The small one, the “double headed snake”

Area = 59

New try, utilizing the identical diagonals as Weather Vane and shuffling round 9 of the edges:

Area = 1221

some outcomes for the issue typically

Proof:

Since the n-gons cannot be deformed, now we have to step away from the (0,0) place of the primary nook each in x and y path on the grid. But since now we have to shut the form too, we should find yourself at (0,0) once more. As such, each step in x or y path should be “undone”, and their sum due to this fact a good quantity. In the orthogonal case, this solely permits $4 | n$ or $4 | n + 1$, for the reason that sum of the aspect lengths, $frac{n(n+1)}{2}$ is just even in these instances.

Regarding diagonals:

Diagonal sides contribute to each the x path and the y path. But since they type a Pythagorean triple, one of many legs are all the time a good quantity. Therefore, the sum of the legs all the time have the identical parity because the hypotenuse, and the parity argument within the earlier paragraph isn’t affected.

But that does not say something about whether or not the allowed values of $n$ even have a geometrically legitimate realisation. I’ll deal with this later

The following assemble (“snake physique”) goes to be the core of all later reasoning:

It consists of a terminating tail, which makes use of the quantity “1” and a pair of small integers, adopted by some variety of sides constructed by teams of 4 consecutive integers, collectively forming a spiral.

If one can discover a “snake head” of fastened measurement, so the remaining integers can be utilized to construct a “snake physique”, one has an express building for infinitely many values of $n$

A easy snake head:

For the case $4 | n$,  $ n geq 8$

The different easy snake head required:

For the case $4 | n + 1$,  $ n geq 15$

This solely leaves just a few stray instances, which we are able to take care of one after the other:

• 3: Trivially inconceivable
• 4: Trivially inconceivable
• 7: ${(0,0),(7,0),(7,-2),(10,-2),(6,-5),(6,-1),(0,-1)}$
• 11: ${(0,0),(11,0),(15,-3),(15,-5),(18,-5),(10,1),(1,1),(1,5),(8,5),(8,6),(0,6)}$

All potential values of $n$ have then been coated.

The snake heads from the proof above already present a easy higher sure:

$4 | n$

$frac{n(n+1)}{4} – 2$

$4 | n + 1$

$frac{n(n+1)}{4} – 11$

But this may be improved.

The $4 | n + 1$ head has a easy generalization:

Lowering the higher sure to:

$frac{n(n+1)}{4} – 16lfloor{frac{n+5}{20}rfloor} + 5$

For the $4 | n$ case, we’d like a special head to do the identical:

Lowering the higher sure to:

$frac{n(n+1)}{4} – 16lfloor{frac{n}{20}rfloor} + 3$

In the $4 | n + 1$ case, now we have this monstrosity:

Which works out to a sure of

$frac{n(n+1)}{4} – 11p^2 – 23p – 11$, the place $p = lfloor{frac{n – 15}{20}rfloor}$

Asymptotically, that is

$frac{89}{400} cdot n^2$

The $4 | n$ could be very related:

Bound to:

$frac{n(n+1)}{4} – 11p^2 – 11p – 13$, the place $p = lfloor{frac{n – 20}{20}rfloor}$

Which is asymptotically the identical.

First, it could be helpful to get a clearer instinct of the quadratic snake heads already introduced. Let’s check out the quantity line:

Notably, for 3/4 or the vary, we merely have pairs of 3-4-5 multiples interspaced by snake physique segments.

The subsequent trick is to note that for values on the shape $130k + 26$ and $130k + 39$, now we have two multiples of the 5-12-13 triple with only a single pair of 3-4-5 triples in between.

We can merely exchange this area with a barely extra elaborate building, of smaller space.

Of course, when just a few 3-4-5 triples are taken out of circulation, the quantity line is not divided into 1/4 and three/4. The new ratios are 12/51 and 39/51.

The “barely extra elaborate building” is that this one, combining two such 130k areas right into a repeating unit:

Segment lengths to not scale, solely widths.

The purple, inexperienced and pink pairs are scavenged from the $5k + 1$ and $5k – 1$ pairs, since minimizing their lengths could be very a lot helpful resulting from their width. Their authentic use may be changed by stealing extra snake physique legs, at no further value.

Pinning down all of the phrases of the sure can be tedious, however I’ve calculated the asymptotic behaviour, which is:

$frac{657301}{3005600} cdot n^2$

Ever so barely higher than the comparatively much less complicated quadratic constructions, enhancing them by 1.7%

Pick’s theorem is as soon as once more extremely related.

While it is onerous to ensure the need of inside lattice factors (in reality I feel an indefinitely expandable building with solely finitely a lot of them exists), perimeter factors are significantly simpler. For occasion, there are at the very least $n$ perimeter factors, as a result of each vertex is one. That offers $lceil{frac{n}{2} – 1rceil}$

By the generalized model of Euclid’s building of Pythagorean triples, a quantity that is neither composite nor the sum of two squares can solely be positioned orthogonally in a lattice.

By the asymptotically equal variety of Pythagorean and non-Pythagorean primes, and the prime quantity theorem, there are asymptotically $frac{n}{log{n}}$ such numbers.

Applying Pick’s theorem to this reality, the decrease sure is proportional to:

$frac{n^2}{log{n}}$

I might be very if somebody can discover a sure within the type of a relentless issue multiplied by $n^2$

In extra sensible phrases, one takes all aspect lengths, and specific every of them as $ok cdot z$, the place $z$ is the diagonal of a primitive Pythagorean triple (counting $(0,1,1)$ as a primitive triple), as massive as potential. All of those $ok$ values are summed, being the minimal variety of perimeter lattice factors. The sure for this process is sort of presumably higher than the one obtained via the PNT. (Bound requested!)

(work in progress)

An simply obtainable higher sure is:

$lfloor{frac{n^2 (n+1)^2}{16pi}rfloor}$

By noticing that the world cannot presumably be bigger than that of a circle by the given circumference

The same sure is:

$lfloor{frac{1}{4} frac{n(n+1)}{2} cot{left(frac{2pi}{n(n+1)}proper)}rfloor}$

Because

The space cannot be bigger than a $frac{n(n+1)}{2}$ sided common polygon

(work in progress)

In the case of $8|n$, now we have this indefinitely expandable building:

It’s additionally relevant to all $4|n$, since we are able to simply specific this as $12 + 8k$, inserting the edges above into the $n=12$ answer, which is already confirmed to exist.

This offers a sure of:

$frac{1}{128} n^4$, plus some phrases of decrease order

This manner of “mending” the development for $4|n$ could be very highly effective (Do the identical for the $4|n + 1$ case, simply use the 11-gon answer as base). It’s just like the Euclidean division theorem, and since we have already got an existence proof for all potential values of $n$, we are able to now go wild with constructions requiring $n$ to have very constricted varieties, however nonetheless be instantly generalizable for sufficiently massive $n$

Time for an additional repeatable unit:

Grouping 8 consecutive items of those, one aspect of a “massive sq.” can include parts 1 and eight in all these teams, one other aspect 2 and seven, the third 3 and 6, and the final 4 and 5. (all summing to the identical size)

This requires $40|n$, however since now we have proven that any such building may be generalized to all legitimate $n$ of adequate measurement, it is a common answer.

It interprets 5 perimeter items right into a diagonal of size $frac{sqrt{365}}{5}$, in distinction to the answer introduced earlier that merely achieves 2 perimeter items right into a diagonal of $sqrt{2}$

This improves the asymptotic behaviour to:

$frac{73}{8000}n^4$

Including extra primitive Pythagorean triples (and thus making the integers within the fraction horribly massive), one might enhance this barely.

A greater snake head for the $4|n$ case, proven right here enhancing the most effective identified answer for $n=40$ from 375 to 361

Gallery of options for a lot of n