A120247 Triangle of Hankel transforms of binomial(n+k, k).

1, 1, -1, 1, -3, -1, 1, -6, -10, 1, 1, -10, -50, 35, 1, 1, -15, -175, 490, 126, -1, 1, -21, -490, 4116, 5292, -462, -1, 1, -28, -1176, 24696, 116424, -60984, -1716, 1, 1, -36, -2520, 116424, 1646568, -3737448, -736164, 6435, 1, 1, -45, -4950, 457380, 16818516, -133613766, -131589315, 9202050, 24310, -1

Offset: 0

Author

Paul Barry, Jun 12 2006

Keywords

• easy
• sign
• tabl

Comments

Columns include -A000217, -A006542, A107915.

Row k is the Hankel transform of C(n+k, k).

The matrix inverse starts

1;

1, -1;

-2, 3, -1;

-15, 24, -10, 1;

434, -700, 300, -35, 1;

47670, -76950, 33075, -3920, 126, -1;

-19787592, 31943835, -13733720, 1629936, -52920, 462, -1; - R. J. Mathar, Mar 22 2013

Examples

			Triangle begins
  1;
  1,  -1;
  1,  -3,    -1;
  1,  -6,   -10,     1;
  1, -10,   -50,    35,      1;
  1, -15,  -175,   490,    126,     -1;
  1, -21,  -490,  4116,   5292,   -462,    -1;
  1, -28, -1176, 24696, 116424, -60984, -1716,   1;
		

Links

• G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Crossrefs

Columns include: A000217, A006542, A107915.

Cf. A120248.

Programs

Magma

p:= func< m,k | k eq 0 select 1 else (&*[Binomial(m+j, k+1): j in [1..k]]) >;
A120247:= func< n,k | (-1)^Floor((k+1)/2)*p(n,k)/p(k,k) >;
[A120247(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 15 2023

Maple

A120247 := proc(n,k)
    (cos(Pi*k/2)-sin(Pi*k/2))*mul(binomial(n+j+1,k+1),j=0..k-1)/mul(binomial(k+j+1,k+1),j=0..k-1) ;
    simplify(%) ;
end proc: # R. J. Mathar, Mar 22 2013

Mathematica

p[m_, k_]:= Product[Binomial[m+j, k+1], {j,k}];
T[n_, k_]:= (-1)^Floor[(k+1)/2]*p[n,k]/p[k,k];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 15 2023 *)

SageMath

def p(m,k): return product(binomial(m+j+1,k+1) for j in range(k))
def A120247(n,k): return (-1)^((k+1)//2)*p(n,k)/p(k,k)
flatten([[A120247(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Mar 15 2023

Formula

T(n, k) = (cos(pi*k/2) - sin(pi*k/2))*( Product{j=0..k-1} C(n+j+1, k+1)/Product{j=0..k-1} C(k+j+1, k+1) ).