cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A108553 Square array, read by antidiagonals, where row n equals the crystal ball sequence for D_n lattice.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 13, 13, 7, 1, 1, 25, 55, 25, 9, 1, 1, 41, 169, 147, 41, 11, 1, 1, 61, 411, 625, 309, 61, 13, 1, 1, 85, 853, 2051, 1681, 561, 85, 15, 1, 1, 113, 1583, 5577, 6981, 3721, 923, 113, 17, 1, 1, 145, 2705, 13203, 23673, 18733, 7225, 1415, 145, 19, 1
Offset: 0

Views

Author

Paul D. Hanna, Jun 10 2005

Keywords

Comments

Rows 0 and 2 are included by extension since they fit the formula. Row 1 equals the odd numbers in order that triangle A108556 maintains that A108556(n,n-1) = (n/2)*A108556(n,n) for all n>=1, where row n of triangle A108556 equals the inverse binomial transform of row n of this square array.

Examples

			Square array begins:
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...
1,3,5,7,9,11,13,15,17,19,21,23,25,27,...
1,5,13,25,41,61,85,113,145,181,221,265,...
1,13,55,147,309,561,923,1415,2057,2869,...
1,25,169,625,1681,3721,7225,12769,21025,...
1,41,411,2051,6981,18733,42783,86983,...
1,61,853,5577,23673,76389,204205,476113,...
1,85,1583,13203,68853,264825,824083,...
Inverse binomial transform of rows gives
rows of triangle A108556:
1,
1,2,
1,4,4,
1,12,30,20,
1,24,120,192,96,
1,40,330,940,1080,432, ...
Product of the g.f. of row n and (1-x)^(n+1)
generates the symmetric triangle A108558:
1;
1,1;
1,2,1;
1,9,9,1;
1,20,54,20,1;
1,35,180,180,35,1; ...
The row sums of triangle A108558 equals the
main diagonal of triangle A108556.

Crossrefs

Cf. A108554 (diagonal), A108555 (antidiagonal sums), A108556, A108558, 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).

Programs

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

Formula

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

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.

Original entry on oeis.org

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

Views

Author

Paul D. Hanna, Jun 10 2005

Keywords

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)

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

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.

Programs

  • 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 *)
  • 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)

A108556 Triangle, read by rows, where row n equals the inverse binomial transform of the crystal ball sequence for D_n lattice.

Original entry on oeis.org

1, 1, 2, 1, 4, 4, 1, 12, 30, 20, 1, 24, 120, 192, 96, 1, 40, 330, 940, 1080, 432, 1, 60, 732, 3200, 6240, 5568, 1856, 1, 84, 1414, 8708, 25200, 37184, 27104, 7744, 1, 112, 2480, 20352, 80960, 173824, 206080, 126976, 31744, 1, 144, 4050, 42588, 221544, 643824, 1096032, 1085760, 579456, 128768
Offset: 0

Views

Author

Paul D. Hanna, Jun 10 2005

Keywords

Comments

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

Examples

			Triangle begins:
1;
1,2;
1,4,4;
1,12,30,20;
1,24,120,192,96;
1,40,330,940,1080,432;
1,60,732,3200,6240,5568,1856;
1,84,1414,8708,25200,37184,27104,7744;
1,112,2480,20352,80960,173824,206080,126976,31744; ...

Links

Crossrefs

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).
Cf. A063007, A127674.

Programs

  • Mathematica
    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}]];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 26 2018, from PARI *)
  • 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)

Formula

Main diagonal equals A008353: 2^(n-1)*(2^n-n) for n>1.
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
Showing 1-3 of 3 results.

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