(2*n, k - 1)//n
    for n in range(1, 21): print([T(n, k) for k in range(1, n + 1)]) # Indranil Ghosh, Nov 07 2017
    
  • R
    T <- function(n, k) {
      choose(n, k)*choose(2*n, k - 1)/n
    } # Indranil Ghosh, Nov 07 2017
    
  • Sage
    def A108767(n,k,m): return binomial(n,k)*binomial(m*n,k-1)/n
    flatten([[A108767(n,k,2) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Feb 20 2021
    

Formula

T(n, k) = binomial(n, k)*binomial(2*n, k-1)/n.
T(n, n) = A000108(n) (the Catalan numbers).
Sum_{k=1..n} k*T(n,k) = A025174(n).
G.f.: T-1, where T = T(t, z) satisfies T = 1 + z*T^2*(T - 1 + t).
From Peter Bala, Oct 22 2008: (Start)
Define a functional I on formal power series of the form f(x) = 1 + ax + bx^2 + ... by the following iterative process. Define inductively f^(1)(x) = f(x) and f^(n+1)(x) = f(x*f^(n)(x)) for n >= 1. Then set I(f(x)) = Limit_{n -> oo} f^(n)(x) in the x-adic topology on the ring of formal power series; the operator I may also be defined by I(f(x)) := 1/x*series reversion of x/f(x).
The o.g.f. for the array of Narayana numbers A001263 is I(1 + t*x + t*x^2 + t*x^3 + ...) = 1 + t*x + (t + t^2)*x^2 + (t + 3*t^2 + t^3)*x^3 + ... . The o.g.f. for the current array is IoI(1 + t*x + t*x^2 + t*x^3 + ...) = 1 + t*x + (t + 2*t^2)*x^2 + (t + 6*t^2 + 5*t^3)*x^3 + ... . Cf. A132081 and A141618. Alternatively, the o.g.f. of this array can be obtained from a single application of I, namely, form I(1 + t*x^2 + t*x^4 + t*x^6 + ...) = 1 + t*x^2 + (t + 2*t^2)*x^4 + (t + 6*t^2 + 5*t^3)*x^6 + ... and then replace x by sqrt(x). This is a particular case of the general result that forming the n-fold composition I^(n)(f(x)) and then replacing x with x^n produces the same result as I(f(x^n)). (End)
O.g.f. is series reversion with respect to x of x/((1+x)*(1+x*u)^2). Cf. A173020. - Peter Bala, Sep 12 2012
n-th row polynomial = x * hypergeom([1 - n, -2*n], [2], x). - Peter Bala, Aug 30 2023
Showing 1-6 of 6 results.