A230211 Trapezoid of dot products of row 8 (signs alternating) with sequential 9-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 9-tuples (C(8,0), -C(8,1), ..., -C(8,7), C(8,8)) and (C(n-1,k-8), C(n-1,k-7), ..., C(n-1,k)), n >= 1, 0 <= k <= n+7.

1, -8, 28, -56, 70, -56, 28, -8, 1, 1, -7, 20, -28, 14, 14, -28, 20, -7, 1, 1, -6, 13, -8, -14, 28, -14, -8, 13, -6, 1, 1, -5, 7, 5, -22, 14, 14, -22, 5, 7, -5, 1, 1, -4, 2, 12, -17, -8, 28, -8, -17, 12, 2, -4, 1, 1, -3, -2, 14, -5, -25, 20, 20, -25, -5, 14

Offset

1,2

Comments

The array is trapezoidal rather than triangular because C(n,k) is not uniquely defined for all negative n and negative k.

Row sums are 0.

Coefficients of ((x-1)^8)(x+1)^(n-1), n > 0.

Links

G. C. Greubel, Rows n=1..50 of trapezoid, flattened

Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.

Formula

T(n,k) = Sum_{i=0..n+m-1} (-1)^(i+m)*C(m,i)*C(n-1,k-i), n >= 1, with T(n,0) = (-1)^m and m=8.

Example

Trapezoid begins:
  1, -8, 28, -56,  70, -56,  28,  -8,   1;
  1, -7, 20, -28,  14,  14, -28,  20,  -7,   1;
  1, -6, 13,  -8, -14,  28, -14,  -8,  13,  -6,  1;
  1, -5,  7,   5, -22,  14,  14, -22,   5,   7, -5,  1;
  1, -4,  2,  12, -17,  -8,  28,  -8, -17,  12,  2, -4,  1;
  1, -3, -2,  14,  -5, -25,  20,  20, -25,  -5, 14, -2, -3, 1;
  1, -2, -5,  12,   9, -30,  -5,  40,  -5, -30,  9, 12, -5, -2, 1;
etc.

Mathematica

Flatten[Table[CoefficientList[(x - 1)^8 (x + 1)^n, x], {n, 0, 7}]] (* T. D. Noe, Oct 25 2013 *)
m=8; Table[If[k == 0, (-1)^m, Sum[(-1)^(j+m)*Binomial[m, j]*Binomial[n-1, k-j], {j, 0, n+m-1}]], {n, 1, 10}, {k, 0, n+m-1}]//Flatten (* G. C. Greubel, Nov 28 2018 *)

Prog

(PARI) m=8; for(n=1, 10, for(k=0, n+m-1, print1(if(k==0, (-1)^m, sum(j=0, n+m-1, (-1)^(j+m)*binomial(m, j)*binomial(n-1, k-j))), ", "))) \\ G. C. Greubel, Nov 28 2018
(Magma) m:=8; [[k le 0 select (-1 )^m else (&+[(-1)^(j+m)* Binomial(m, j) *Binomial(n-1, k-j): j in [0..(n+m-1)]]): k in [0..(n+m-1)]]: n in [1..10]]; // G. C. Greubel, Nov 28 2018
(Sage) m=8; [[sum((-1)^(j+m)*binomial(m, j)*binomial(n-1, k-j) for j in range(n+m)) for k in range(n+m)] for n in (1..10)] # G. C. Greubel, Nov 28 2018

Crossrefs

Using row j of the alternating Pascal triangle as generator: A007318 (j=0), A008482 and A112467 (j=1 after the first term in each), A182533 (j=2 after the first two rows), A230206-A230210 (j=3 to j=7), A230212 (j=9).

Sequence in context: A131850 A009504 A001486 * A229393 A173681 A045850

Adjacent sequences: A230208 A230209 A230210 * A230212 A230213 A230214

Keyword

easy,sign,tabf

Author

Dixon J. Jones, Oct 12 2013

Status

approved