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 1-2 of 2 results.

A131402 2*A007318 - (A046854 + A065941 - A000012).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 4, 4, 1, 1, 6, 7, 6, 1, 1, 7, 14, 14, 7, 1, 1, 9, 20, 33, 20, 9, 1, 1, 10, 31, 56, 56, 31, 10, 1, 1, 12, 40, 97, 111, 97, 40, 12, 1, 1, 13, 55, 142, 217, 217, 142, 55, 13, 1, 1, 15, 67, 213, 358, 463, 358, 213, 67, 15, 1, 1, 16, 86, 287, 590, 841, 841, 590, 287, 86, 16, 1
Offset: 0

Views

Author

Gary W. Adamson, Jul 07 2007

Keywords

Comments

Row sums = A131403: (1, 2, 5, 10, 21, 44, 93, ...).

Examples

			First few rows of the triangle are:
  1;
  1,  1;
  1,  3,  1;
  1,  4,  4,  1;
  1,  6,  7,  6,  1;
  1,  7, 14, 14,  7,  1;
  1,  9, 20, 33, 20,  9,  1;
  1, 10, 31, 56, 56, 31, 10,  1;
  ...
		

Crossrefs

Row sums are A131403.

Programs

  • PARI
    T(n,k) = if(k <= n, 2*binomial(n, k) + 1 - binomial((n + k)\2, k) - binomial(n-(k+1)\2, k\2), 0) \\ Andrew Howroyd, Aug 09 2018

Formula

2*A007318 - (A046854 + A065941 - A000012) as infinite lower triangular matrices.
T(n,k) = 2*binomial(n, k) + 1 - binomial(floor((n + k)/2), k) - binomial(n-floor((k+1)/2), floor(k/2)). - Andrew Howroyd, Aug 09 2018

Extensions

Missing terms inserted and a(55) and beyond from Andrew Howroyd, Aug 09 2018

A131405 Row sums of triangle A131404.

Original entry on oeis.org

1, 2, 7, 16, 37, 82, 179, 384, 813, 1702, 3531, 7272, 14889, 30342, 61603, 124700, 251825, 507582, 1021535, 2053388, 4123481, 8274002, 16591767, 33254356, 66623317, 133432082, 267164239, 534814024, 1070413693, 2142098602, 4286254091, 8575836312, 17157057669
Offset: 0

Views

Author

Gary W. Adamson, Jul 07 2007

Keywords

Examples

			a(4) = 37 = sum of row 4 terms of triangle A131404: (1 + 11 + 13 + 11 + 1).
		

Crossrefs

Row sums of A131404.
Cf. A131403.

Programs

  • PARI
    Vec((1 - 3*x + 5*x^2 - 6*x^3 + 4*x^4)/((1 - x)^2*(1 - 2*x)*(1 - x - x^2)) + O(x^40)) \\ Andrew Howroyd, Aug 09 2018

Formula

From Andrew Howroyd, Aug 09 2018: (Start)
a(n) = 5*a(n-1) - 8*a(n-2) + 3*a(n-3) + 3*a(n-4) - 2*a(n-5).
G.f.: (1 - 3*x + 5*x^2 - 6*x^3 + 4*x^4)/((1 - x)^2*(1 - 2*x)*(1 - x - x^2)).
(End)

Extensions

Terms a(10) and beyond from Andrew Howroyd, Aug 09 2018
Showing 1-2 of 2 results.