A108556 - OEIS

%I #12 Jul 26 2018 06:35:33

%S 1,1,2,1,4,4,1,12,30,20,1,24,120,192,96,1,40,330,940,1080,432,1,60,

%T 732,3200,6240,5568,1856,1,84,1414,8708,25200,37184,27104,7744,1,112,

%U 2480,20352,80960,173824,206080,126976,31744,1,144,4050,42588,221544,643824,1096032,1085760,579456,128768

%N Triangle, read by rows, where row n equals the inverse binomial transform of the crystal ball sequence for D_n lattice.

%C Row n equals the inverse binomial transform of row n of the square array A108553.

%C Array of f-vectors for type D root polytopes [Ardila et al.]. See A063007 and A127674 for the arrays of f-vectors for type A and type C root polytopes respectively. - _Peter Bala_, Oct 23 2008

%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.

%F Main diagonal equals A008353: 2^(n-1)*(2^n-n) for n>1.

%F O.g.f. : rational function N(x,z)/D(x,z), where N(x,z) = 1 - 3*(1 + 2*x)*z + (3 + 8*x + 8*x^2)*z^2 - (1 + 2*x)*(1 - 6*x - 6*x^2)z^3 - 8*x*(1 + x)(1 + 2*x + 2*x^2)*z^4 + 2*x*(1 + x)*(1 + 2*x)*z^5 and D(x,z) = ((1-z)^2 - 4*x*z)*(1 - z*(1 + 2*x))^2. - _Peter Bala_, Oct 23 2008

%e Triangle begins:

%e 1;

%e 1,2;

%e 1,4,4;

%e 1,12,30,20;

%e 1,24,120,192,96;

%e 1,40,330,940,1080,432;

%e 1,60,732,3200,6240,5568,1856;

%e 1,84,1414,8708,25200,37184,27104,7744;

%e 1,112,2480,20352,80960,173824,206080,126976,31744; ...

%t T[n_, k_] := Module[{A}, A = Table[Table[If[r - 1 == 0 || c - 1 == 0, 1, If[r - 1 == 1, 2c - 1, Sum[Binomial[r + c - j - 2, c - j - 1] (Binomial[2r - 2, 2j] - 2(r - 1) Binomial[r - 3, j - 1]), {j, 0, c - 1}]]], {c, 1, n + 1}], {r, 1, n + 1}]; SeriesCoefficient[((A[[n + 1]]. x^Range[0, n]) /. x -> x/(1 + x))/(1 + x), {x, 0, k}]];

%t Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 26 2018, from PARI *)

%o (PARI) T(n,k)=local(A=vector(n+1,r,vector(n+1,c,if(r-1==0 || c-1==0,1,if(r-1==1,2*c-1, sum(j=0,c-1,binomial(r+c-j-2,c-j-1)*(binomial(2*r-2,2*j)-2*(r-1)*binomial(r-3,j-1)))))))); polcoeff(subst(Ser(A[n+1]),x,x/(1+x))/(1+x),k)

%Y Cf. A108553, A108557 (row sums), A108558, Rows are inverse binomial transforms 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 Cf. A063007, A127674.

%K nonn,tabl

%O 0,3

%A _Paul D. Hanna_, Jun 10 2005