A108558 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.

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, 54, 1, 1, 77, 931, 2863, 2863, 931, 77, 1, 1, 104, 1724, 7768, 12550, 7768, 1724, 104, 1, 1, 135, 2934, 18186, 43128, 43128, 18186, 2934, 135, 1, 1, 170, 4685, 38200, 124850, 183356, 124850, 38200, 4685, 170, 1

Offset

0,5

Comments

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.

From Peter Bala, Oct 23 2008: (Start)

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.

The Hilbert transform of this array (as defined in A145905) equals A108553.

(End)

Links

Seiichi Manyama, Rows n = 0..139, flattened

F. Ardila, M. Beck, S. Hosten, J. Pfeifle and K. Seashore, Root polytopes and growth series of root lattices, arXiv:0809.5123 [math.CO], 2008.

J. H. Conway and N. J. A. Sloane, Low-dimensional lattices. VII Coordination sequences, Proc. R. Soc. Lond. A (1997) 453, 2369-2389.

Formula

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.

From Peter Bala, Oct 23 2008: (Start)

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.

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).

(End)

Example

G.f.s of initial rows of square array A108553 are:
  (1)/(1-x),
  (1 + x)/(1-x)^2,
  (1 + 2*x + x^2)/(1-x)^3,
  (1 + 9*x + 9*x^2 + x^3)/(1-x)^4,
  (1 + 20*x + 54*x^2 + 20*x^3 + x^4)/(1-x)^5,
  (1 + 35*x + 180*x^2 + 180*x^3 + 35*x^4 + x^5)/(1-x)^6.
Triangle begins:
  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,     54,      1;
  1,  77,  931,  2863,   2863,    931,     77,     1;
  1, 104, 1724,  7768,  12550,   7768,   1724,   104,    1;
  1, 135, 2934, 18186,  43128,  43128,  18186,  2934,  135,   1;
  1, 170, 4685, 38200, 124850, 183356, 124850, 38200, 4685, 170, 1;
  ...

Mathematica

T[1, 0] = T[1, 1]=1; T[n_, k_] := Binomial[2n, 2k] - 2n Binomial[n-2, k-1];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 25 2018 *)

Prog

(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)))

Crossrefs

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).

T(2n,n) gives A305693.

Sequence in context: A246664 A229962 A141601 * A128434 A176417 A119731

Adjacent sequences: A108555 A108556 A108557 * A108559 A108560 A108561

Keyword

nonn,tabl

Author

Paul D. Hanna, Jun 10 2005

Status

approved