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.

A089627 T(n,k) = binomial(n,2*k)*binomial(2*k,k) for 0 <= k <= n, triangle read by rows.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 6, 0, 0, 1, 12, 6, 0, 0, 1, 20, 30, 0, 0, 0, 1, 30, 90, 20, 0, 0, 0, 1, 42, 210, 140, 0, 0, 0, 0, 1, 56, 420, 560, 70, 0, 0, 0, 0, 1, 72, 756, 1680, 630, 0, 0, 0, 0, 0, 1, 90, 1260, 4200, 3150, 252, 0, 0, 0, 0, 0, 1, 110, 1980, 9240, 11550, 2772, 0, 0, 0, 0, 0, 0, 1, 132, 2970, 18480, 34650, 16632, 924, 0, 0, 0, 0, 0, 0, 1, 156, 4290, 34320, 90090, 72072, 12012, 0, 0, 0, 0, 0, 0, 0, 1, 182, 6006, 60060, 210210, 252252, 84084, 3432, 0, 0, 0, 0, 0, 0, 0
Offset: 0

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Author

Philippe Deléham, Dec 31 2003

Keywords

Comments

The rows of this triangle are the gamma vectors of the n-dimensional type B associahedra (Postnikov et al., p.38 ). Cf. A055151 and A101280. - Peter Bala, Oct 28 2008
T(n,k) is the number of Grand Motzkin paths of length n having exactly k upsteps (1,1). Cf. A109189, A055151. - Geoffrey Critzer, Feb 05 2014
The result Sum_{k = 0..floor(n/2)} C(n,2*k)*C(2*k,k)*x^k = (sqrt(1 - 4*x))^n* P(n,1/sqrt(1 - 4*x)) expressing the row polynomials of this triangle in terms of the Legendre polynomials P(n,x) is due to Catalan. See Laden, equation 7.10, p. 56. - Peter Bala, Mar 18 2018

Examples

			Triangle begins:
  1
  1,   0
  1,   2,    0
  1,   6,    0,     0
  1,  12,    6,     0,     0
  1,  20,   30,     0,     0,     0
  1,  30,   90,    20,     0,     0,   0
  1,  42,  210,   140,     0,     0,   0, 0
  1,  56,  420,   560,    70,     0,   0, 0, 0
  1,  72,  756,  1680,   630,     0,   0, 0, 0, 0
  1,  90, 1260,  4200,  3150,   252,   0, 0, 0, 0, 0
  1, 110, 1980,  9240, 11550,  2772,   0, 0, 0, 0, 0, 0
  1, 132, 2970, 18480, 34650, 16632, 924, 0, 0, 0, 0, 0, 0
Relocating the zeros to be evenly distributed and interpreting the triangle as the coefficients of polynomials
                     1
                     1
                 1 + 2 q^2
                 1 + 6 q^2
            1 + 12 q^2 +  6 q^4
            1 + 20 q^2 + 30 q^4
       1 + 30 q^2 +  90 q^4 +  20 q^6
       1 + 42 q^2 + 210 q^4 + 140 q^6
  1 + 56 q^2 + 420 q^4 + 560 q^6 + 70 q^8
then the substitution q^k -> 1/(floor(k/2)+1) gives the Motzkin numbers A001006.
- _Peter Luschny_, Aug 29 2011
		

Crossrefs

Row sums A002426. Antidiagonal sums A098479.

Programs

  • Maple
    for i from 0 to 12 do seq(binomial(i, j)*binomial(i-j, j), j=0..i) od; # Zerinvary Lajos, Jun 07 2006
    # Alternatively:
    R := (n, x) -> simplify(hypergeom([1/2 - n/2, -n/2], [1], 4*x)):
    Trow := n -> seq(coeff(R(n,x), x, j), j=0..n):
    seq(print(Trow(n)), n=0..9); # Peter Luschny, Mar 18 2018
  • Mathematica
    nn=15;mxy=(1-x-(1-2x+x^2-4x^2y)^(1/2))/(2x^2 y);Map[Select[#,#>0&]&, CoefficientList[Series[1/(1-x-2y x^2mxy),{x,0,nn}],{x,y}]]//Grid (* Geoffrey Critzer, Feb 05 2014 *)
  • PARI
    T(n,k) = binomial(n,2*k)*binomial(2*k,k);
    concat(vector(15, n, vector(n, k, T(n-1, k-1)))) \\ Gheorghe Coserea, Sep 01 2018

Formula

T(n,k) = n!/((n-2*k)!*k!*k!).
E.g.f.: exp(x)*BesselI(0, 2*x*sqrt(y)). - Vladeta Jovovic, Apr 07 2005
O.g.f.: ( 1 - x - sqrt(1 - 2*x + x^2 - 4*x^2*y))/(2*x^2*y). - Geoffrey Critzer, Feb 05 2014
R(n, x) = hypergeom([1/2 - n/2, -n/2], [1], 4*x) are the row polynomials. - Peter Luschny, Mar 18 2018
From Peter Bala, Jun 23 2023: (Start)
T(n,k) = Sum_{i = 0..k} (-1)^i*binomial(n, i)*binomial(n-i, k-i)^2. Cf. A063007(n,k) = Sum_{i = 0..k} binomial(n, i)^2*binomial(n-i, k-i).
T(n,k) = A063007(n-k,k); that is, the diagonals of this table are the rows of A063007. (End)