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.

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A366998 a(n) is the numerator of the maximum expected number of steps of a random walk on the square lattice until it lands on a mined lattice point, given that mines are placed on all but n points.

Original entry on oeis.org

0, 1, 4, 12, 28, 8, 124, 128, 263, 9, 1303, 519707, 380, 3435
Offset: 0

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Author

Pontus von Brömssen, Nov 01 2023

Keywords

Comments

For all n <= 13 except n = 3, the optimal placement of the mine-free points is unique, up to rotations and reflections with respect to the starting point. See linked illustration.

Examples

			For n = 0, the random walk stops before it can take any step, so a(0) = 0.
For n = 1, only the mine at the starting point can be swept, so the random walk always stops after 1 step and a(1) = 1.
For n = 2, the starting point and one adjacent point can be swept. The random walk then has probability 1/4 of surviving at each step, which implies that the expected number of steps is 4/3, so a(2) = 4. (The number of steps follows a geometric distribution.)
		

Crossrefs

Cf. A365964, A366999 (denominators), A369368 (hexagonal lattice), A369370 (triangular lattice).

A365963 Let I(n) be the moment of inertia of the polyomino with binary code A246521(n+1) about an axis through its center of mass perpendicular to the plane of the polyomino, the polyomino having a unit point mass in the center of each of its cells. a(n) is I(n) times the number of cells of the polyomino.

Original entry on oeis.org

0, 1, 4, 6, 14, 8, 11, 12, 20, 20, 32, 28, 38, 30, 32, 26, 26, 24, 30, 20, 50, 40, 49, 69, 61, 33, 49, 37, 46, 41, 52, 41, 53, 42, 61, 53, 52, 61, 34, 41, 50, 57, 85, 70, 73, 65, 69, 65, 60, 53, 56, 69, 49, 44, 45, 105, 82, 58, 64, 88, 86, 76, 74, 94, 86, 82
Offset: 1

Views

Author

Pontus von Brömssen, Sep 23 2023

Keywords

Comments

If the cells have a uniform density of 1 instead of point masses in the centers, the moment of inertia is I(n) + k/6 = a(n)/k + k/6, where k is the number of cells.

Examples

			As an irregular triangle:
  0;
  1;
  4, 6;
  14, 8, 11, 12, 20;
  20, 32, 28, 38, 30, 32, 26, 26, 24, 30, 20, 50;
  ...
The five tetrominoes have moments of inertia 7/2, 2, 11/4, 3, 5 (in the order they appear in A246521). Multiplying these numbers by 4, we obtain the 4th row.
The last term of the k-th row of the irregular triangle corresponds to the straight k-omino, whose moment of inertia is k*(k^2-1)/12, so the last term of the k-th row is k^2*(k^2-1)/12 = A002415(k). (This ought to be the largest term of the k-th row.)
		

Crossrefs

Cf. A000105 (row lengths), A002415, A246521, A365964 (row minima).

Formula

If the centers of the cells of the polyomino have coordinates (x_i,y_i), 1 <= i <= k, its moment of inertia is Sum_{i=1..k} x_i^2+y_i^2 - (Sum_{i=1..k} x_i)^2/k - (Sum_{i=1..k} y_i)^2/k.
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