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Area: 8.
Perimeter: 18.
Size: 2x7.
Is rectangular: no.
Is convex: yes.
Holes: 0.
Order: 2.
Square order: 8.
Prime rectangles: ≥ 4.
Smallest rectangle (2x8):
Smallest square (8x8):
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.
Smallest prime reptile (8L1x4):
$N(w; h)$ - 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 $A(w; h)$
$N(n; m) = T(n; m) = 0, \qquad 16\nmid nm \tag{1}$
Assume L1 octomino tiles $n\times m$ rectangles for $nm\not\equiv 0\pmod{16}$.
Place numbers in rectangles' cells according to function $F(x,y)\equiv8x+1\pmod{16}$, where $x$ and $y$ are cells' coordinates (zero-based). On the one hand, L1 octomino, no matter how placed, covers sum congruent to $0\pmod{16}$. Then sum covered by all octominoes is also congruent to $0\pmod{16}$. On the other hand, rectangle covers sum congruent to $\sum_{x=0}^{n-1}\sum_{y=0}^{m-1}\left(8x+1\right)$, which is not congruent to $0\pmod{16}$ for $nm\not\equiv 0\pmod{16}$. Contradiction, as octomino tiles this rectangle and thus sum covered by all octominoes should be equal to sum covered by rectangle. Thus only assumption we made is false - L1 octomino doesn't tile $n\times m$ rectangles for $nm\not\equiv 0\pmod{16}$. Q.E.D.
