A181543 Triangle of cubed binomial coefficients, T(n,k) = C(n,k)^3, read by rows.

1, 1, 1, 1, 8, 1, 1, 27, 27, 1, 1, 64, 216, 64, 1, 1, 125, 1000, 1000, 125, 1, 1, 216, 3375, 8000, 3375, 216, 1, 1, 343, 9261, 42875, 42875, 9261, 343, 1, 1, 512, 21952, 175616, 343000, 175616, 21952, 512, 1, 1, 729, 46656, 592704, 2000376, 2000376, 592704, 46656, 729, 1

Offset: 0

Author

Paul D. Hanna, Oct 30 2010

Keywords

• nonn
• tabl

Comments

Diagonal of rational function R(x,y,z,t) = 1/(1 + y + z + x*y + y*z + t*x*z + (t+1)*x*y*z) with respect to x, y, z, i.e., T(n,k) = [(xyz)^n*t^k] R(x,y,z,t). - Gheorghe Coserea, Jul 01 2018

Examples

			Triangle begins:
  1;
  1,   1;
  1,   8,     1;
  1,  27,    27,      1;
  1,  64,   216,     64,       1;
  1, 125,  1000,   1000,     125,       1;
  1, 216,  3375,   8000,    3375,     216,      1;
  1, 343,  9261,  42875,   42875,    9261,    343,     1;
  1, 512, 21952, 175616,  343000,  175616,  21952,   512,   1;
  1, 729, 46656, 592704, 2000376, 2000376, 592704, 46656, 729, 1;
  ...
		

Links

• Indranil Ghosh, Rows 0..120 of triangle, flattened
• Jeffrey S. Geronimo, Hugo J. Woerdeman, and Chung Y. Wong, The autoregressive filter problem for multivariable degree one symmetric polynomials, arXiv:2101.00525 [math.CA], 2021.

Crossrefs

Cf. A000172 (row sums), A181545 (antidiagonal sums), A002897, A181544, A248658.

Variants: A008459, A007318.

Programs

Maple

T:= (n, k)-> binomial(n, k)^3:
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Jan 06 2021

Mathematica

Flatten[Table[Binomial[n,k]^3,{n,0,10},{k,0,n}]] (* Harvey P. Dale, May 23 2011 *)

PARI

T(n,k)=binomial(n,k)^3
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print())

PARI

T(n,k)=polcoeff(polcoeff(sum(m=0,n,(3*m)!/m!^3*x^(2*m)*y^m/(1-x-x*y+x*O(x^n))^(3*m+1)),n,x),k,y)
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print()) \\ Paul D. Hanna, Nov 04 2010

PARI

diag(expr, N=22, var=variables(expr)) = {
  my(a = vector(N));
  for (k = 1, #var, expr = taylor(expr, var[#var - k + 1], N));
  for (n = 1, N, a[n] = expr;
    for (k = 1, #var, a[n] = polcoeff(a[n], n-1)));
  return(a);
};
x='x; y='y; z='z; t='t;
concat(apply(Vec, diag(1/(1 + y + z + x*y + y*z + t*x*z + (t+1)*x*y*z), 10, [x, y, z]))) \\ Gheorghe Coserea, Jul 01 2018

Formula

Row sums equal A000172, the Franel numbers.

Central terms are A002897(n) = C(2n,n)^3.

Antidiagonal sums equal A181545;

The g.f. of the antidiagonal sums is Sum_{n>=0} (3n)!/(n!)^3 * x^(3n)/(1-x-x^2)^(3n+1).

G.f. for column k: [Sum_{j=0..2k} A181544(k,j)*x^j]/(1-x)^(3k+1), where the row sums of A181544 equals De Bruijn's s(3,n) = (3n)!/(n!)^3.

G.f.: A(x,y) = Sum_{n>=0} (3n)!/n!^3 * x^(2n)*y^n/(1-x-x*y)^(3n+1). - Paul D. Hanna, Nov 04 2010