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.

A306189 Number of minimum dominating sets in the n-Sierpinski gasket graph.

Original entry on oeis.org

3, 6, 2, 392, 1656976026, 122836566640423857273582993856, 50043395758253154294953783566500246788902420299683914045600060272160541415159062540151890
Offset: 1

Views

Author

Eric W. Weisstein, Jan 28 2019

Keywords

Crossrefs

Cf. A347508, A347515, A376478 (domination number).

Programs

  • PARI
    a(n)={my(s=[[1+O(x),0,0,0],[0,0,x+O(x^2)],[0,x^2+O(x^3)],[x^3+O(x^4)]]);for(k=2,n,s=vector(4,i,vector(5-i,j,sum(xy=0,3,sum(xz=0,3,sum(yz=0,3,s[1+(i>1)+!xy+!xz][1+(j>3)+(xy%2)+(xz%2)]*s[1+(i>2)+!xy+!yz][1+(j>2)+(xy\2)+(yz%2)]*s[1+(i>3)+!xz+!yz][1+(j>1)+(xz\2)+(yz\2)]/x^(!xy+!xz+!yz)))))));pollead([1,3,3,1]*vectorv(4,i,s[i][5-i]))} \\ Christian Sievers, Jul 21 2024, improved Jul 25 2024

Extensions

a(5) and beyond from Christian Sievers, Jul 21 2024

A377657 Array read by ascending antidiagonals: A(n, k) = Sum_{j=0..k} tan(j*Pi/(1 + 2*k))^(2*n).

Original entry on oeis.org

1, 0, 2, 0, 3, 3, 0, 9, 10, 4, 0, 27, 90, 21, 5, 0, 81, 850, 371, 36, 6, 0, 243, 8050, 7077, 1044, 55, 7, 0, 729, 76250, 135779, 33300, 2365, 78, 8, 0, 2187, 722250, 2606261, 1070244, 113311, 4654, 105, 9, 0, 6561, 6841250, 50028755, 34420356, 5476405, 312390, 8295, 136, 10
Offset: 0

Views

Author

Peter Luschny, Nov 10 2024

Keywords

Comments

Based on an observation made by Fredrik Johansson about A376777, which is the main diagonal of this array.

Examples

			Array begins
  [0] 1,    2,       3,         4,           5,            6, ... A000027
  [1] 0,    3,      10,        21,          36,           55, ... A014105
  [2] 0,    9,      90,       371,        1044,         2365, ... A377858
  [3] 0,   27,     850,      7077,       33300,       113311, ... A376778
  [4] 0,   81,    8050,    135779,     1070244,      5476405, ...
  [5] 0,  243,   76250,   2606261,    34420356,    264893255, ...
  [6] 0,  729,  722250,  50028755,  1107069876,  12813875437, ...
  [7] 0, 2187, 6841250, 960335173, 35607151476, 619859803695, ...
  .
Seen as a triangle T(n, k) = A(n-k, k):
  [0] 1;
  [1] 0,    2;
  [2] 0,    3,      3;
  [3] 0,    9,     10,       4;
  [4] 0,   27,     90,      21,       5;
  [5] 0,   81,    850,     371,      36,      6;
  [6] 0,  243,   8050,    7077,    1044,     55,    7;
  [7] 0,  729,  76250,  135779,   33300,   2365,   78,   8;
  [8] 0, 2187, 722250, 2606261, 1070244, 113311, 4654, 105, 9;
		

Crossrefs

Columns: A376478.
Cf. A376777 (main diagonal), A377658 (antidiagonal sums).
Cf. A091042.

Programs

  • Maple
    A := (n, k) -> add(tan(j*Pi/(1 + 2*k))^(2*n), j = 0..k):
    seq(print(seq(round(evalf(A(n, k), 32)), k = 0..6)), n = 0..7);
  • PARI
    A(n, k) = {trace(matcompanion(sum(m=0, k, x^m*binomial(2*k+1, 2*(k-m))*(-1)^(m+1)))^n)+(n==0) } \\ Thomas Scheuerle, Nov 11 2024

Formula

Row n of A091042(n, k) = binomial(2*n+1, 2*k) gives the polynomial Pe(n, x), with zeros in -tan(Pi/2*n+1)^2, -tan(2*Pi/2*n+1)^2, ..., -tan(n*Pi/2*n+1)^2. Let Pm(n, k, x) be the polynomial with zeros in (-tan(Pi/2*n+1)^2)^k, (-tan(2*Pi/2*n+1)^2)^k, ..., (-tan(n*Pi/2*n+1)^2)^k, then A(k, n) is the coefficient of X^(n-1) in the polynomial Pm(n, k, x). A way to do this calculation without evaluation of irrational numbers is to obtain the companion matrix M of the polynomial Pe(n, x), then A(k, n) = tr(M^k) (the trace of M^k). - Thomas Scheuerle, Nov 11 2024
Showing 1-2 of 2 results.