A230208 Trapezoid of dot products of row 5 (signs alternating) with sequential 6-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 6-tuples (C(5,0), -C(5,1), ..., -C(5,5)) and (C(n-1,k-5), C(n-1,k-4), ..., C(n-1,k)), n >= 1, 0 <= k <= n+4.

-1, 5, -10, 10, -5, 1, -1, 4, -5, 0, 5, -4, 1, -1, 3, -1, -5, 5, 1, -3, 1, -1, 2, 2, -6, 0, 6, -2, -2, 1, -1, 1, 4, -4, -6, 6, 4, -4, -1, 1, -1, 0, 5, 0, -10, 0, 10, 0, -5, 0, 1, -1, -1, 5, 5, -10, -10, 10, 10, -5, -5, 1, 1, -1, -2, 4, 10, -5, -20, 0, 20, 5

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)^5 (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, and 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=5.

Example

Trapezoid begins:
  -1,  5, -10, 10,  -5,   1;
  -1,  4,  -5,  0,   5,  -4,  1;
  -1,  3,  -1, -5,   5,   1, -3,  1;
  -1,  2,   2, -6,   0,   6, -2, -2,  1;
  -1,  1,   4, -4,  -6,   6,  4, -4, -1,  1;
  -1,  0,   5,  0, -10,   0, 10,  0, -5,  0, 1;
  -1, -1,   5,  5, -10, -10, 10, 10, -5, -5, 1, 1;
etc.

Mathematica

Flatten[Table[CoefficientList[(x - 1)^5 (x + 1)^n, x], {n, 0, 7}]] (* T. D. Noe, Oct 25 2013 *)
m=5; 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 29 2018 *)

Prog

(PARI) m=5; 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 29 2018
(Magma) m:=5; [[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 29 2018
(Sage) m=5; [[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 29 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-A230207 (j=3 and j=4), A230209-A230212 (j=6 to j=9).

Sequence in context: A198286 A001483 A173679 * A168228 A277950 A087109

Adjacent sequences: A230205 A230206 A230207 * A230209 A230210 A230211

Keyword

easy,sign,tabf

Author

Dixon J. Jones, Oct 12 2013

Status

approved