A230208 -id:A230208 - cp's OEIS frontend

A230207 Trapezoid of dot products of row 4 (signs alternating) with sequential 5-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 5-tuples (C(4,0), -C(4,1), C(4,2), -C(4,3), C(4,4)) and (C(n-1,k-4), C(n-1,k-3), C(n-1,k-2), C(n-1,k-1), C(n-1,k)), n >= 1, 0 <= k <= n+3.

1, -4, 6, -4, 1, 1, -3, 2, 2, -3, 1, 1, -2, -1, 4, -1, -2, 1, 1, -1, -3, 3, 3, -3, -1, 1, 1, 0, -4, 0, 6, 0, -4, 0, 1, 1, 1, -4, -4, 6, 6, -4, -4, 1, 1, 1, 2, -3, -8, 2, 12, 2, -8, -3, 2, 1, 1, 3, -1, -11, -6, 14, 14, -6, -11, -1, 3, 1, 1, 4, 2, -12, -17, 8

Offset: 1

Dixon J. Jones, Oct 12 2013

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)^4 (x+1)^(n-1) for n > 0.

			Trapezoid begins:
  1, -4,  6, -4,  1;
  1, -3,  2,  2, -3,  1;
  1, -2, -1,  4, -1, -2,  1;
  1, -1, -3,  3,  3, -3, -1,  1;
  1,  0, -4,  0,  6,  0, -4,  0,  1;
  1,  1, -4, -4,  6,  6, -4, -4,  1, 1;
  1,  2, -3, -8,  2, 12,  2, -8, -3, 2, 1;
etc.

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.

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 (j=3), A230208-A230212 (j=5 to j=9).

m:=4; [[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

Flatten[Table[CoefficientList[(x - 1)^4 (x + 1)^n, x], {n, 0, 7}]] (* T. D. Noe, Oct 25 2013 *)
m=4; 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 *)

m=4; 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

m=4; [[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

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=4.

A230209 Trapezoid of dot products of row 6 (signs alternating) with sequential 7-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 7-tuples (C(6,0), -C(6,1), ..., -C(6,5), C(6,6)) and (C(n-1,k-6), C(n-1,k-5), ..., C(n-1,k)), n >= 1, 0 <= k <= n+5.

Original entry on oeis.org

1, -6, 15, -20, 15, -6, 1, 1, -5, 9, -5, -5, 9, -5, 1, 1, -4, 4, 4, -10, 4, 4, -4, 1, 1, -3, 0, 8, -6, -6, 8, 0, -3, 1, 1, -2, -3, 8, 2, -12, 2, 8, -3, -2, 1, 1, -1, -5, 5, 10, -10, -10, 10, 5, -5, -1, 1, 1, 0, -6, 0, 15, 0, -20, 0, 15, 0, -6, 0, 1, 1, 1, -6

Offset: 1

Dixon J. Jones, Oct 12 2013

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)^6 (x+1)^(n-1).

			Trapezoid begins:
  1, -6, 15, -20,  15,  -6,   1;
  1, -5,  9,  -5,  -5,   9,  -5,  1;
  1, -4,  4,   4, -10,   4,   4, -4,  1;
  1, -3,  0,   8,  -6,  -6,   8,  0, -3,  1;
  1, -2, -3,   8,   2, -12,   2,  8, -3, -2,  1;
  1, -1, -5,   5,  10, -10, -10, 10,  5, -5, -1, 1;
  1,  0, -6,   0,  15,   0, -20,  0, 15,  0, -6, 0, 1;
etc.

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.

m:=6; [[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

Flatten[Table[CoefficientList[(x - 1)^6 (x + 1)^n, x], {n, 0, 7}]] (* T. D. Noe, Oct 25 2013 *)
m=6; 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 *)

m=6; 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

m=6; [[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

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=6.

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