Inspired by Polyomino T hexomino and rectangle packing into rectangle
See also series Tiling rectangles with F pentomino plus rectangles and Tiling rectangles with Hexomino plus rectangle #1
Previous puzzle in this series Tiling rectangles with Heptomino plus rectangle #4
Next puzzle in this series Tiling rectangles with Heptomino plus rectangle #7
The goal is to tile rectangles as small as possible with the given heptomino, in this case number 6 of the 108 heptominoes. We allow the addition of copies of a rectangle. For each rectangle $a\times b$, find the smallest area larger rectangle that copies of $a\times b$ plus at least one of the given heptomino will tile.
Example with the $1\times 1$ you can tile a $2\times 5$ as follows:
Now we don't need to consider $1\times 1$ further as we have found the smallest rectangle tilable with copies of the heptomino plus copies of $1\times 1$.
I found 14 more. I considered component rectangles of width 1 through 11 and length to 31 but my search was not complete.
List of known sizes:
- Width 1: Lengths 1 to 8, 10 to 12
- Width 2: Lengths 2, 3, 5
- Width 3: Length 5
Most of these could be tiled by hand using logic rather than just trial and error.
Answer
Finally, a more interesting heptomino :) (in the sense that previous ones all had generalizable solutions who looked very much like this hexomino)
Here's the minimal solution for $1 \times 2$:
and for $2 \times 2$:
For $3 \times 5$:
My program found another one for $2 \times 7$:
a very narrow one for $1 \times 10$:
another one for $1 \times 11$:
and another one for $1 \times 12$:
This is probably the $1 \times 8$ solution you're looking for:
I like how this one and Jaap's attempt are fundamentally different; this one is 'chaos' and the other one 'order'. It's asymmetric but it can be turned in a symmetric one; there are two ways to tile the irregular shape formed (twice) by the darker shaded polyominos. If you use the same one for both, you get a symmetric solution.
Here is the minimal solution for $1 \times 9$:
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