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A089627
T(n,k) = binomial(n,2*k)*binomial(2*k,k) for 0 <= k <= n, triangle read by rows.
17
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
0,5
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
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.
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).
T(n,k) = A063007(n-k,k); that is, the diagonals of this table are the rows of A063007. (End)
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
Row sums A002426. Antidiagonal sums A098479.
Sequence in context: A151860 A338774 A330891 * A306534 A344392 A331787
KEYWORD
easy,nonn,tabl
AUTHOR
Philippe Deléham, Dec 31 2003
STATUS
approved