A262767 -id:A262767 - cp's OEIS frontend

A375673 n and a(n) (with a(n) >= n) are the edges of the minimum-area rectangle such that its area is an integer multiple of its perimeter.

6, 4, 20, 12, 42, 8, 18, 15, 110, 12, 156, 35, 30, 16, 272, 36, 342, 20, 28, 99, 506, 24, 100, 143, 54, 28, 812, 45, 930, 32, 66, 255, 140, 36, 1332, 323, 78, 40, 1640, 56, 1806, 44, 90, 483, 2162, 48, 294, 75, 102, 52, 2756, 108, 66, 56, 114, 783, 3422, 60, 3660, 899

Offset: 3

Paolo Xausa, Aug 25 2024

nonn

No such rectangle exists for n = 1 or n = 2.

			The first rectangles are listed below.
.
     |           | area/per. |    area   | perimeter
   n |    a(n)   | (A375674) | (A375675) | (A375676)
  ---------------------------------------------------
   3 |      6    |     1     |     18    |     18
   4 |      4    |     1     |     16    |     16
   5 |     20    |     2     |    100    |     50
   6 |     12    |     2     |     72    |     36
   7 |     42    |     3     |    294    |     98
   8 |      8    |     2     |     64    |     32
   9 |     18    |     3     |    162    |     54
  10 |     15    |     3     |    150    |     50
  ...
For n = 9, two rectangles exist with the area being an integer multiple of the perimeter: one with sides (9, 18) and one with sides (9, 74). a(9) is the smaller one.

Paolo Xausa, Table of n, a(n) for n = 3..1000

Cf. A375674 (area/perimeter), A375675 (area), A375676 (perimeter).

Cf. A262767, A329402, A306841.

A375673[n_] := Module[{b, r}, SolveValues[2*r == n*b/(n+b) && b >= n, {b, r}, Integers, MaxRoots -> 1][[1,1]]];
Array[A375673, 100, 3]

a(n) = A375675(n)/n.
a(n) = (A375676(n) - 2*n)/2.
a(n) = n for n = 4*k (k >= 1).

A375674 a(n) is the area/perimeter ratio of the rectangle whose edges are n and A375673(n).

Original entry on oeis.org

1, 1, 2, 2, 3, 2, 3, 3, 5, 3, 6, 5, 5, 4, 8, 6, 9, 5, 6, 9, 11, 6, 10, 11, 9, 7, 14, 9, 15, 8, 11, 15, 14, 9, 18, 17, 13, 10, 20, 12, 21, 11, 15, 21, 23, 12, 21, 15, 17, 13, 26, 18, 15, 14, 19, 27, 29, 15, 30, 29, 18, 16, 20, 22, 33, 17, 23, 21, 35, 18, 36, 35, 25, 19

Offset: 3

Paolo Xausa, Aug 25 2024

nonn

Paolo Xausa, Table of n, a(n) for n = 3..1000

Cf. A375673 (edges), A375675 (area), A375676 (perimeter).

Cf. A262767, A329402, A306841.

A375674[n_] := Module[{b, r}, SolveValues[2*r == n*b/(n+b) && b >= n, {b, r}, Integers, MaxRoots -> 1][[1,2]]];
Array[A375674, 100, 3]

a(n) = A375675(n)/A375676(n).

A375675 a(n) is the area of the rectangle whose edges are n and A375673(n).

Original entry on oeis.org

18, 16, 100, 72, 294, 64, 162, 150, 1210, 144, 2028, 490, 450, 256, 4624, 648, 6498, 400, 588, 2178, 11638, 576, 2500, 3718, 1458, 784, 23548, 1350, 28830, 1024, 2178, 8670, 4900, 1296, 49284, 12274, 3042, 1600, 67240, 2352, 77658, 1936, 4050, 22218, 101614, 2304, 14406

Offset: 3

Paolo Xausa, Aug 25 2024

nonn

Paolo Xausa, Table of n, a(n) for n = 3..1000

Cf. A375673 (edges), A375674 (area/perimeter), A375676 (perimeter).

Cf. A262767, A329402, A306841.

A375675[n_] := Module[{b, r}, SolveValues[2*r == n*b/(n+b) && b >= n, {b, r}, Integers, MaxRoots -> 1][[1,1]]*n];
Array[A375675, 100, 3]

a(n) = n*A375673(n).

a(n) = A375674(n)*A375676(n).

A375676 a(n) is the perimeter of the rectangle whose edges are n and A375673(n).

Original entry on oeis.org

18, 16, 50, 36, 98, 32, 54, 50, 242, 48, 338, 98, 90, 64, 578, 108, 722, 80, 98, 242, 1058, 96, 250, 338, 162, 112, 1682, 150, 1922, 128, 198, 578, 350, 144, 2738, 722, 234, 160, 3362, 196, 3698, 176, 270, 1058, 4418, 192, 686, 250, 306, 208, 5618, 324, 242, 224, 342

Offset: 3

Paolo Xausa, Aug 25 2024

nonn

Paolo Xausa, Table of n, a(n) for n = 3..1000

Cf. A375673 (edges), A375674 (area/perimeter) A375675 (area).

Cf. A262767, A329402, A306841.

A375676[n_] := Module[{b, r}, 2*(SolveValues[2*r == n*b/(n+b) && b >= n, {b, r}, Integers, MaxRoots -> 1][[1,1]] + n)];
Array[A375676, 100, 3]

a(n) = 2*(n + A375673(n)).

a(n) = A375675(n)/A375674(n).

A283056 Size of the smallest polyomino that admits a hole of size n.

Original entry on oeis.org

0, 7, 9, 11, 11, 13, 13

Offset: 0

Dmitry Kamenetsky, Feb 27 2017

The task here is to surround a hole of size n with the least number of squares. The hole is another polyomino, so we can obtain a lower bound using A027709: minimal perimeter of polyomino with n square cells. We need an extra 3 (or more) diagonal cells to surround any hole. Hence a(n) >= A027709(n) + 3 = 2*ceiling(2*sqrt(n)) + 3.
For rectangular holes we can obtain an upper bound using A262767: minimum perimeter of a rectangle with area n and integer sides. Hence a(n) <= A262767(n) + 3.
Perhaps a(n) is actually equal to A027709(n)+3?

			For n=1, we have a single square hole, so a(1)=7.
For n=2, we have a domino hole, so a(2)=9.
For n=3, we can use either an L or V tromino hole, so a(3)=11.
For n=4, we use the square tetromino hole, so a(4)=11.
For n=5, we use the P pentomino hole, so a(5)=13.
For n=6, we use the 2 X 3 rectangle hole, so a(6)=13.

Cf. A027709.

cp's OEIS Frontend

A375673 n and a(n) (with a(n) >= n) are the edges of the minimum-area rectangle such that its area is an integer multiple of its perimeter.

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A375674 a(n) is the area/perimeter ratio of the rectangle whose edges are n and A375673(n).

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A375675 a(n) is the area of the rectangle whose edges are n and A375673(n).

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A375676 a(n) is the perimeter of the rectangle whose edges are n and A375673(n).

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Programs

Mathematica

Formula

A283056 Size of the smallest polyomino that admits a hole of size n.

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Keywords

Comments

Examples

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