Tiling rectangles with Heptomino plus rectangle #3

Question

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

Next puzzle in this series: Tiling rectangles with Heptomino plus rectangle #4

...up to 100 to come... I'll post them a few at a time. Why is this first one #3: Numbering is as per my heptomino data file and will skip rectifiable and uninteresting heptominoes. Some of them will be posed as no-computer hand-tiling only puzzles.

The goal is to tile rectangles as small as possible with the given heptomino, in this case number 3 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 $2\times 6$ 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 87 more but lots of them can be found by 'expansion rules'. I considered component rectangles of width 1 through 11 and length to 31 but my search was far from complete.

List of known sizes:

• Width 1: Lengths 1 to 20, 22 to 25, 29 to 30
• Width 2: Lengths 2 to 18, 22 to 24, 29 to 31
• Width 3: Lengths 3 to 8, 14 to 15
• Width 4: Lengths 4 to 25, 27, 29 to 31
• Width 5: Lengths 7 to 8
• Width 7: Length 8
• Width 8: Lengths 9 to 10

Many of them could be tiled by hand fairly easily.

Answer 1

Let's get this party started with $1 \times 2$:

3x7=21

_{(that wasn't too obvious, see the edit ...)}

and $2 \times 2$, which also works for $2 \times 4$:

3x10=30

Like here, there are again some generalizable solutions for $1 \times n$. Which one is smaller depends on $n$:

Left: if $n-1$ is divisible by 7, size: $(n+1) \times (n+5)$.
If $2n-1$ is divisible by 7, size: $(n+1) \times (2n+5)$.
If $3n-1$ is divisible by 7, size: $(n+1) \times (3n+5)$. (and so on:
If $kn-1$ is divisible by 7, size: $(n+1) \times (kn+5)$. )

Right: if $n$ is divisible by 7, size: $(n+1) \times (n+7)$. Otherwise, the size is $(n+1) \times 8n$.

They can probably be extended to $2 \times n$ ($n$ odd) in a similar way as with the hexomino puzzle. For example, here is $2 \times 7$:

14x11=154