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.

User: Juan Pablo Herrera P.

Juan Pablo Herrera P.'s wiki page.

Juan Pablo Herrera P. has authored 3 sequences.

A277951 Triangle read by rows, in which row n gives coefficients in expansion of ((x^n - 1)/(x - 1))^6.

Original entry on oeis.org

1, 1, 6, 15, 20, 15, 6, 1, 1, 6, 21, 50, 90, 126, 141, 126, 90, 50, 21, 6, 1, 1, 6, 21, 56, 120, 216, 336, 456, 546, 580, 546, 456, 336, 216, 120, 56, 21, 6, 1, 1, 6, 21, 56, 126, 246, 426, 666, 951, 1246, 1506, 1686, 1751, 1686, 1506, 1246, 951, 666, 426, 246, 126, 56, 21, 6, 1
Offset: 1

Views

Author

Juan Pablo Herrera P., Nov 18 2016

Keywords

Comments

Sum of n-th row is n^6. The n-th row contains 6n-5 entries. Largest coefficients of each row are listed in A071816.
The n-th row is the sixth row of the n-nomial triangle. For example, row 2 (1,6,15,20,15,6,1) is the sixth row in the binomial triangle
T(n,k) gives the number of possible ways of randomly selecting k cards from n-1 sets, each with six different playing cards. It is also the number of lattice paths from (0,0) to (6,k) using steps (1,0), (1,1), (1,2), ..., (1,n-1).

Examples

			Triangle starts:
1;
1, 6, 15, 20, 15, 6, 1;
1, 6, 21, 50, 90, 126, 141, 126, 90, 50, 21, 6, 1;
1, 6, 21, 56, 120, 216, 336, 456, 546, 580, 546, 456, 336, 216, 120, 56, 21, 6, 1.

Links

Crossrefs

Cf. A000332, A004737, A071816, A109439, A154286, A277949, A277950.
Mentioned in: A273975

Programs

  • Mathematica
    Table[CoefficientList[Series[((x^n - 1)/(x - 1))^6, {x, 0, 6 n}], x], {n, 10}] // Flatten
  • PARI
    row(n) = Vec(((1 - x^n)/(1 - x))^6);
tabf(nn) = for (n=1, nn, print(row(n)));

Formula

T(n,k) = Sum_{i=k-n+1..k} A277950(T(n,i))

A277950 Triangle read by rows, in which row n gives coefficients in expansion of ((x^n - 1)/(x - 1))^5.

Original entry on oeis.org

1, 1, 5, 10, 10, 5, 1, 1, 5, 15, 30, 45, 51, 45, 30, 15, 5, 1, 1, 5, 15, 35, 65, 101, 135, 155, 155, 135, 101, 65, 35, 15, 5, 1, 1, 5, 15, 35, 70, 121, 185, 255, 320, 365, 381, 365, 320, 255, 185, 121, 70, 35, 15, 5, 1
Offset: 1

Views

Author

Juan Pablo Herrera P., Nov 05 2016

Keywords

Comments

Sum of n-th row is n^5. The n-th row contains 5n-4 entries. Largest coefficients of each row are listed in A077044.
The n-th row is the fifth row of the n-nomial triangle. For example, row 2 (1,5,10,10,5,1) is the fifth row in the binomial triangle.
T(n,k) gives the number of possible ways of randomly selecting k cards from n-1 sets, each with five different playing cards. It is also the number of lattice paths from (0,0) to (5,k) using steps (1,0), (1,1), (1,2), ..., (1,n-1).

Examples

			Triangle starts:
1;
1, 5, 10, 10, 5, 1;
1, 5, 15, 30, 45, 51, 45, 30, 15, 5, 1;
1, 5, 15, 35, 65, 101, 135, 155, 155, 135, 101, 65, 35, 15, 5, 1;
1, 5, 15, 35, 70, 121, 185, 255, 320, 365, 381, 365, 320, 255, 185, 121, 70, 35, 15, 5, 1;

Links

Crossrefs

Cf. A000332, A004737, A077044, A109439, A154286, A277949, A277951.
Mentioned in: A273975.

Programs

  • Mathematica
    Table[CoefficientList[Series[((x^n - 1)/(x - 1))^5, {x, 0, 5 n}], x], {n, 10}] // Flatten
  • PARI
    row(n) = Vec(((1 - x^n)/(1 - x))^5); tabf(nn) = for (n=1, nn, print(row(n)));

Formula

T(n,k) = Sum_{i=k-n+1..k} A277949(T(n,i)).
From Juan Pablo Herrera P., Dec 20 2016: (Start)
T(n,k) = A000332(k+4) = (k+4)!/(k!*24) if 0 =< k < n.
T(n,k) = ((k+4)!/k!-5*(k-n+4)!/(k-n)!)/24 if n =< k < 2*n.
T(n,k) = ((k+4)!/k!-5*(k-n+4)!/(k-n)!+10*(k-2*n+4)!/(k-2*n)!)/24 if 2*n =< k < 3*n.
T(n,k) = ((5*n-k-1)!/(5*n-k-5)!-5*(4*n-k-1)!/(4*n-k-5)!)/24 if 3*n-4 =< k < 4*n-4.
T(n,k) = A000332(5*n-k-1) = (5*n-k-1)!/(5*n-k-5)!*24 4*n-4 =< k < 5*n-4. (End)

A277949 Triangle read by rows, in which row n gives coefficients in expansion of ((x^n - 1)/(x - 1))^4.

Original entry on oeis.org

1, 1, 4, 6, 4, 1, 1, 4, 10, 16, 19, 16, 10, 4, 1, 1, 4, 10, 20, 31, 40, 44, 40, 31, 20, 10, 4, 1, 1, 4, 10, 20, 35, 52, 68, 80, 85, 80, 68, 52, 35, 20, 10, 4, 1, 1, 4, 10, 20, 35, 56, 80, 104, 125, 140, 146, 140, 125, 104, 80, 56, 35, 20, 10, 4, 1
Offset: 1

Views

Author

Juan Pablo Herrera P., Nov 05 2016

Keywords

Comments

Sum of n-th row is n^4. The n-th row contains 4n-3 entries. Largest coefficients of each row are listed in A005900.
The n-th row is the fourth row of the n-nomial triangle. For example, row 2 (1,4,6,4,1) is the fourth row in the binomial triangle.
T(n,k) gives the number of possible ways of randomly selecting k cards from n-1 sets, each with four different playing cards. It is also the number of lattice paths from (0,0) to (4,k) using steps (1,0), (1,1), (1,2), ..., (1,n-1).

Examples

			Triangle starts:
1;
1, 4, 6, 4, 1;
1, 4, 10, 16, 19, 16, 10, 4, 1;
1, 4, 10, 20, 31, 40, 44, 40, 31, 20, 10, 4, 1;
1, 4, 10, 20, 35, 52, 68, 80, 85, 80, 68, 52, 35, 20, 10, 4, 1;
1, 4, 10, 20, 35, 56, 80, 104, 125, 140, 146, 140, 125, 104, 80, 56, 35, 20, 10, 4, 1.
...
There are T(3,2) = 10 ways to select 2 cards from two sets of four playing cards ABCD, namely, {AA}, {AB}, {AC}, {AD}, {BB}, {BC}, {BD}, {CC}, {CD}, and {DD}.

Links

Crossrefs

Cf. A000292, A004737, A005900, A109439, A210440, A277950, A277951.
Mentioned in: A273975

Programs

  • Mathematica
    Table[CoefficientList[Series[((x^n - 1)/(x - 1))^4, {x, 0, 4 n}], x], {n, 6}] // Flatten (* Michael De Vlieger, Nov 10 2016 *)
  • PARI
    row(n) = Vec(((1 - x^n)/(1 - x))^4);
tabf(nn) = for (n=1, nn, print(row(n)));

Formula

T(n,k) = Sum_{i=k-n+1..k} A109439(T(n,i)).
T(n,k) = A000292(k+1) = (k+3)!/(k!*6) if 0 =< k < n,
T(n,k) = ((k+3)*(k+2)*(k+1)-4*(k-n+3)*(k-n+2)*(k-n+1))/6 if n =< k < 2*n,
T(n,k) = ((4*n-1-k)*(4*n-2-k)*(4*n-3-k)-4*(3*n-1-k)*(3*n-2-k)*(3*n-3-k))/6 if 2*n-3 =< k < 3*n-3,
T(n,k) = A000292(4*n-3-k) = (4*n-1-k)!/((4*n-4-k)!*6) if 3*n-3 =< k < 4n-3.

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