[seqfan] A006528 and quadrilateres in a square
Jean-Luc Manguin
jean-luc.manguin at unicaen.fr
Thu Sep 3 17:13:08 CEST 2020
Hello,
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
More information about the SeqFan
mailing list