Dixon J. Jones - cp's OEIS frontend

A343446 Coefficients of the series S(p, q) for which -(p^(1/3))S converges to the largest real root of x^4 - px + q, where 0 < p and 0 < q < 3*(p/4)^(4/3).

-1, 1, 4, 40, 648, 14560, 418880, 14696640, 608608000, 29056867840, 1571364748800, 94937979136000, 6337884013260800, 463301182536192000, 36806315255277568000, 3157533815406530560000, 290912372128665391104000, 28648563542097847828480000

Offset: 0

Dixon J. Jones, May 26 2021

Based on formulas for series solutions of trinomials given in Eagle article.
S(p, q) = Sum_{n>=0} (a(n)*q^n)/((3^n)*(p^(4n/3))*n!)
In general, given m > 1, p > 0 and 0 < q < m*(p/(m + 1))^((m + 1)/m), the series S(m, p, q) for which (-p^(1/m))*S converges to the largest real root of x^(m + 1) - p*x + q has coefficients c(n) = m^(n - 1)*((n + m - 1)/m)(n - 1), where (x)_k is the Pochhammer symbol for Gamma(x + k)/Gamma(k), and S(m, p, q) = Sum{n>=0}(c(n)*q^n)/((m^n)*(p^(n*(m + 1)/m)*n!).

Albert Eagle, Series for all the roots of a trinomial equation, Am. Math. Monthly, vol. 46, no. 7 (Aug. - Sep., 1939), pp. 422 - 425.

A343445 relates similarly to the largest real root of x^3 - p*x + q.

A206300 relates similarly to the largest real root of x^3 - 3*u*x + 4*u, u >= 4.

a := proc(n) option remember; if n < 3 then [-1, 1, 4][n+1] else 4*(4*n - 7)*(4*n - 10)*(4*n - 13)*a(n-3) fi; end:
seq(a(n), n = 0..20); # Peter Bala, Jul 23 2024

Clear[a]; a=Table[3^(n - 1) Pochhammer[(n + 2)/3, n - 1], {n, 0, 20}]
(* In general, for the series S(m, p, q) for which (-p^(1/m))*S converges to the largest real root of x^(m + 1) - p*x + q, the first n + 1 coefficients are: *)
Clear[c]; c[m_,n_] := Table[m^(k - 1) Pochhammer[(k + m - 1)/m, k - 1], {k, 0, n}](* and S(m, p, q) to n + 1 terms is given by *)
Clear[s]; s[m_,p_,q_,n_]:= Sum[c[m,n][[k + 1]]*q^k/((m^k)*(p^(k (m + 1)/m))*k!), {k, 0, n}]

a(n) = 3^(n - 1)*((n + 2)/3)_(n - 1), where (x)_k is the Pochhammer symbol for Gamma(x + k)/Gamma(x).

a(n) = 4*(4*n - 7)*(4*n - 10)*(4*n - 13)*a(n-3) with a(0) = -1, a(1) = 1 and a(2) = 4. - Peter Bala, Jul 23 2024

A343445 Coefficients of the series S(p, q) for which (-sqrt(p))S converges to the largest real root of x^3 - px + q for 0 < p and 0 < q < 2*(p/3)^(3/2).

Original entry on oeis.org

-1, 1, 3, 24, 315, 5760, 135135, 3870720, 130945815, 5109350400, 225881530875, 11158821273600, 609202488769875, 36422392637030400, 2366751668870964375, 166086110424858624000, 12517749576658530579375, 1008474862499741564928000, 86485131825133787772901875

Offset: 0

Dixon J. Jones, Apr 15 2021

Based on formulas for series solutions of trinomials given in Eagle article.

Albert Eagle, Series for all the roots of a trinomial equation, Am. Math. Monthly, 46, no. 7 (Aug. - Sep., 1939), pp. 422-425.

Cf. A113551, A343446.

a := proc(n) option remember; if n = 1 then 1 elif n = 2 then 3 else 3*(3*n - 5)*(3*n - 7)*a(n-2) fi; end:
seq(a(n), n = 1..20); # Peter Bala, Jul 23 2024

Clear[a]; a = Table[2^(n - 1)Gamma[(3*n - 1)/2]/Gamma[(n + 1)/2], {n, 0, 20}] (* or equivalently *)
Clear[a]; a = Table[2^(n - 1)Pochhammer[(n + 1)/2, n - 1], {n, 0, 20}]

a(n) = 2^(n - 1) * Gamma((3*n - 1)/2) / Gamma((n + 1)/2).
a(n) = 2^(n - 1) * ((n + 1)/2)_(n - 1), where (x)_k is the Pochhammer symbol for Gamma(x + k) / Gamma(k).
a(n) = 3*A113551(n-1) for n>=2. - Hugo Pfoertner, Apr 16 2021
E.g.f.: (sqrt(3)*sin(arcsin(3*sqrt(3)*x)/3) - 3*cos(arcsin(3*sqrt(3)*x)/3))/3. - Stefano Spezia, May 23 2021
a(n) = 3*(3*n - 5)*(3*n - 7)*a(n-2) with a(0) = -1, a(1) = 1 and a(2) = 3. - Peter Bala, Jul 23 2024
a(n) ~ 3^(3*n/2-1) * n^(n-1) / exp(n). - Amiram Eldar, Sep 02 2025

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.

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.

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.

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.

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.

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.

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.

Original entry on oeis.org

-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.

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User: Dixon J. Jones

A343446 Coefficients of the series S(p, q) for which -(p^(1/3))*S converges to the largest real root of x^4 - p*x + q, where 0 < p and 0 < q < 3*(p/4)^(4/3).

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A343445 Coefficients of the series S(p, q) for which (-sqrt(p))*S converges to the largest real root of x^3 - p*x + q for 0 < p and 0 < q < 2*(p/3)^(3/2).

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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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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.

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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.

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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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A343446 Coefficients of the series S(p, q) for which -(p^(1/3))S converges to the largest real root of x^4 - px + q, where 0 < p and 0 < q < 3*(p/4)^(4/3).

A343445 Coefficients of the series S(p, q) for which (-sqrt(p))S converges to the largest real root of x^3 - px + q for 0 < p and 0 < q < 2*(p/3)^(3/2).