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.

A316675 Triangle read by rows: T(n,k) gives the number of ways to stack n triangles in a valley so that the right wall has k triangles for n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 2, 1, 1, 1, 0, 0, 1, 1, 3, 2, 1, 1, 1, 0, 0, 1, 1, 3, 3, 2, 1, 1, 1, 0, 0, 1, 1, 3, 3, 3, 2, 1, 1, 1, 0, 0, 1, 1, 4, 3, 4, 3, 2, 1, 1, 1, 0, 0, 1, 1, 5, 4, 5, 4, 3, 2, 1, 1, 1
Offset: 0

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Author

Seiichi Manyama, Jul 10 2018

Keywords

Examples

			T(8,4) = 3.
    *                             *
   / \                           / \
  *---*   *     *---*---*       *---*
   \ / \ / \     \ / \ / \     / \ / \
    *---*---*     *---*---*   *---*---*
     \ / \ /       \ / \ /     \ / \ /
      *---*         *---*       *---*
       \ /           \ /         \ /
        *             *           *
Triangle begins:
  1;
  0, 1;
  0, 0, 1;
  0, 0, 1, 1;
  0, 0, 1, 1, 1;
  0, 0, 1, 1, 1, 1;
  0, 0, 1, 1, 1, 1,  1;
  0, 0, 1, 1, 2, 1,  1,  1;
  0, 0, 1, 1, 3, 2,  1,  1,  1;
  0, 0, 1, 1, 3, 3,  2,  1,  1,  1;
  0, 0, 1, 1, 3, 3,  3,  2,  1,  1,  1;
  0, 0, 1, 1, 4, 3,  4,  3,  2,  1,  1, 1;
  0, 0, 1, 1, 5, 4,  5,  4,  3,  2,  1, 1, 1;
  0, 0, 1, 1, 5, 5,  6,  5,  4,  3,  2, 1, 1, 1;
  0, 0, 1, 1, 5, 5,  8,  6,  5,  4,  3, 2, 1, 1, 1;
  0, 0, 1, 1, 6, 5, 10,  8,  7,  5,  4, 3, 2, 1, 1, 1;
  0, 0, 1, 1, 7, 6, 11, 10, 10,  7,  5, 4, 3, 2, 1, 1, 1;
  0, 0, 1, 1, 7, 7, 13, 11, 12, 10,  7, 5, 4, 3, 2, 1, 1, 1;
  0, 0, 1, 1, 7, 7, 16, 13, 14, 12, 10, 7, 5, 4, 3, 2, 1, 1, 1;
  ...
		

Crossrefs

Row sums give A006950.
Sums of even columns give A059777.
Cf. A072233.

Formula

For m >= 0,
Sum_{n>=2m} T(n,2m) *x^n = x^(2m) * Product_{j=1..m} (1+x^(2j-1))/(1-x^(2j)).
Sum_{n>=2m+1} T(n,2m+1)*x^n = x^(2m+1) * Product_{j=1..m} (1+x^(2j-1))/(1-x^(2j)).