A357613 Triangle read by rows T(n, k) = binomial(2*n, k) * binomial(3*n - k, 2*n).

1, 3, 2, 15, 20, 6, 84, 168, 105, 20, 495, 1320, 1260, 504, 70, 3003, 10010, 12870, 7920, 2310, 252, 18564, 74256, 120120, 100100, 45045, 10296, 924, 116280, 542640, 1058148, 1113840, 680680, 240240, 45045, 3432

Offset

0,2

Comments

Each line should be the f-vector of a cellular complex. The sequence seems to give the coefficients in a binomial basis of the integer-valued polynomials (x+1)*(x+2)*...*(x+2*n)*(x+1)*(x+2)*...*(x+n)/(n!*(2n)!).

The precise expansion is (x+1)*(x+2)*...*(x+2*n)*(x+1)*(x+2)*...*(x+n)/(n!*(2*n)!) = Sum_{k = 0..n} (-1)^k*T(n,k)*binomial(x+3*n-k, 3*n-k), as can be verified using the WZ algorithm. For example, n = 3 gives (x+1)^2*(x+2)^2*(x+3)^2*(x+4)*(x+5)*(x+6)/(3!*6!) = 84*binomial(x+9, 9) - 168*binomial(x+8, 8) + 105*binomial(x+7, 7) - 20*binomial(x+6, 6). - Peter Bala, Jun 25 2023

Links

Table of n, a(n) for n=0..35.

Formula

T(n,k) = binomial(2*n, k) * binomial(3*n - k, 2*n) for 0 <= k <= n

Example

As a triangle of numbers, this starts with
  1;
  3, 2;
  15, 20, 6;
  84, 168, 105, 20;
  495, 1320, 1260, 504, 70.
Here is an example for n=1 as coefficients (up to sign) in the binomial basis of integer-valued polynomials:
(x+1)*(x+2)*(x+1)/2 = 3*binomial(x+3,3)-2*binomial(x+2,2).

Maple

A357613 := proc(n, k)
    binomial(2*n, k)*binomial(3*n-k, 2*n) ;
end proc:
seq(seq(A357613(n, k), k=0..n), n=0..10) ; # R. J. Mathar, Jul 06 2023

Mathematica

Table[Binomial[2n, k]Binomial[3n-k, 2n], {n, 0, 10}, {k, 0, n}]//Flatten (* Harvey P. Dale, Oct 11 2023 *)

Prog

(SageMath)
def a(n):
    return [binomial(2 * n, k) * binomial(3 * n - k, 2 * n)
            for k in range(n + 1)]

Crossrefs

Row sums A026000. Cf. A000984, A005809 (k=0), A144485 (k=1), A033282, A110608, A243660.

Sequence in context: A302845 A291251 A223523 * A133932 A111999 A286947

Adjacent sequences: A357610 A357611 A357612 * A357614 A357615 A357616

Keyword

nonn,tabl,easy

Author

F. Chapoton, Oct 06 2022

Status

approved