[seqfan] Re: Splitting the square into equal parts
primeness at borve.org
Tue Nov 10 14:31:32 CET 2020
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?
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 :
>Mine is :
>I am now progressing on 15, soon 16.
>Thanks again ; best regards.
>----- 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?
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