[seqfan] Re: Splitting the square into equal parts
Jack Grahl
jack.grahl at gmail.com
Tue Nov 10 14:45:28 CET 2020
I may have misunderstood but it seems to me that the side length ratio
doesn't affect the question at all?
We should be able to take a solution for a square and stretch the square
along one axis without affecting the relative areas of the rectangles, and
the same to transform a rectangle into a square.
On Tue, 10 Nov 2020, 13:31 Neil Fernandez, <primeness at borve.org> wrote:
> Hi Jean-Luc,
>
> This is a lovely idea and makes me think of paintings by Piet Mondrian.
> I'll be interested to see whether a nice constant can be got out of it.
>
> I suggest that the definition might be clearer if phrased as follows:
>
> "Starting with a square, we reduce it (n-1) times by cutting off and
> setting aside a rectangle and leaving a rectangle behind, such that the
> rectangle left at the end is equal in area to each rectangle cut off.
> a(n) is the number of ways of doing this for each n."
>
> Another way of getting constants might be to free up the side-length
> ratio R:1 of the starting rectangle, such that R doesn't have to be 1,
> and then to require a value for the limit L of a(n+1)/a(n). What values
> of L give what values of R?
>
> Best regards
>
> Neil
>
> In message <686122479.43113475.1604997949618.JavaMail.zimbra at unicaen.fr>
> , Jean-Luc Manguin <jean-luc.manguin at unicaen.fr> writes
> >Thank you very much Elijah, you have perfectly understood my idea.
> >
> >The difference you noticed comes from this condition I put in my message :
> >
> >"Every new segment must lean on two existing segments, or in other words
> its
> >begin and end points must be on existing segments (of course, the first
> segment
> >leans on 2 opposite edges of the square)."
> >
> >If you look at the illustration of A108066 here
> https://oeis.org/A108066/a108066
> >.pdf, you can see in the 5-parts division that number 15 do not match
> this
> >condition. In the 6-part division, numbers 52, 53 and 54 do not match.
> This
> >explains the differences.
> >
> >Here is A108066 sequence :
> >
> >1 1
> >2 1
> >3 2
> >4 6
> >5 18
> >6 65
> >7 281
> >8 1343
> >9 6953
> >10 38023
> >
> >Mine is :
> >
> >1 1
> >2 1
> >3 2
> >4 6
> >5 17
> >6 62
> >7 247
> >8 1075
> >9 4912
> >10 23458
> >11 115924
> >12 575678
> >
> >I am now progressing on 15, soon 16.
> >
> >Thanks again ; best regards.
> >
> >JLM
> >
> >----- Mail original -----
> >De: "Elijah Beregovsky" <elijah.beregovsky at gmail.com>
> >À: "seqfan" <seqfan at list.seqfan.eu>
> >Envoyé: Lundi 9 Novembre 2020 21:12:34
> >Objet: [seqfan] Re: Splitting the square into equal parts
> >
> >Could you elaborate on the procedure you're investigating?
> >If I understood you correctly, you're counting partitions of a square into
> >rectangles of equal area (you cannot end up with any other shapes if you
> >disallow "hanging" edges, ends of which do not lie on existing line
> >segments). It is a nice problem and it already has a sequence in OEIS -
> >https://oeis.org/A108066. However, its terms do not equal yours. A
> closely
> >related sequence https://oeis.org/A189243 treats symmetrical solutions as
> >different. But it also does not have the same terms as your sequences.
> Have
> >I misunderstood your idea?
> >Best wishes,
> >Elijah Beregovsky
>
> --
> Neil Fernandez
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>
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