A110522 Riordan array (1/(1+x), x*(1-2*x)/(1+x)^2).

1, -1, 1, 1, -5, 1, -1, 12, -9, 1, 1, -22, 39, -13, 1, -1, 35, -115, 82, -17, 1, 1, -51, 270, -344, 141, -21, 1, -1, 70, -546, 1106, -773, 216, -25, 1, 1, -92, 994, -2954, 3199, -1466, 307, -29, 1, -1, 117, -1674, 6888, -10791, 7461, -2487, 414, -33, 1, 1, -145, 2655, -14484, 31179, -30645, 15060, -3900, 537, -37, 1

Offset: 0

Author

Paul Barry, Jul 24 2005

Keywords

• easy
• sign
• tabl

Comments

Inverse of A110519.

Product of inverse binomial transform matrix (1/(1+x), x/(1+x)) and (1, x*(1-3*x)) (A110517).

Examples

			Rows begin
   1;
  -1,    1;
   1,   -5,    1;
  -1,   12,   -9,    1;
   1,  -22,   39,  -13,    1;
  -1,   35, -115,   82,  -17,    1;
		

Links

• G. C. Greubel, Rows n = 0..50 of the triangle, flattened
• Paul Barry, A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays, J. Int. Seq. 16 (2013) #13.5.4.

Crossrefs

Cf. A110519 (inverse), A110523 (row sums), A110524 (diagonal sums).

Cf. A000326, A016813, A033999, A052924, A078005, A110517, A236267.

Programs

Magma

A110522:= func< n,k | (-1)^(n+k)*(&+[ 3^(j-k)*Binomial(k,j-k)*Binomial(n,j) : j in [0..n]] ) >;
[A110522(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Dec 28 2023

Mathematica

T[n_,k_]:= Sum[(-1)^(n-j)*(-3)^(j-k)*Binomial[k, j- k]*Binomial[n, j], {j,0,n}];
Table[T[n,k], {n,0,20}, {k,0,n}]//Flatten (* G. C. Greubel, Aug 30 2017 *)

PARI

A110522(n,k) = if(n==0, 1, sum(j=0,n, (-1)^(n-j)*(-3)^(j-k)*binomial(n,j)*binomial(k, j-k)));
for(n=0,12, for(k=0,n, print1(A110522(n,k), ", "))) \\ G. C. Greubel, Aug 30 2017; Dec 28 2023

SageMath

def A110522(n,k): return (-1)^(n+k)*sum(3^(j-k)*binomial(k,j-k)*binomial(n,j) for j in range(n+1))
flatten([[A110522(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Dec 28 2023

Formula

Number triangle T(n, k) = Sum_{j=0..n} (-1)^(n-j)*C(n, j)*(-3)^(j-k)*C(k, j-k).

T(n, k) = Sum_{j=0..n} Sum_{i=0..k} C(k, i)*C(n+k-i-j-1, n-k-i-j)*(-1)^(n-k)*2^i.

Sum_{k=0..n} T(n, k) = A110523(n) (row sums).

Sum_{k=0..floor(n/2)} T(n-k, k) = A110524(n) (diagonal sums).

T(n,k) = T(n-1,k-1) - 2*T(n-1,k) - T(n-2,k) - 2*T(n-2,k-1), T(0,0) = 1, T(1,0) = -1, T(1,1) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Jan 12 2014

From G. C. Greubel, Dec 28 2023: (Start)

T(n, 0) = A033999(n).

T(n, 1) = (-1)^(n-1)*A000326(n), n >= 1.

T(n, n) = 1.

T(n, n-1) = -A016813(n-1), n >= 1.

T(n, n-2) = A236267(n-2), n >= 2.

Sum_{k=0..n} (-1)^k*T(n, k) = (-1)^n*A052924(n).

Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = (-1)^n*A078005(n). (End)