%I #37 Aug 06 2024 09:42:36
%S 1,1,1,1,2,1,1,9,9,1,1,20,54,20,1,1,35,180,180,35,1,1,54,447,852,447,
%T 54,1,1,77,931,2863,2863,931,77,1,1,104,1724,7768,12550,7768,1724,104,
%U 1,1,135,2934,18186,43128,43128,18186,2934,135,1,1,170,4685,38200,124850,183356,124850,38200,4685,170,1
%N Symmetric triangle, read by rows, where row n equals the (n+1)-th differences of the crystal ball sequence for D_n lattice, for n>=0.
%C Row n equals the (n+1)-th differences of row n of the square array A108553. G.f. of row n equals: (1-x)^(n+1)*CBD_n(x), where CBD_n denotes the g.f. of the crystal ball sequence for D_n lattice.
%C From _Peter Bala_, Oct 23 2008: (Start)
%C Let D_n be the root lattice generated as a monoid by {+-e_i +- e_j: 1 <= i not equal to j <= n}. Let P(D_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the h-vectors of a unimodular triangulation of P(D_n) [Ardila et al.]. See A108556 for the corresponding array of f-vectors for these type D_n polytopes. See A008459 for the array of h-vectors for type A_n polytopes and A086645 for the array of h-vectors associated with type C_n polytopes.
%C The Hilbert transform of this array (as defined in A145905) equals A108553.
%C (End)
%H Seiichi Manyama, <a href="/A108558/b108558.txt">Rows n = 0..139, flattened</a>
%H F. Ardila, M. Beck, S. Hosten, J. Pfeifle and K. Seashore, <a href="http://arxiv.org/abs/0809.5123">Root polytopes and growth series of root lattices</a>, arXiv:0809.5123 [math.CO], 2008.
%H J. H. Conway and N. J. A. Sloane, <a href="https://citeseerx.ist.psu.edu/pdf/59eb4e0f4a4c5af6dbc8e9445fff1ed4a2a24c8a">Low-dimensional lattices. VII Coordination sequences</a>, Proc. R. Soc. Lond. A (1997) 453, 2369-2389.
%F T(n, k) = C(2*n, 2*k) - 2*n*C(n-2, k-1) for n>1, with T(0, 0)=1, T(1, 0)=T(1, 1)=1. Row sums equal A008353: 2^(n-1)*(2^n-n) for n>1.
%F From _Peter Bala_, Oct 23 2008: (Start)
%F O.g.f. : rational function N(x,z)/D(x,z), where N(x,z) = 1 - 3*(1 + x)*z + (3 + 2*x + 3*x^2)*z^2 - (1 + x)*(1 - 8*x + x^2)z^3 - 8*x*(1 + x^2)*z^4 + 2*x*(1 + x)*(1 - x)^2*z^5 and D(x,z) = ((1 - z)^2 - 2*x*z*(1 + z) + x^2*z^2)*(1 - z*(1 + x))^2.
%F For n >= 2, the row n generating polynomial equals 1/2*[(1 + sqrt(x))^(2n) + (1 - sqrt(x))^(2n)] - 2*n*x*(1 + x)^(n-2).
%F (End)
%e G.f.s of initial rows of square array A108553 are:
%e (1)/(1-x),
%e (1 + x)/(1-x)^2,
%e (1 + 2*x + x^2)/(1-x)^3,
%e (1 + 9*x + 9*x^2 + x^3)/(1-x)^4,
%e (1 + 20*x + 54*x^2 + 20*x^3 + x^4)/(1-x)^5,
%e (1 + 35*x + 180*x^2 + 180*x^3 + 35*x^4 + x^5)/(1-x)^6.
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 2, 1;
%e 1, 9, 9, 1;
%e 1, 20, 54, 20, 1;
%e 1, 35, 180, 180, 35, 1;
%e 1, 54, 447, 852, 447, 54, 1;
%e 1, 77, 931, 2863, 2863, 931, 77, 1;
%e 1, 104, 1724, 7768, 12550, 7768, 1724, 104, 1;
%e 1, 135, 2934, 18186, 43128, 43128, 18186, 2934, 135, 1;
%e 1, 170, 4685, 38200, 124850, 183356, 124850, 38200, 4685, 170, 1;
%e ...
%t T[1, 0] = T[1, 1]=1; T[n_, k_] := Binomial[2n, 2k] - 2n Binomial[n-2, k-1];
%t Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 25 2018 *)
%o (PARI) T(n,k)=if(n<k || k<0,0,if(n==0 || n==1,1,binomial(2*n,2*k)-2*n*binomial(n-2,k-1)))
%Y Cf. A108553, A008353, A108558, A008459, A086645, A108556. Row n equals (n+1)-th differences of: A001844 (row 2), A005902 (row 3), A007204 (row 4), A008356 (row 5), A008358 (row 6), A008360 (row 7), A008362 (row 8), A008377 (row 9), A008379 (row 10).
%Y T(2n,n) gives A305693.
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Jun 10 2005