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
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
KEYWORD
AUTHOR
F. Chapoton, Oct 06 2022
STATUS
approved