A108998 Square array, read by antidiagonals, where row n equals the coordination sequence of B_n lattice, for n >= 0.

1, 1, 0, 1, 2, 0, 1, 8, 2, 0, 1, 18, 16, 2, 0, 1, 32, 74, 24, 2, 0, 1, 50, 224, 170, 32, 2, 0, 1, 72, 530, 768, 306, 40, 2, 0, 1, 98, 1072, 2562, 1856, 482, 48, 2, 0, 1, 128, 1946, 6968, 8130, 3680, 698, 56, 2, 0, 1, 162, 3264, 16394, 28320, 20082, 6432, 954, 64, 2, 0

Offset: 0

Author

Paul D. Hanna, Jun 17 2005

Keywords

• nonn
• tabl

Comments

Compare with A108553, where row n equals the crystal ball sequence for D_n lattice.

Examples

			Square array begins:
  1,  0,    0,     0,     0,      0,      0,      0, ...
  1,  2,    2,     2,     2,      2,      2,      2, ...
  1,  8,   16,    24,    32,     40,     48,     56, ...
  1, 18,   74,   170,   306,    482,    698,    954, ...
  1, 32,  224,   768,  1856,   3680,   6432,  10304, ...
  1, 50,  530,  2562,  8130,  20082,  42130,  78850, ...
  1, 72, 1072,  6968, 28320,  85992, 214864, 467544, ...
  1, 98, 1946, 16394, 83442, 307314, 907018, ...
Product of the g.f. of row n and (1-x)^n generates the rows of triangle A109001:
  1;
  1,  1;
  1,  6,   1;
  1, 15,  23,    1;
  1, 28, 102,   60,    1;
  1, 45, 290,  402,  125,   1;
  1, 66, 655, 1596, 1167, 226, 1; ...
		

Links

• Muniru A Asiru, Rows n=0..110 of antidiagonals, flattened
• R. Bacher, P. de la Harpe and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.

Crossrefs

Cf. A108999 (main diagonal), A109000 (antidiagonal sums), A109001, A022144 (row 2), A022145 (row 3), A022146 (row 4), A022147 (row 5), A022148 (row 6), A022149 (row 7), A022150 (row 8), A022151 (row 9), A022152 (row 10), A022153 (row 11), A022154 (row 12).

Programs

PARI

T(n,k)=if(n<0 || k<0,0,sum(j=0,k, binomial(n+k-j-1,k-j)*(binomial(2*n+1,2*j)-2*n*binomial(n-1,j-1))))

Formula

T(n, k) = Sum_{j=0..k} C(n+k-j-1, k-j)*(C(2*n+1, 2*j)-2*n*C(n-1, j-1)) for n >= k >= 0.

G.f. for coordination sequence of B_n lattice: ((Sum_{i=0..n} binomial(2*n+1, 2*i)*z^i)-2*n*z*(1+z)^(n-1))/(1-z)^n. [Bacher et al.]