cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

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

Views

Author

Dixon J. Jones, Oct 12 2013

Keywords

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

Examples

			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.

Links

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-A230211 (j=3 to j=8).

Programs

  • Magma
    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
  • Mathematica
    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 *)
  • PARI
    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
  • Sage
    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

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=9.

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

Views

Author

Dixon J. Jones, Oct 12 2013

Keywords

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

Examples

			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.

Links

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

Programs

  • Magma
    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
  • Mathematica
    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 *)
  • PARI
    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
  • Sage
    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

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=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

Views

Author

Dixon J. Jones, Oct 12 2013

Keywords

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.

Examples

			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.

Links

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).

Programs

  • 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
  • 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 *)
  • 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
  • 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

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.

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

Views

Author

Dixon J. Jones, Oct 12 2013

Keywords

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

Examples

			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.

Links

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

Programs

  • Magma
    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
  • Mathematica
    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 *)
  • PARI
    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
  • Sage
    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

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=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

Views

Author

Dixon J. Jones, Oct 12 2013

Keywords

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.

Examples

			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.

Links

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).

Programs

  • 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
  • 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 *)
  • 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
  • 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

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.

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

Views

Author

Dixon J. Jones, Oct 12 2013

Keywords

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

Examples

			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.

Links

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

Programs

  • Magma
    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
  • Mathematica
    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 *)
  • PARI
    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
  • Sage
    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

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=8.
Showing 1-6 of 6 results.

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