Select polyominoes for a set (currently 1 or 2), for which tilings should be shown.

Then click "Show" button.

You may also see list of all polyomino sets for which data is available here.

Area: 6.

Size: 3x3.

Holes: 0.

Order: 2.

Square order: 24.

Prime rectangles: ≥ 1.

Smallest rectangle (3x4):

Smallest square (12x12):

No odd rectangles exist.

Blue number - strongly prime rectangle (which cannot be divided into two or more number of rectangles tileable by this set).

Green number - weakly prime rectangle (which cannot be divided into two rectangles tileable by this set, but which can be divided into three or more rectangles).

Purple number - prime rectangle (unknown if weakly or strongly prime).

Red number - composite rectangle (which can be divided into two rectangles tileable by this set).

Gray number - it is unknown whether rectangle is prime or composite.

Question mark (?) - solution count is unknown.

Click on underlined numbers to view picture with one solution.

w \ h

1-2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

N>0

1-2

0

3

0

0

4

0

0

5

0

0

0

0

6

0

0

0

0

7

0

0

0

0

0

0

8

0

0

0

0

0

9

0

0

0

0

0

0

10

0

0

0

0

0

0

0

0

0

11

0

0

0

0

0

0

0

0

0

0

12

0

0

13

0

0

0

0

0

0

0

0

0

0

0

14

0

0

0

0

0

0

0

0

0

0

0

0

15

0

0

0

0

0

0

0

0

0

0

0

16

0

0

0

0

0

0

0

0

0

0

17

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

18

0

0

0

0

0

0

0

0

0

0

0

0

0

19

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

?

20

0

0

0

0

0

0

0

0

0

0

0

?

?

21

0

0

0

0

0

0

0

0

0

0

0

0

0

?

?

N>0

x

4k

3k

x

4k

12k

3k

4k

12k

12k

all

12k

12k

4k

3k

12k

4k

?

?

Smallest prime reptile (6Jx6):

$N(w; h)$ - number of ways to tile $w\times h$ rectangle (including symmetric solutions)

$T(w; h) = \begin{cases} 1, & N(w; h) \geq 1 \\ 0, & \text{else} \end{cases}$ - tileability function, $1$ if tiles rectangle, $0$ otherwise

$A(w; h) = \left(N(w; h)\right)^{\frac{1}{wh}}$ - average number of ways to tile cell in $w\times h$ rectangle (including symmetric solutions)

$G(T; x; y) = \sum_{w=1}^{\infty}\sum _{h=1}^{\infty}T(w; h)x^wy^h$ - bivariate generating function of $T(w; h)$

$G(A; x; y) = \sum_{w=1}^{\infty}\sum _{h=1}^{\infty}A(w; h)x^wy^h$ - bivariate generating function of $A(w; h)$

$N(1; n) = T(1; n) = 0, \qquad n \geq 1 \tag{1}$

$N(2; n) = T(2; n) = 0, \qquad n \geq 1 \tag{2}$

$N(3; n) = 2 \times N(3; n - 4), \qquad n \geq 5 \tag{3}$

$N(4; n) = 2 \times N(4; n - 3), \qquad n \geq 4 \tag{4}$

$N(5; n) = T(5; n) = 0, \qquad n \geq 1 \tag{5}$

$N(6; n) = 4 \times N(6; n - 4), \qquad n \geq 5 \tag{6}$

$N(7; n) = 256 \times N(7; n - 12), \qquad n \geq 13 \tag{7}$

$N(8; n) = 4 \times N(8; n - 3), \qquad n \geq 4 \tag{8}$

$N(9; n) = 8 \times N(9; n - 4), \qquad n \geq 5 \tag{9}$

$N(10; n) = 3072 \times N(10; n - 12), \qquad n \geq 13 \tag{10}$

$N(11; n) = 6144 \times N(11; n - 12), \qquad n \geq 13 \tag{11}$

$N(12; n) = 8 \times N(12; n - 3) + 16 \times N(12; n - 4), \qquad n \geq 5 \tag{12}$

$N(13; n) = 32768 \times N(13; n - 12), \qquad n \geq 13 \tag{13}$

$N(n; m) = T(n; m) = 0, \qquad 3\nmid n,3\nmid m,4\nmid n,4\nmid m \tag{14}$

First, note that J hexominoes always come in following pairs, because there is only two ways to fill concave area:

Second, assume J hexomino tiles $n\times m$ rectangles for $3\nmid n,3\nmid m,4\nmid n,4\nmid m$.

Place numbers in rectangles' cells according to function $F(x,y)\equiv 1+12\left(\left\lfloor\frac{x}{4}\right\rfloor - \left\lfloor\frac{x-1}{4}\right\rfloor + \left\lfloor\frac{y}{4}\right\rfloor - \left\lfloor\frac{y-1}{4}\right\rfloor\right)\pmod{24}$, where $x$ and $y$ are cells' coordinates (zero-based). On the one hand, each J hexomino pair, no matter how placed, covers sum congruent to $0\pmod{24}$. Then sum covered by all hexominoes is also congruent to $0\pmod{24}$. On the other hand, rectangle covers sum congruent to $\sum_{x=0}^{n-1}\sum_{y=0}^{m-1}F(x,y)$, which is not congruent to $0\pmod{24}$ for $3\nmid n,3\nmid m,4\nmid n,4\nmid m$. Contradiction, as hexomino tiles this rectangle and thus sum covered by all hexominoes should be equal to sum covered by rectangle. Thus only assumption we made is false - J hexomino doesn't tile $n\times m$ rectangles for $3\nmid n,3\nmid m,4\nmid n,4\nmid m$. Q.E.D.