A182533 -id:A182533 - cp's OEIS frontend

A230206 Trapezoid of dot products of row 3 (signs alternating) with sequential 4-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 4-tuples (C(3,0), -C(3,1), C(3,2), -C(3,3)) and (C(n-1,k-3), C(n-1,k-2), C(n-1,k-1), C(n-1,k)), n >= 1, 0 <= k <= n+2.

-1, 3, -3, 1, -1, 2, 0, -2, 1, -1, 1, 2, -2, -1, 1, -1, 0, 3, 0, -3, 0, 1, -1, -1, 3, 3, -3, -3, 1, 1, -1, -2, 2, 6, 0, -6, -2, 2, 1, -1, -3, 0, 8, 6, -6, -8, 0, 3, 1, -1, -4, -3, 8, 14, 0, -14, -8, 3, 4, 1, -1, -5, -7, 5, 22, 14, -14, -22, -5, 7

Offset: 1

Dixon J. Jones, Oct 11 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)^3 (x+1)^(n-1) for n > 0.

			Trapezoid begins
  -1,  3, -3,  1;
  -1,  2,  0, -2,  1;
  -1,  1,  2, -2, -1,  1;
  -1,  0,  3,  0, -3,  0,  1;
  -1, -1,  3,  3, -3, -3,  1, 1;
  -1, -2,  2,  6,  0, -6, -2, 2, 1;
  -1, -3,  0,  8,  6, -6, -8, 0, 3, 1;

Dixon J. Jones, Rows n = 1..100 for irregular triangle, 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.

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), A230207-A230212 (j=4 to j=9).

m:=3; [[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)^3 (x + 1)^n, x], {n, 0, 7}]] (* T. D. Noe, Oct 25 2013 *)
m=3; 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=3; 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=3; [[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=3.

A230212 Trapezoid of dot products of row 9 (signs alternating) with sequential 10-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 10-tuples (C(9,0), -C(9,1), ..., C(9,8), -C(9,9)) and (C(n-1,k-9), C(n-1,k-8), ..., C(n-1,k)), n >= 1, 0 <= k <= n+8.

Original entry on oeis.org

-1, 9, -36, 84, -126, 126, -84, 36, -9, 1, -1, 8, -27, 48, -42, 0, 42, -48, 27, -8, 1, -1, 7, -19, 21, 6, -42, 42, -6, -21, 19, -7, 1, -1, 6, -12, 2, 27, -36, 0, 36, -27, -2, 12, -6, 1, -1, 5, -6, -10, 29, -9, -36, 36, 9, -29, 10, 6, -5, 1, -1, 4, -1, -16

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)^9 (x+1)^(n-1), n > 0.

			Trapezoid begins:
  -1, 9, -36,  84, -126, 126, -84,  36,  -9,   1;
  -1, 8, -27,  48,  -42,   0,  42, -48,  27,  -8,   1;
  -1, 7, -19,  21,    6, -42,  42,  -6, -21,  19,  -7,  1;
  -1, 6, -12,   2,   27, -36,   0,  36, -27,  -2,  12, -6,  1;
  -1, 5,  -6, -10,   29,  -9, -36,  36,   9, -29,  10,  6, -5,  1;
  -1, 4,  -1, -16,   19,  20, -45,   0,  45, -20, -19, 16,  1, -4,  1;
  -1, 3,   3, -17,    3,  39, -25, -45,  45,  25, -39, -3, 17, -3, -3, 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-A230211 (j=3 to j=8).

m:=9; [[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)^9 (x + 1)^n, x], {n, 0, 7}]] (* T. D. Noe, Oct 25 2013 *)
m=9; 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=9; 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=9; [[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=9.

A130713 a(0)=a(2)=1, a(1)=2, a(n)=0 for n > 2.

Original entry on oeis.org

1, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Paul Curtz and Tanya Khovanova, Jul 01 2007

Self-convolution of A019590. Up to a sign the convolutional inverse of the natural numbers sequence. - Tanya Khovanova, Jul 14 2007

Iterated partial sums give the chain A130713 -> A113311 -> A008574 -> A001844 -> A005900 -> A006325 -> A033455 -> A259181, up to index. The k-th term of the n-th partial sums is (n^2-7n+14 + 4k(k+n-4))(k+n-4)!/(k-1)!/(n-1)!, for k > 3-n. Iterating partial sums in reverse (n-th differences with n zeros prepended) gives row (n+3) of A182533, modulo signs and trailing zeros. - Travis Scott, Feb 19 2023

Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
Dominika Závacká, Cristina Dalfó, and Miquel Angel Fiol, Integer sequences from k-iterated line digraphs, CEUR: Proc. 24th Conf. Info. Tech. - Appl. and Theory (ITAT 2024) Vol 3792, 156-161. See p. 161, Table 2.

A130713:=n->binomial(2*n, n^2); seq(A130713(n), n=0..100); # Wesley Ivan Hurt, Mar 08 2014

Table[Binomial[2 n, n^2], {n, 0, 100}] (* Wesley Ivan Hurt, Mar 08 2014 *)

G.f.: 1 + 2*x + x^2.

a(n) = binomial(2n, n^2). - Wesley Ivan Hurt, Mar 08 2014

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.

Original entry on oeis.org

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.

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.

Original entry on oeis.org

-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

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)^5 (x-1)^(n-1), n > 0.

			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.

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.

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).

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

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 *)

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

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

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.

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.

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-A230208 (j=3 to j=5), A230210-A230212 (j=7 to j=9).

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.

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.

Original entry on oeis.org

-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

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)^7 (x+1)^(n-1), n > 0.

			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.

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-A230209 (j=3 to j=6), A230211-A230212 (j=8 and j=9).

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

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 *)

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

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

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.

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.

Original entry on oeis.org

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

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

			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.

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-A230210 (j=3 to j=7), A230212 (j=9).

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

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 *)

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

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

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.

cp's OEIS Frontend

A230206 Trapezoid of dot products of row 3 (signs alternating) with sequential 4-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 4-tuples (C(3,0), -C(3,1), C(3,2), -C(3,3)) and (C(n-1,k-3), C(n-1,k-2), C(n-1,k-1), C(n-1,k)), n >= 1, 0 <= k <= n+2.

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A230212 Trapezoid of dot products of row 9 (signs alternating) with sequential 10-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 10-tuples (C(9,0), -C(9,1), ..., C(9,8), -C(9,9)) and (C(n-1,k-9), C(n-1,k-8), ..., C(n-1,k)), n >= 1, 0 <= k <= n+8.

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A130713 a(0)=a(2)=1, a(1)=2, a(n)=0 for n > 2.

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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.

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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.

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