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.

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;
		

Crossrefs

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)