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

1,3

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.

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.

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.

Example

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).

Sequence in context: A240198 A016072 A335163 * A095867 A253535 A349171

Adjacent sequences: A365960 A365961 A365962 * A365964 A365965 A365966

Keyword

nonn,tabf

Author

Pontus von Brömssen, Sep 23 2023

Status

approved