A178301 Triangle T(n,k) = binomial(n,k)*binomial(n+k+1,n+1) read by rows, 0 <= k <= n.

1, 1, 3, 1, 8, 10, 1, 15, 45, 35, 1, 24, 126, 224, 126, 1, 35, 280, 840, 1050, 462, 1, 48, 540, 2400, 4950, 4752, 1716, 1, 63, 945, 5775, 17325, 27027, 21021, 6435, 1, 80, 1540, 12320, 50050, 112112, 140140, 91520, 24310, 1, 99, 2376, 24024, 126126, 378378, 672672, 700128, 393822, 92378

Offset: 0

Author

Alford Arnold, May 30 2010

Keywords

• easy
• nonn
• tabl

Comments

Antidiagonal sums are given by A113682. - Johannes W. Meijer, Mar 24 2013

The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial binomial(x+n,n)*binomial(x+n,n-1) in the basis made of the binomial(x+i,i). - F. Chapoton, Nov 01 2022

Chapoton's observation above is correct: the precise expansion is binomial(x+n,n)*binomial(x+n,n-1) = Sum_{k = 0..n-1} (-1)^k*T(n-1,n-1-k)*binomial(x+2*n-1-k,2*n-1-k), as can be verified using the WZ algorithm. For example, n = 4 gives binomial(x+4,4)*binomial(x+4,3) = 35*binomial(x+7,7) - 45*binomial(x+6,6) + 15*binomial(x+5,5) - binomial(x+4,4). - Peter Bala, Jun 24 2023

Examples

			n=0: 1;
n=1: 1,  3;
n=2: 1,  8,  10;
n=3: 1, 15,  45,   35;
n=4: 1, 24, 126,  224,   126;
n=5: 1, 35, 280,  840,  1050,   462;
n=6: 1, 48, 540, 2400,  4950,  4752,  1716;
n=7: 1, 63, 945, 5775, 17325, 27027, 21021, 6435;
		

Links

• David A. Corneth, Table of n, a(n) for n = 0..10010
• Author?, Norm of a continuous function, dxdy.ru (in Russian).

Crossrefs

Cf. A007318, A047781 (row sums), A178300, A182626, A123160, A132813.

Programs

Maple

A178301 := proc(n,k)
        binomial(n,k)*binomial(n+k+1,n+1) ;
end proc: # R. J. Mathar, Mar 24 2013
R := proc(n) add((-1)^(n+k)*(2*k+1)*orthopoly:-P(k,2*x+1)/(n+1), k=0..n) end:
for n from 0 to 6 do seq(coeff(R(n), x, k), k=0..n) od; # Peter Luschny, Aug 25 2021

Mathematica

Flatten[Table[Binomial[n,k]Binomial[n+k+1,n+1],{n,0,10},{k,0,n}]] (* Harvey P. Dale, Aug 23 2014 *)

Maxima

create_list(binomial(n,k)*binomial(n+k+1,n+1),n,0,12,k,0,n); /* Emanuele Munarini, Dec 16 2016 */

PARI

R(n,x) = sum(k=0,n, (-1)^(n+k) * (2*k+1) * pollegendre(k,2*x+1)) / (n+1); \\ Max Alekseyev, Aug 25 2021

Formula

T(n,k) = A007318(n,k) * A178300(n+1,k+1).

From Peter Bala, Jun 18 2015: (Start)

n-th row polynomial R(n,x) = Sum_{k = 0..n} binomial(n,k)*binomial(n+k+1,n+1)*x^k = Sum_{k = 0..n} (-1)^(n+k)*binomial(n+1,k+1)*binomial(n+k+1,n+1)*(1 + x)^k.

Recurrence: (2*n - 1)*(n + 1)*R(n,x) = 2*(4*n^2*x + 2*n^2 - x - 1)*R(n-1,x) - (2*n + 1)(n - 1)*R(n-2,x) with R(0,x) = 1, R(1,x) = 1 + 3*x.

A182626(n) = -R(n-1,-2) for n >= 1. (End)

From Peter Bala, Jul 20 2015: (Start)

n-th row polynomial R(n,x) = Jacobi_P(n,0,1,2*x + 1).

(1 + x)*R(n,x) gives the row polynomials of A123160. (End)

G.f.: (1+x-sqrt(1-2*x+x^2-4*x*y))/(2*(1+y)*x*sqrt(1-2*x+x^2-4*x*y)). - Emanuele Munarini, Dec 16 2016

R(n,x) = Sum_{k=0..n} (-1)^(n+k)*(2*k+1)*P(k,2*x+1)/(n+1), where P(k,x) is the k-th Legendre polynomial (cf. A100258) and P(k,2*x+1) is the k-th shifted Legendre polynomial (cf. A063007). - Max Alekseyev, Jun 28 2018; corrected by Peter Bala, Aug 08 2021

Polynomial g(n,x) = R(n,-x)/(n+1) delivers the maximum of f(1)^2/(Integral_{x=0..1} f(x)^2 dx) over all polynomials f(x) with real coefficients and deg(f(x)) <= n. This maximum equals (n+1)^2. See dxdy.ru link. - Max Alekseyev, Jun 28 2018