polyomino colinearity - estimating peak

[h...@crypt.org]

I'd like to evaluate a_5(n), the sequence equivalent to a_4(n) = A380990
for (at most) 5 colinear points. I have an approach that would work (given
enough time and diskspace), but I'm worried the diskspace requirements may
prove to be so unreasonable that it would be foolish to try.

The main unknown for the diskspace requirement is M = max a_5(n).
I think it should be possible to get a rough estimate for M given
an appropriate set of assumptions, but I'm not sure how to go about it.
Can someone suggest how to reach such an estimate? Is there an obvious
or standard way to do so?

Assumptions:
- the first half of the log graph for a_5(n) will have essentially the same
shape as that for a_4(n);
- max(a_5(n)) will occur at n=56.

Known initial values:
1, 1, 2, 5, 12, 34, 104, 339, 1133, 3845, 13080, 44374, 149495, 498659,
1641870, 5324339, 16973955

a_4(n+1)/a_4(n) up to max:
1.000 2.000 2.500 2.200 2.818 2.742 3.082 2.916 2.952 2.812
2.716 2.555 2.414 2.254 2.100 1.945 1.812 1.687 1.568 1.455
1.354 1.259 1.164 1.077 1.005

a_5(n+1)/a_5(n) up to known:
1.000 2.000 2.500 2.400 2.833 3.059 3.260 3.342 3.394 3.402
3.393 3.369 3.336 3.293 3.243 3.188

Thanks in advance for any suggestions,

Hugo van der Sanden

[Pontus von Brömssen]

Do you know for sure that a_5 is a finite sequence, i.e., that A380991(5) exists? Or could there exist arbitrarily large polyominoes with no 6 collinear cells?

/Pontus

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[h...@crypt.org]

No I don't. I strongly suspect that A380991(n) is defined and finite
for all n >= 1, but I don't know of any proof.

Hugo

=?UTF-8?Q?Pontus_von_Br=C3=B6mssen?= <pontus.vo...@gmail.com> wrote:
:Do you know for sure that a_5 is a finite sequence, i.e., that A380991
:<https://oeis.org/A380991>(5) exists? Or could there exist arbitrarily

:large polyominoes with no 6 collinear cells?
:
:/Pontus
:

:On Thu, Feb 12, 2026 at 7:26=E2=80=AFPM <h...@crypt.org> wrote:
:
:> I'd like to evaluate a_5(n), the sequence equivalent to a_4(n) =3D A38099=
:0
:> for (at most) 5 colinear points. I have an approach that would work (give=
:n
:> enough time and diskspace), but I'm worried the diskspace requirements ma=
:y
:> prove to be so unreasonable that it would be foolish to try.
:>
:> The main unknown for the diskspace requirement is M =3D max a_5(n).

:> I think it should be possible to get a rough estimate for M given
:> an appropriate set of assumptions, but I'm not sure how to go about it.
:> Can someone suggest how to reach such an estimate? Is there an obvious
:> or standard way to do so?
:>
:> Assumptions:

:> - the first half of the log graph for a_5(n) will have essentially the sa=
:me
:> shape as that for a_4(n);
:> - max(a_5(n)) will occur at n=3D56.
:>
:> Known initial values: