polyomino colinearity - estimating peak
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h...@crypt.org
Feb 12, 2026, 1:26:54 PMFeb 12
to seq...@googlegroups.com, 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
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
Feb 13, 2026, 6:55:35 AMFeb 13
to seq...@googlegroups.com
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
Feb 13, 2026, 7:58:21 AMFeb 13
to seq...@googlegroups.com, 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
:
:> 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).
:me
:> shape as that for a_4(n);
:> - max(a_5(n)) will occur at n=3D56.
:>
:> Known initial values:
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:
:
:/Pontus
:
:
:> 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=
:> 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:
:me
:> shape as that for a_4(n);
:> - max(a_5(n)) will occur at n=3D56.
:>
:> Known initial values:
h...@crypt.org
Feb 20, 2026, 11:34:14 AMFeb 20
to seq...@googlegroups.com, h...@crypt.org
Here is a handwavy argument that could perhaps be firmed up to form part
of a proof:
An infinitely growing polyomino with finite colinearity can start off
going in a straight line, but it must eventually start to turn.
Consider the quarter-circle of radius r, from (r, 0) to (0, r).
We can consider a "best-fit" polyomino that approximates this; such
a polyomino would start off going vertically from (r, 0), and for some
k would first move left at (r - 1, k).
Assuming the closeness-of-fit function c(x, y) looks like
sqrt((x^2 + y^2) / r^2), we can solve the quartic c(r - 1, k) c(r, k + 1) = 1
to find k in terms of r. I haven't solved that, but I think it gives
something close to k = 2 sqrt(r). Regardless of the precise relationship,
it clearly increases with r.
Thus a best-fit quarter-circle polyomino has unbounded colinearity as
the radius grows.
For an infinitely growing polyomino to avoid turning back on itself,
it would have to turn at a rate slower than a circle; it must therefore
have colinearity at least as great as that of a circle with the corresponding
radius, and its colinearity is therefore unbounded.
A couple of obvious flaws that would need firming up:
- need to prove that more creative shapes cannot do better than a slow
spiral;
- the vertical is a bit of a special case, need to prove that other parts
of the arc also generate unbounded colinearity.
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
:
:> 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).
:me
:> shape as that for a_4(n);
:> - max(a_5(n)) will occur at n=3D56.
:>
:> Known initial values:
:t.org
:> .
:>
:
:--=20
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:
:--0000000000003de6ab064ab3486e
:Content-Type: text/html; charset="UTF-8"
:Content-Transfer-Encoding: quoted-printable
:
:<div dir=3D"ltr">Do you know for sure that a_5 is a finite sequence,=C2=A0i=
:.e., that <a href=3D"https://oeis.org/A380991" target=3D"_blank">A380991</a=
:>(5) exists? Or could there exist arbitrarily large polyominoes with no 6 c=
:ollinear cells?<div><br></div><div>/Pontus</div></div><br><div class=3D"gma=
:il_quote"><div dir=3D"ltr" class=3D"gmail_attr">On Thu, Feb 12, 2026 at 7:2=
:6=E2=80=AFPM <<a href=3D"mailto:h...@crypt.org" target=3D"_blank">hv@crypt=
:.org</a>> wrote:<br></div><blockquote class=3D"gmail_quote" style=3D"mar=
:gin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1=
:ex">I'd like to evaluate a_5(n), the sequence equivalent to a_4(n) =3D =
:A380990<br>
:for (at most) 5 colinear points. I have an approach that would work (given<=
:br>
:enough time and diskspace), but I'm worried the diskspace requirements =
:may<br>
:prove to be so unreasonable that it would be foolish to try.<br>
:<br>
:The main unknown for the diskspace requirement is M =3D max a_5(n). <br>
:I think it should be possible to get a rough estimate for M given<br>
:an appropriate set of assumptions, but I'm not sure how to go about it.=
:<br>
:Can someone suggest how to reach such an estimate? Is there an obvious<br>
:or standard way to do so?<br>
:<br>
:Assumptions:<br>
:- the first half of the log graph for a_5(n) will have essentially the same=
:<br>
:=C2=A0 shape as that for a_4(n);<br>
:- max(a_5(n)) will occur at n=3D56.<br>
:<br>
:Known initial values:<br>
:=C2=A0 1, 1, 2, 5, 12, 34, 104, 339, 1133, 3845, 13080, 44374, 149495, 4986=
:59,<br>
:=C2=A0 1641870, 5324339, 16973955<br>
:<br>
:a_4(n+1)/a_4(n) up to max:<br>
:=C2=A0 1.000 2.000 2.500 2.200 2.818 2.742 3.082 2.916 2.952 2.812<br>
:=C2=A0 2.716 2.555 2.414 2.254 2.100 1.945 1.812 1.687 1.568 1.455<br>
:=C2=A0 1.354 1.259 1.164 1.077 1.005<br>
:<br>
:a_5(n+1)/a_5(n) up to known:<br>
:=C2=A0 1.000 2.000 2.500 2.400 2.833 3.059 3.260 3.342 3.394 3.402<br>
:=C2=A0 3.393 3.369 3.336 3.293 3.243 3.188<br>
:<br>
:Thanks in advance for any suggestions,<br>
:<br>
:Hugo van der Sanden<br>
:<br>
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:_blank">https://groups.google.com/d/msgid/seqfan/202602121805.61CI5pM24888%=
:40crypt.org</a>.<br>
:</blockquote></div>
:
:<p></p>
:
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:seqfan/CAND8vCXmF0BsnyA2vgHOp7bjJuWkZcNERzUH2Noi5TYAd%3DjMHw%40mail.gmail.c=
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:.com</a>.<br />
:
:--0000000000003de6ab064ab3486e--
of a proof:
An infinitely growing polyomino with finite colinearity can start off
going in a straight line, but it must eventually start to turn.
Consider the quarter-circle of radius r, from (r, 0) to (0, r).
We can consider a "best-fit" polyomino that approximates this; such
a polyomino would start off going vertically from (r, 0), and for some
k would first move left at (r - 1, k).
Assuming the closeness-of-fit function c(x, y) looks like
sqrt((x^2 + y^2) / r^2), we can solve the quartic c(r - 1, k) c(r, k + 1) = 1
to find k in terms of r. I haven't solved that, but I think it gives
something close to k = 2 sqrt(r). Regardless of the precise relationship,
it clearly increases with r.
Thus a best-fit quarter-circle polyomino has unbounded colinearity as
the radius grows.
For an infinitely growing polyomino to avoid turning back on itself,
it would have to turn at a rate slower than a circle; it must therefore
have colinearity at least as great as that of a circle with the corresponding
radius, and its colinearity is therefore unbounded.
A couple of obvious flaws that would need firming up:
- need to prove that more creative shapes cannot do better than a slow
spiral;
- the vertical is a bit of a special case, need to prove that other parts
of the arc also generate unbounded colinearity.
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:
:
:/Pontus
:
:
:> 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=
:> 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:
:me
:> shape as that for a_4(n);
:> - max(a_5(n)) will occur at n=3D56.
:>
:> 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
:>
:> --
:> You received this message because you are subscribed to the Google Groups
:> "SeqFan" group.
:> To unsubscribe from this group and stop receiving emails from it, send an
:> email to seqfan+un...@googlegroups.com.
:> To view this discussion visit
:> https://groups.google.com/d/msgid/seqfan/202602121805.61CI5pM24888%40cryp=
:> 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
:>
:> --
:> You received this message because you are subscribed to the Google Groups
:> "SeqFan" group.
:> To unsubscribe from this group and stop receiving emails from it, send an
:> email to seqfan+un...@googlegroups.com.
:> To view this discussion visit
:t.org
:> .
:>
:
:--=20
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:--0000000000003de6ab064ab3486e
:Content-Type: text/html; charset="UTF-8"
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:
:<div dir=3D"ltr">Do you know for sure that a_5 is a finite sequence,=C2=A0i=
:.e., that <a href=3D"https://oeis.org/A380991" target=3D"_blank">A380991</a=
:>(5) exists? Or could there exist arbitrarily large polyominoes with no 6 c=
:ollinear cells?<div><br></div><div>/Pontus</div></div><br><div class=3D"gma=
:il_quote"><div dir=3D"ltr" class=3D"gmail_attr">On Thu, Feb 12, 2026 at 7:2=
:6=E2=80=AFPM <<a href=3D"mailto:h...@crypt.org" target=3D"_blank">hv@crypt=
:.org</a>> wrote:<br></div><blockquote class=3D"gmail_quote" style=3D"mar=
:gin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1=
:ex">I'd like to evaluate a_5(n), the sequence equivalent to a_4(n) =3D =
:A380990<br>
:for (at most) 5 colinear points. I have an approach that would work (given<=
:br>
:enough time and diskspace), but I'm worried the diskspace requirements =
:may<br>
:prove to be so unreasonable that it would be foolish to try.<br>
:<br>
:The main unknown for the diskspace requirement is M =3D max a_5(n). <br>
:I think it should be possible to get a rough estimate for M given<br>
:an appropriate set of assumptions, but I'm not sure how to go about it.=
:<br>
:Can someone suggest how to reach such an estimate? Is there an obvious<br>
:or standard way to do so?<br>
:<br>
:Assumptions:<br>
:- the first half of the log graph for a_5(n) will have essentially the same=
:<br>
:=C2=A0 shape as that for a_4(n);<br>
:- max(a_5(n)) will occur at n=3D56.<br>
:<br>
:Known initial values:<br>
:=C2=A0 1, 1, 2, 5, 12, 34, 104, 339, 1133, 3845, 13080, 44374, 149495, 4986=
:59,<br>
:=C2=A0 1641870, 5324339, 16973955<br>
:<br>
:a_4(n+1)/a_4(n) up to max:<br>
:=C2=A0 1.000 2.000 2.500 2.200 2.818 2.742 3.082 2.916 2.952 2.812<br>
:=C2=A0 2.716 2.555 2.414 2.254 2.100 1.945 1.812 1.687 1.568 1.455<br>
:=C2=A0 1.354 1.259 1.164 1.077 1.005<br>
:<br>
:a_5(n+1)/a_5(n) up to known:<br>
:=C2=A0 1.000 2.000 2.500 2.400 2.833 3.059 3.260 3.342 3.394 3.402<br>
:=C2=A0 3.393 3.369 3.336 3.293 3.243 3.188<br>
:<br>
:Thanks in advance for any suggestions,<br>
:<br>
:Hugo van der Sanden<br>
:<br>
:-- <br>
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:40crypt.org</a>.<br>
:</blockquote></div>
:
:<p></p>
:
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:.com</a>.<br />
:
:--0000000000003de6ab064ab3486e--
Pontus von Brömssen
Feb 21, 2026, 4:23:27 PMFeb 21
to seq...@googlegroups.com
I once tried to get a counterexample (arbitrarily large polyominoes with a bounded number of collinear cells) by constructing polyominoes that approximates a line whose slope is the golden ratio, but it didn't work out.
/Pontus
To view this discussion visit https://groups.google.com/d/msgid/seqfan/202602201612.61KGCBM19825%40crypt.org.
h...@crypt.org
Feb 24, 2026, 9:03:50 AMFeb 24
to seq...@googlegroups.com, h...@crypt.org
Just to close this out: I spent some time with Claude AI on this, and it
came to the conclusion that with the given values I should expect a peak
at around n=78 with a peak value around 2 x 10^25.
As such I consider a_5() and A380991(5) firmly out of reach for any approach
I can think of.
Hugo
I wrote:
:I'd like to evaluate a_5(n), the sequence equivalent to a_4(n) = A380990
came to the conclusion that with the given values I should expect a peak
at around n=78 with a peak value around 2 x 10^25.
As such I consider a_5() and A380991(5) firmly out of reach for any approach
I can think of.
Hugo
I wrote:
:I'd like to evaluate a_5(n), the sequence equivalent to a_4(n) = A380990
David desJardins
Feb 24, 2026, 3:06:28 PM (14 days ago) Feb 24
to seq...@googlegroups.com
Your a_5() seems hard, but A380991(5) seems probably tractable with some kind of branching approach.
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h...@crypt.org
Feb 24, 2026, 6:58:27 PM (14 days ago) Feb 24
to seq...@googlegroups.com, h...@crypt.org
I don't know what you mean by "some kind of branching approach", but
I'd like to.
I guess that must involve generating (or proving the (non-)existence
of) qualifying size-n polyominoes without having to consider all the
possible size-{n-1} precursors.
Hugo
David desJardins <da...@desjardins.org> wrote:
:Your a_5() seems hard, but A380991(5) seems probably tractable with some
:kind of branching approach.
:
:90
:> :for (at most) 5 colinear points. I have an approach that would work (giv=
:en
:> :enough time and diskspace), but I'm worried the diskspace requirements m=
:ay
:> :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).
:> :
:> :Known initial values:
:> : 1, 1, 2, 5, 12, 34, 104, 339, 1133, 3845, 13080, 44374, 149495, 498659=
:,
:> : 1641870, 5324339, 16973955
:t.org
:> .
:>
:
:--=20
:You received this message because you are subscribed to the Google Groups "=
:SeqFan" group.
:To unsubscribe from this group and stop receiving emails from it, send an e=
:mail to seqfan+un...@googlegroups.com.
:To view this discussion visit https://groups.google.com/d/msgid/seqfan/CAP%=
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:
:--000000000000efd54b064b976bb0
:able with some kind of branching=C2=A0approach.</div><br><div class=3D"gmai=
:l_quote gmail_quote_container"><div dir=3D"ltr" class=3D"gmail_attr">On Tue=
:, Feb 24, 2026 at 6:03=E2=80=AFAM <<a href=3D"mailto:h...@crypt.org">hv@cr=
:ypt.org</a>> wrote:<br></div><blockquote class=3D"gmail_quote" style=3D"=
:margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-lef=
:t:1ex">Just to close this out: I spent some time with Claude AI on this, an=
:d it<br>
:came to the conclusion that with the given values I should expect a peak<br=
:>
:at around n=3D78 with a peak value around 2 x 10^25.<br>
:<br>
:As such I consider a_5() and A380991(5) firmly out of reach for any approac=
:h<br>
:I can think of.<br>
:<br>
:Hugo<br>
:<br>
:I wrote:<br>
::I'd like to evaluate a_5(n), the sequence equivalent to a_4(n) =3D A38=
:0990<br>
::for (at most) 5 colinear points. I have an approach that would work (given=
:<br>
::enough time and diskspace), but I'm worried the diskspace requirements=
: may<br>
::prove to be so unreasonable that it would be foolish to try.<br>
::<br>
::The main unknown for the diskspace requirement is M =3D max a_5(n). <br>
::I think it should be possible to get a rough estimate for M given<br>
::an appropriate set of assumptions, but I'm not sure how to go about it=
:.<br>
::Can someone suggest how to reach such an estimate? Is there an obvious<br>
::or standard way to do so?<br>
::<br>
::Assumptions:<br>
::- the first half of the log graph for a_5(n) will have essentially the sam=
:e<br>
::=C2=A0 shape as that for a_4(n);<br>
::- max(a_5(n)) will occur at n=3D56.<br>
:659,<br>
::=C2=A0 2.716 2.555 2.414 2.254 2.100 1.945 1.812 1.687 1.568 1.455<br>
::=C2=A0 1.354 1.259 1.164 1.077 1.005<br>
::=C2=A0 3.393 3.369 3.336 3.293 3.243 3.188<br>
::<br>
::Thanks in advance for any suggestions,<br>
:seqfan/202602241341.61ODfCM06770%40crypt.org" rel=3D"noreferrer" target=3D"=
:_blank">https://groups.google.com/d/msgid/seqfan/202602241341.61ODfCM06770%=
:seqfan/CAP%3DxTqO%3DcTyunykaOKWFK3z7vFFaDL59f%2BzJ2uvW_fKNyW%2Bjog%40mail.g=
:mail.com?utm_medium=3Demail&utm_source=3Dfooter">https://groups.google.com/=
:d/msgid/seqfan/CAP%3DxTqO%3DcTyunykaOKWFK3z7vFFaDL59f%2BzJ2uvW_fKNyW%2Bjog%=
:40mail.gmail.com</a>.<br />
:
:--000000000000efd54b064b976bb0--
I'd like to.
I guess that must involve generating (or proving the (non-)existence
of) qualifying size-n polyominoes without having to consider all the
possible size-{n-1} precursors.
Hugo
David desJardins <da...@desjardins.org> wrote:
:Your a_5() seems hard, but A380991(5) seems probably tractable with some
:kind of branching approach.
:
:On Tue, Feb 24, 2026 at 6:03=E2=80=AFAM <h...@crypt.org> wrote:
:
:> Just to close this out: I spent some time with Claude AI on this, and it
:> came to the conclusion that with the given values I should expect a peak
:> at around n=3D78 with a peak value around 2 x 10^25.
:
:> Just to close this out: I spent some time with Claude AI on this, and it
:> came to the conclusion that with the given values I should expect a peak
:>
:> As such I consider a_5() and A380991(5) firmly out of reach for any
:> approach
:> I can think of.
:>
:> Hugo
:>
:> I wrote:
:> :I'd like to evaluate a_5(n), the sequence equivalent to a_4(n) =3D A3809=
:> As such I consider a_5() and A380991(5) firmly out of reach for any
:> approach
:> I can think of.
:>
:> Hugo
:>
:> I wrote:
:90
:> :for (at most) 5 colinear points. I have an approach that would work (giv=
:en
:> :enough time and diskspace), but I'm worried the diskspace requirements m=
:ay
:> :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
:> same
:> : shape as that for a_4(n);
:> :- max(a_5(n)) will occur at n=3D56.
:> :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);
:> :
:> :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
:>
:> --
:> You received this message because you are subscribed to the Google Groups
:> "SeqFan" group.
:> To unsubscribe from this group and stop receiving emails from it, send an
:> email to seqfan+un...@googlegroups.com.
:> To view this discussion visit
:> https://groups.google.com/d/msgid/seqfan/202602241341.61ODfCM06770%40cryp=
:> :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
:>
:> --
:> You received this message because you are subscribed to the Google Groups
:> "SeqFan" group.
:> To unsubscribe from this group and stop receiving emails from it, send an
:> email to seqfan+un...@googlegroups.com.
:> To view this discussion visit
:t.org
:> .
:>
:
:--=20
:You received this message because you are subscribed to the Google Groups "=
:SeqFan" group.
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:mail to seqfan+un...@googlegroups.com.
:To view this discussion visit https://groups.google.com/d/msgid/seqfan/CAP%=
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:
:--000000000000efd54b064b976bb0
:Content-Type: text/html; charset="UTF-8"
:Content-Transfer-Encoding: quoted-printable
:
:<div dir=3D"ltr">Your a_5() seems hard, but A380991(5) seems probably tract=
:Content-Transfer-Encoding: quoted-printable
:
:able with some kind of branching=C2=A0approach.</div><br><div class=3D"gmai=
:l_quote gmail_quote_container"><div dir=3D"ltr" class=3D"gmail_attr">On Tue=
:, Feb 24, 2026 at 6:03=E2=80=AFAM <<a href=3D"mailto:h...@crypt.org">hv@cr=
:ypt.org</a>> wrote:<br></div><blockquote class=3D"gmail_quote" style=3D"=
:margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-lef=
:t:1ex">Just to close this out: I spent some time with Claude AI on this, an=
:d it<br>
:came to the conclusion that with the given values I should expect a peak<br=
:>
:at around n=3D78 with a peak value around 2 x 10^25.<br>
:<br>
:As such I consider a_5() and A380991(5) firmly out of reach for any approac=
:h<br>
:I can think of.<br>
:<br>
:Hugo<br>
:<br>
:I wrote:<br>
::I'd like to evaluate a_5(n), the sequence equivalent to a_4(n) =3D A38=
:0990<br>
::for (at most) 5 colinear points. I have an approach that would work (given=
:<br>
::enough time and diskspace), but I'm worried the diskspace requirements=
: may<br>
::prove to be so unreasonable that it would be foolish to try.<br>
::<br>
::The main unknown for the diskspace requirement is M =3D max a_5(n). <br>
::I think it should be possible to get a rough estimate for M given<br>
::an appropriate set of assumptions, but I'm not sure how to go about it=
:.<br>
::Can someone suggest how to reach such an estimate? Is there an obvious<br>
::or standard way to do so?<br>
::<br>
::Assumptions:<br>
::- the first half of the log graph for a_5(n) will have essentially the sam=
:e<br>
::=C2=A0 shape as that for a_4(n);<br>
::- max(a_5(n)) will occur at n=3D56.<br>
::<br>
::Known initial values:<br>
::=C2=A0 1, 1, 2, 5, 12, 34, 104, 339, 1133, 3845, 13080, 44374, 149495, 498=
::Known initial values:<br>
:659,<br>
::=C2=A0 1641870, 5324339, 16973955<br>
::<br>
::a_4(n+1)/a_4(n) up to max:<br>
::=C2=A0 1.000 2.000 2.500 2.200 2.818 2.742 3.082 2.916 2.952 2.812<br>
::<br>
::a_4(n+1)/a_4(n) up to max:<br>
::=C2=A0 2.716 2.555 2.414 2.254 2.100 1.945 1.812 1.687 1.568 1.455<br>
::=C2=A0 1.354 1.259 1.164 1.077 1.005<br>
::<br>
::a_5(n+1)/a_5(n) up to known:<br>
::=C2=A0 1.000 2.000 2.500 2.400 2.833 3.059 3.260 3.342 3.394 3.402<br>
::a_5(n+1)/a_5(n) up to known:<br>
::=C2=A0 3.393 3.369 3.336 3.293 3.243 3.188<br>
::<br>
::Thanks in advance for any suggestions,<br>
::<br>
::Hugo van der Sanden<br>
:<br>
:-- <br>
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::Hugo van der Sanden<br>
:<br>
:-- <br>
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:40crypt.org</a>.<br>
:</blockquote></div>
:
:<p></p>
:
:-- <br />
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:</blockquote></div>
:
:<p></p>
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