[seqfan] Re: Splitting the square into equal parts

Jean-Luc Manguin jean-luc.manguin at unicaen.fr
Tue Nov 10 09:45:49 CET 2020

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.


----- 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

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