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.

A338076 Diagonal terms in the expansion of 1/(1-x-2*y-3*z).

Original entry on oeis.org

1, 36, 3240, 362880, 44906400, 5884534656, 800296713216, 111714888130560, 15898425017080320, 2296439169133824000, 335647548960599715840, 49531592018516268810240, 7367824312754294985523200, 1103342589983347322447462400, 166176904368920474278821888000
Offset: 0

Views

Author

N. J. A. Sloane, Oct 22 2020

Keywords

Comments

Expand the rational function 1/(1-x-2*y-3*z) as Sum_i Sum_j Sum_k c(i,j,k)*x^i*y^j*z^k; a(n) = c(n,n,n).

Crossrefs

Programs

  • Maple
    N:= 25: # for a(0)..a(N)
    F:= 1/(1-x-2*y-3*z):
    S1:= series(F,x,N+1):
    L1:= [seq(coeff(S1,x,i),i=0..N)]:
    L2:= [seq(coeff(series(L1[i+1],y,i+1),y,i),i=0..N)]:
    seq(coeff(series(L2[i+1],z,i+1),z,i),i=0..N); # Robert Israel, Oct 24 2020
  • Mathematica
    nmax = 20; Flatten[{1, Table[Coefficient[Series[1/(1-x-2*y-3*z), {x, 0, n}, {y, 0, n}, {z, 0, n}], x^n*y^n*z^n], {n, 1, nmax}]}] (* Vaclav Kotesovec, Oct 23 2020 *)

Formula

Conjectures from Robert Israel, Oct 25 2020: (Start)
a(n+1) = 18*(3*n+1)*(3*n+2)*a(n)/(n+1)^2.
G.f.: hypergeom([1/3, 2/3], [1], 162*x). (End)
a(n) = 6^n * (3*n)! / n!^3. - Vaclav Kotesovec, Oct 28 2020

Extensions

More terms from Vaclav Kotesovec, Oct 23 2020

A338337 Coefficient of x^(6*n)*y^(6*n)*z^(6*n) in the expansion of 1/(1-x-y^2-z^3).

Original entry on oeis.org

1, 4620, 135795660, 5190977391600, 221838126928317900, 10086906430733029017120, 477156732636269771364879600, 23199870600247661786357661924000, 1150983828787218131441395889200471500, 57991163446756752913635026142306805792320, 2957727121295876265116937111814024549631408160
Offset: 0

Views

Author

N. J. A. Sloane, Oct 22 2020

Keywords

Comments

The other diagonal coefficients are zero.

Crossrefs

Programs

  • Maple
    a:= proc(n) local h; 1/(1-x-y^2-z^3); for h in [x, y, z]
          do coeff(series(%, h, 1+6*n), h, 6*n) od
        end:
    seq(a(n), n=0..10);  # Alois P. Heinz, Oct 23 2020
  • Mathematica
    nmax = 10; Flatten[{1, Table[Coefficient[Series[1/(1 - x - y^2 - z^3), {x, 0, 6*n}, {y, 0, 6*n}, {z, 0, 6*n}], x^(6*n)*y^(6*n)*z^(6*n)], {n, 1, nmax}]}] (* Vaclav Kotesovec, Oct 23 2020 *)

Formula

a(n) = (11*n)! / ((2*n)! * (3*n)! * (6*n)!). - Vaclav Kotesovec, Oct 28 2020

Extensions

More terms from Alois P. Heinz, Oct 23 2020

A338077 Diagonal terms in the expansion of (1+x*y+y*z+z*x)/(1-x-y-z).

Original entry on oeis.org

1, 9, 126, 2310, 47250, 1027026, 23207184, 538748496, 12757863690, 306752696250, 7465133615940, 183458150153460, 4545211223957040, 113378500045162800, 2844670649392440000, 71731904712206892480, 1816739665054871280570, 46189610653753780435530, 1178358502858339948645500
Offset: 0

Views

Author

N. J. A. Sloane, Oct 22 2020

Keywords

Comments

Expand that rational function as Sum_i Sum_j Sum_k c(i,j,k)*x^i*y^j*z^k; then a(n) = c(n,n,n).

Crossrefs

Programs

  • Mathematica
    nmax = 20; Flatten[{1, Table[Coefficient[Series[(1+x*y+y*z+z*x)/(1-x-y-z), {x, 0, n}, {y, 0, n}, {z, 0, n}], x^n*y^n*z^n], {n, 1, nmax}]}] (* Vaclav Kotesovec, Oct 23 2020 *)

Formula

a(n) = (4*n - 1) * (3*n)! / ((3*n - 1) * n!^3). - Vaclav Kotesovec, Oct 28 2020

Extensions

More terms from Vaclav Kotesovec, Oct 23 2020
Showing 1-3 of 3 results.