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.

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

Author

Pontus von Brömssen, Sep 23 2023

Keywords

• nonn
• tabf

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.)
		

Links

• Pontus von Brömssen, Table of n, a(n) for n = 1..6473 (rows 1..10).
• Index entries for sequences related to moment of inertia.
• Index entries for sequences related to polyominoes.

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.