OFFSET
0,5
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
LINKS
Alois P. Heinz, Rows n = 0..140, flattened
Rui Duarte and António Guedes de Oliveira, A Famous Identity of Hajós in Terms of Sets, Journal of Integer Sequences, Vol. 17 (2014), #14.9.1.
Hyman N. Laden, An historical, and critical development of the theory of Legendre polynomials before 1900, Master of Arts Thesis, University of Maryland 1938.
Shi-Mei Ma, On gamma-vectors and the derivatives of the tangent and secant functions, arXiv:1304.6654 [math.CO], 2013.
A. Postnikov, V. Reiner, and L. Williams, Faces of generalized permutohedra, arXiv:math/0609184 [math.CO], 2006-2007.
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).
EXAMPLE
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
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 *)
PROG
(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
CROSSREFS
KEYWORD
AUTHOR
Philippe Deléham, Dec 31 2003
STATUS
approved