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
Examples
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; ...
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.
Crossrefs
Programs
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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 *)
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PARI
T(n,k)=if(n
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)
Comments