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.

A109439 Triangle read by rows, in which row n gives coefficients in expansion of ((1 - x^n)/(1 - x))^3.

Original entry on oeis.org

1, 1, 3, 3, 1, 1, 3, 6, 7, 6, 3, 1, 1, 3, 6, 10, 12, 12, 10, 6, 3, 1, 1, 3, 6, 10, 15, 18, 19, 18, 15, 10, 6, 3, 1, 1, 3, 6, 10, 15, 21, 25, 27, 27, 25, 21, 15, 10, 6, 3, 1, 1, 3, 6, 10, 15, 21, 28, 33, 36, 37, 36, 33, 28, 21, 15, 10, 6, 3, 1, 1, 3, 6, 10, 15, 21, 28, 36, 42, 46, 48, 48
Offset: 1

Views

Author

Labos Elemer, Jun 30 2005

Keywords

Comments

Sum of n-th row is n^3. The n-th row contains 3n-2 entries. Largest coefficients in rows are listed in A077043. The 255th row describes the distribution of color lattice points in the 765 r+g+b=k planes of the 24-bit RGB-cube with 256^3 points.
Also, the number of cubes of dimension 1 X 1 X 1 needed to build a cube by layers perpendicular to the main diagonal. Each layer is made up of regular triangular numbers T near the summits and truncated T's in the middle. E.g., cube 3^3 is made of layers 1, 3, 6, 7, 6, 3, 1, using T1, T2, T3 and a regularly truncated T4, 7 instead of 10. - M. Dauchez (mdzzdm(AT)yahoo.fr), Aug 31 2005
The n-th row is the third row of the (n+1)-nomial triangle. For example, row 1 (1,3,3,1) is the third row in the binomial triangle; row 5 is the third row of the 6-nomial triangle. - Bob Selcoe, Feb 18 2014
It appears that T(n,k) gives the number of possible ways of randomly selecting k cards from n-1 sets, each with three different playing cards. - Juan Pablo Herrera P., Nov 04 2016

Examples

			Triangle starts:
  1;
  1, 3, 3, 1;
  1, 3, 6, 7, 6, 3, 1;
  1, 3, 6,10,12,12,10, 6, 3, 1;
  1, 3, 6,10,15,18,19,18,15,10, 6, 3, 1;
  1, 3, 6,10,15,21,25,27,27,25,21,15,10, 6, 3, 1;
  1, 3, 6,10,15,21,28,33,36,37,36,33,28,21,15,10, 6, 3, 1.
		

Crossrefs

Programs

  • Mathematica
    Flatten[Table[CoefficientList[Series[((1-x^n)/(1-x))^3,{x,1,3*n}],x], {n,1,100}],1]
  • PARI
    row(n) = Vec(((1 - x^n)/(1 - x))^3);
    tabf(nn) = for (n=1, nn, print(row(n))); \\ Michel Marcus, Oct 12 2016

Formula

From Juan Pablo Herrera P., Oct 17 2016: (Start)
T(n,k) = A000217(k+1) = (k+2)!/(k!*2) if 0 <= k < n,
T(n,k) = (9*n-3*n^2+6*k*n-6*k-2*k^2-4)/2 if n-3 < k < 2*n,
T(n,k) = A000217(3n-k-2) = (3*n-k-1)!/((3*(n-1)-k)!*2) if 2*n-3 < k < 3*n-2.
T(n,k) = Sum_{i=k-n+1..k} A004737(T(n,i)),
T(n,k) = Sum_{i=k-n+1..k} (n-|n-i-1|) if n <= k <= 2*n+1. (End)

Extensions

Offset corrected by Joerg Arndt at the suggestion of Michel Marcus, Oct 12 2016