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