A230210 Trapezoid of dot products of row 7 (signs alternating) with sequential 8-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 8-tuples (C(7,0), -C(7,1), ..., C(7,6), -C(7,7)) and (C(n-1,k-7), C(n-1,k-6), ..., C(n-1,k)), n >= 1, 0 <= k <= n+6.

-1, 7, -21, 35, -35, 21, -7, 1, -1, 6, -14, 14, 0, -14, 14, -6, 1, -1, 5, -8, 0, 14, -14, 0, 8, -5, 1, -1, 4, -3, -8, 14, 0, -14, 8, 3, -4, 1, -1, 3, 1, -11, 6, 14, -14, -6, 11, -1, -3, 1, -1, 2, 4, -10, -5, 20, 0, -20, 5, 10, -4, -2, 1, -1, 1, 6, -6, -15

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)^7 (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=7.

Example

Trapezoid begins:
  -1, 7, -21,  35, -35,  21,  -7,   1;
  -1, 6, -14,  14,   0, -14,  14,  -6,   1;
  -1, 5,  -8,   0,  14, -14,   0,   8,  -5,  1;
  -1, 4,  -3,  -8,  14,   0, -14,   8,   3, -4,  1;
  -1, 3,   1, -11,   6,  14, -14,  -6,  11, -1, -3,  1;
  -1, 2,   4, -10,  -5,  20,   0, -20,   5, 10, -4, -2,  1;
  -1, 1,   6,  -6, -15,  15,  20, -20, -15, 15,  6, -6, -1, 1;
etc.

Mathematica

Flatten[Table[CoefficientList[(x - 1)^7 (x + 1)^n, x], {n, 0, 7}]] (* T. D. Noe, Oct 25 2013 *)
m=7; 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=7; 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:=7; [[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=7; [[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-A230209 (j=3 to j=6), A230211-A230212 (j=8 and j=9).

Sequence in context: A015729 A271972 A001485 * A087111 A173676 A131893

Adjacent sequences: A230207 A230208 A230209 * A230211 A230212 A230213

Keyword

easy,sign,tabf

Author

Dixon J. Jones, Oct 12 2013

Status

approved