# Tiling rectangles with Heptomino plus rectangle #3

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.

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

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

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

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 $2 \times 7$:

• 1x2 is optimal, as is 2x2 (also the 2x4 that gives you), also the 2x7 and the 1x7 and the 1x6. Your notation "n|7k+1, size: n+1x7k+6" is not clear to me, it appears to not follow standard rules of precedence of operators, and has a redundant "1x" which I don't follow. – theonetruepath Jun 9 '18 at 19:43
• $n|7k+1$ means $n$ divides $7k+1$. The $\times$ is only used to multiply the x and the y dimension, it never means multiplying numbers. It therefore has a lower precedence than addition. – Glorfindel Jun 9 '18 at 19:45
• I would have understood "n | (7k+1)" and "(n+1) x (7k+6)" a bit more easily. I still have trouble with "n divides (7k+1)" as I still can't get from there to the size of the component rectangle... and I can see the rectangle in the diagram... – theonetruepath Jun 9 '18 at 20:00
• Yeah, I've gotten better at that. Let me make an edit ... – Glorfindel Jun 9 '18 at 20:01
• OK I follow now. Eg if I decode that for '1x9' I get '10x32' which works... but there is a better generalisation for that case, ie one that results in a 10x10... and in 10x17 for the 1x16, and so on. These appear to be optimal for larger cases too, as well as for some cases where (3*2n-1) is divisible by 7, eg it works for both 1x30 and 1x15 (both size 10x31) – theonetruepath Jun 9 '18 at 20:41