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Area: 4.
Perimeter: 10.
Size: 1x4.
Is rectangular: yes.
Is convex: yes.
Holes: 0.
Order: 1.
Square order: 4.
Odd order: 1.
Prime rectangles: 1.
Smallest rectangle and smallest odd rectangle (1x4):
Smallest square (4x4):
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.
Smallest prime reptile (4Ix2):
Smallest torus and smallest odd torus (1x4):
Smallest square torus (4x4):
Smallest Baiocchi figure (area 16):
$N(w; h)$ - 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 $A(w; h)$
$N(1; 4n) = T(1; 4n) = 1, \qquad n \geq 1 \tag{1}$
$N(2; 4n) = T(2; 4n) = 1, \qquad n \geq 1 \tag{2}$
$N(3; 4n) = T(3; 4n) = 1, \qquad n \geq 1 \tag{3}$
"I" tetrominoes can be placed in $1\times 4n$, $2\times 4n$ and $3\times 4n$ rectangles only in one orientation, thus there is maximum one way to place them. One side is divisible by 4, therefore tetrominoes fit perfectly. Q.E.D.
$N(n; m) = T(n; m) = 0, \qquad 4\nmid n,4\nmid m \tag{4}$
Assume I tetromino tiles $n\times m$ rectangles for $4\nmid n,4\nmid m$.
Place numbers in rectangles' cells according to function $F(x,y)\equiv 2x^2+2y^2+12xy+1\pmod{16}$, where $x$ and $y$ are cells' coordinates (zero-based). On the one hand, I tetromino, no matter how placed, covers sum congruent to $0\pmod{16}$. Then sum covered by all tetrominoes is also congruent to $0\pmod{16}$. On the other hand, rectangle covers sum congruent to $\sum_{x=0}^{n-1}\sum_{y=0}^{m-1}\left(2x^2+2y^2+12xy+1\right)$, which is not congruent to $0\pmod{16}$ for $4\nmid n,4\nmid m$. Contradiction, as tetromino tiles this rectangle and thus sum covered by all tetrominoes should be equal to sum covered by rectangle. Thus only assumption we made is false - I tetromino doesn't tile $n\times m$ rectangles for $4\nmid n,4\nmid m$. Q.E.D.
$T(4n; m) = 1 \tag{5}$
$G(T; x; y) = \frac{xy\left(x^3 y^3+x^3 y^2+x^3 y+x^3+x^2 y^3+x y^3+y^3\right)}{\left(1-x^4\right)\left(1-y^4\right)} \tag{6}$
