[seqfan] A006528 and quadrilateres in a square

Jean-Luc Manguin jean-luc.manguin at unicaen.fr
Thu Sep 3 17:13:08 CEST 2020


The sequence A006528 can give the solution to this interesting (at least for me) problem : 
- put N points on each edge of a square ; the points are equally spaced. 
- you can draw a quadrilatere by drawing a line from a point of an edge to another point of the neighbour edge. 
- how many quadrilateres can you draw, taking the rotations in account ? 

The answer is given by the sequence above, and the formula (N^4 + N^2 + 2*N)/4 is easy to understand because N quadrilateres are squares and are invariant for any 90 degrees rotation, and N^2 are rectangles that are invariant for a 180 degrees rotation. 

Now, the interesting question for geometry lovers is : how many quadrilateres have a surface which is half of the surface of the bounding square ? The first numbers of this sequence (not in OEIS) are (0), 1, 3, 13, 28, 59 .... 

The theoretical formula is welcome ! 
Best regards, 

JL Manguin 

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