OFFSET
1,3
COMMENTS
Terms were calculated by hand by the author. Terms should be verified.
On the infinite square grid we can draw every square anywhere, on any of the four quadrants or on several of them.
Note that distinct arrangements could give the same result for some values of n.
a(7) >= 24. - Hugo Pfoertner, based on data from Hermann Jurksch, Aug 02 2020
From Hugo Pfoertner, Aug 28 2020: (Start)
Hermann Jurksch (private communication) has provided a proof that a(n) <= floor((n^2 + n + 1)*(3/5)).
I myself conjecture that a(n) <= floor((n^2 - n)*(3/5)) for n >= 4.
Lower bounds based on numerical results for the terms a(12)-a(15) are 76, 88, 105, 118. (End)
REFERENCES
Rodolfo Kurchan, Mesmerizing Math Puzzles, Sterling Publications, (2000), problem 87, Square Stamps, p. 57.
LINKS
Rodolfo Kurchan, Problem 578, Puzzle Fun, December 1997
Rodolfo Kurchan, Solutions to problem 578, Puzzle Fun, December 1998
EXAMPLE
From Omar E. Pol, Jul 30 2020: (Start)
For n = 1 we draw a 1 X 1 square, so a(1) = 1.
For n = 2 we draw a 2 X 2 square and then a 1 X 1 square, both with a vertex at the point (0,0) as shown below:
_ _
|_ |
|_|_|
.
We can see only one 1 X 1 square in the arrangement, so a(2) = 1.
For n = 3 first we draw a 3 X 3 square with a vertex at the point (0,0) and then we draw a 2 X 2 square centered at the point (3,3) as shown below:
. _ _
_ _|_ |
| |_|_|
| |
|_ _ _|
.
Note that on the cell (3,3) we have formed a 1 X 1 square.
Finally we draw a 1 X 1 square on the cell (4,4) as shown below:
. _ _
_ _|_|_| <-- cell (4,4)
| |_|_|
| |
|_ _ _|.
Note that on the cells (3,4) and (4,3) we have formed two new 1 X 1 squares, hence the total number of 1 X 1 squares in the arrangement is equal to 4, so a(3) = 4. (End)
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Rodolfo Kurchan, Jul 28 2020
EXTENSIONS
a(7)-a(11) from Hermann Jurksch and Hugo Pfoertner, Aug 28 2020
STATUS
approved