A357613 - OEIS

%I #26 Oct 11 2023 11:04:19

%S 1,3,2,15,20,6,84,168,105,20,495,1320,1260,504,70,3003,10010,12870,

%T 7920,2310,252,18564,74256,120120,100100,45045,10296,924,116280,

%U 542640,1058148,1113840,680680,240240,45045,3432

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

%C 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)!).

%C 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

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

%e As a triangle of numbers, this starts with

%e   1;

%e   3, 2;

%e   15, 20, 6;

%e   84, 168, 105, 20;

%e   495, 1320, 1260, 504, 70.

%e Here is an example for n=1 as coefficients (up to sign) in the binomial basis of integer-valued polynomials:

%e (x+1)*(x+2)*(x+1)/2 = 3*binomial(x+3,3)-2*binomial(x+2,2).

%p A357613 := proc(n,k)

%p     binomial(2*n,k)*binomial(3*n-k,2*n) ;

%p end proc:

%p seq(seq(A357613(n,k),k=0..n),n=0..10) ; # _R. J. Mathar_, Jul 06 2023

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

%o (SageMath)

%o def a(n):

%o     return [binomial(2 * n, k) * binomial(3 * n - k, 2 * n)

%o             for k in range(n + 1)]

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

%K nonn,tabl,easy

%O 0,2

%A _F. Chapoton_, Oct 06 2022