4x
n is about as tight as you can squeeze hexominoes.
3x
n looks like it's pretty much impossible. Consider the blue pieces in the following image:
All four of these need to be in the final construction (and will only fit horizontally) but between them they create six 3-cell deep wells at the edge of the construction. And there's only five hexominoes (the red ones) that could fill those gaps. The I-hexomino could theoretically fill two 3-cell deep wells, but
I think in every possible case the space between the two blue hexominoes either end of it would be less than 6 and therefore unfillable. Unless we're going for constructions with holes permitted, in which case that would be fine after all. But exceedingly difficult.
Then, check out these green cases that create two adjacent 2-cell deep wells. In filling one of them, either you have to use one of the red pieces from above, or you use a piece that has a 2-cell extension, which would then cover the square marked by the red 'X' and create a new 3 (or more) cell deep well.
I know this isn't a rigorous, mathematically watertight proof but it's enough of a deterrent to stop me spending ages looking for 3x
n solutions.
So, back to 4-cell high...
Similar to how it is with pentominoes (the narrower the rectangle, the fewer solutions there are), finding 4x
n rectangles with hexominoes has proven to be surprisingly challenging. The few search programs I know how to use don't seem to like really narrow rectangles very much either, which left me doubtful I'd be able to do much better by hand. (Although in hindsight it's more likely I just don't know how to use the programs as well as I think I do, or how to set them up so that they search efficiently.)
A few months back I'd found the solution below using the combined set of hexominoes and pentominoes. It's not really what I'm aiming for though; the addition of the smaller pentominoes makes this about a hundred times easier, and I was able to place the holes symmetrically as a result.
On 31/12/19 I had another crack at this, aiming for a 4x53 grid using just the hexominoes, but with no constraints on where the two holes would be. Just proving that a 4x
n solution exists would be enough for the first step; making it all pretty could come later.
After far too long (about an hour, maybe? It's hard to say because I tend to lose track of time when doing things like this) I found a solution. This one:
It's butt ugly though; not only are the two holes not placed in any kind of order but one is on the edge of the rectangle too, which just doesn't look right to me. It's not really a hole now, is it? It's just some weird notch out of the side of the puzzle. Oh well, it's a start.
Side note, that 'T' piece near the right-hand end was the absolute worst piece to place. Had I used it up right near the start it might not have been such a pain in the arse, but somehow it escaped my attention until there wasn't a lot of long skinny pieces left that worked well with it. In order to place it vertically rather than horizontally, it needed to have one of the holes either side of it too, since it partitions the rectangle into two parts, both of odd size.
But, it's a proof of concept at least. A symmetrical 4x53 rectangle seems way more possible now than it did before. (Edit 06/09/2020: I found one!)