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%I #66 Jun 23 2023 15:55:55
%S 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,
%T 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,
%U 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
%N T(n,k) = binomial(n,2*k)*binomial(2*k,k) for 0 <= k <= n, triangle read by rows.
%C 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
%C 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
%C 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
%H Alois P. Heinz, <a href="/A089627/b089627.txt">Rows n = 0..140, flattened</a>
%H Rui Duarte and António Guedes de Oliveira, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Duarte/duarte7.html">A Famous Identity of Hajós in Terms of Sets</a>, Journal of Integer Sequences, Vol. 17 (2014), #14.9.1.
%H Hyman N. Laden, <a href="https://drum.lib.umd.edu/handle/1903/16857">An historical, and critical development of the theory of Legendre polynomials before 1900</a>, Master of Arts Thesis, University of Maryland 1938.
%H Shi-Mei Ma, <a href="http://arxiv.org/abs/1304.6654">On gamma-vectors and the derivatives of the tangent and secant functions</a>, arXiv:1304.6654 [math.CO], 2013.
%H A. Postnikov, V. Reiner, and L. Williams, <a href="http://arxiv.org/abs/math/0609184">Faces of generalized permutohedra</a>, arXiv:math/0609184 [math.CO], 2006-2007.
%F T(n,k) = n!/((n-2*k)!*k!*k!).
%F E.g.f.: exp(x)*BesselI(0, 2*x*sqrt(y)). - _Vladeta Jovovic_, Apr 07 2005
%F O.g.f.: ( 1 - x - sqrt(1 - 2*x + x^2 - 4*x^2*y))/(2*x^2*y). - _Geoffrey Critzer_, Feb 05 2014
%F R(n, x) = hypergeom([1/2 - n/2, -n/2], [1], 4*x) are the row polynomials. - _Peter Luschny_, Mar 18 2018
%F From _Peter Bala_, Jun 23 2023: (Start)
%F 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).
%F T(n,k) = A063007(n-k,k); that is, the diagonals of this table are the rows of A063007. (End)
%e Triangle begins:
%e 1
%e 1, 0
%e 1, 2, 0
%e 1, 6, 0, 0
%e 1, 12, 6, 0, 0
%e 1, 20, 30, 0, 0, 0
%e 1, 30, 90, 20, 0, 0, 0
%e 1, 42, 210, 140, 0, 0, 0, 0
%e 1, 56, 420, 560, 70, 0, 0, 0, 0
%e 1, 72, 756, 1680, 630, 0, 0, 0, 0, 0
%e 1, 90, 1260, 4200, 3150, 252, 0, 0, 0, 0, 0
%e 1, 110, 1980, 9240, 11550, 2772, 0, 0, 0, 0, 0, 0
%e 1, 132, 2970, 18480, 34650, 16632, 924, 0, 0, 0, 0, 0, 0
%e Relocating the zeros to be evenly distributed and interpreting the triangle as the coefficients of polynomials
%e 1
%e 1
%e 1 + 2 q^2
%e 1 + 6 q^2
%e 1 + 12 q^2 + 6 q^4
%e 1 + 20 q^2 + 30 q^4
%e 1 + 30 q^2 + 90 q^4 + 20 q^6
%e 1 + 42 q^2 + 210 q^4 + 140 q^6
%e 1 + 56 q^2 + 420 q^4 + 560 q^6 + 70 q^8
%e then the substitution q^k -> 1/(floor(k/2)+1) gives the Motzkin numbers A001006.
%e - _Peter Luschny_, Aug 29 2011
%p for i from 0 to 12 do seq(binomial(i, j)*binomial(i-j, j), j=0..i) od; # _Zerinvary Lajos_, Jun 07 2006
%p # Alternatively:
%p R := (n, x) -> simplify(hypergeom([1/2 - n/2, -n/2], [1], 4*x)):
%p Trow := n -> seq(coeff(R(n,x), x, j), j=0..n):
%p seq(print(Trow(n)), n=0..9); # _Peter Luschny_, Mar 18 2018
%t 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 *)
%o (PARI)
%o T(n,k) = binomial(n,2*k)*binomial(2*k,k);
%o concat(vector(15, n, vector(n, k, T(n-1, k-1)))) \\ _Gheorghe Coserea_, Sep 01 2018
%Y Row sums A002426. Antidiagonal sums A098479.
%Y Cf. A055151, A063007, A109187, A101280.
%K easy,nonn,tabl
%O 0,5
%A _Philippe Deléham_, Dec 31 2003