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%I #50 Oct 05 2020 10:13:12
%S 1,1,4,7,12,17,24,31,42,50,65
%N a(n) is the maximal number of 1 X 1 squares in an arrangement of n squares, from 1 X 1 to n X n.
%C Terms were calculated by hand by the author. Terms should be verified.
%C On the infinite square grid we can draw every square anywhere, on any of the four quadrants or on several of them.
%C Note that distinct arrangements could give the same result for some values of n.
%C a(7) >= 24. - _Hugo Pfoertner_, based on data from Hermann Jurksch, Aug 02 2020
%C From _Hugo Pfoertner_, Aug 28 2020: (Start)
%C Hermann Jurksch (private communication) has provided a proof that a(n) <= floor((n^2 + n + 1)*(3/5)).
%C I myself conjecture that a(n) <= floor((n^2 - n)*(3/5)) for n >= 4.
%C Lower bounds based on numerical results for the terms a(12)-a(15) are 76, 88, 105, 118. (End)
%D Rodolfo Kurchan, Mesmerizing Math Puzzles, Sterling Publications, (2000), problem 87, Square Stamps, p. 57.
%H Rodolfo Kurchan, <a href="http://www.puzzlefun.online/puzzle-fun-18">Problem 578</a>, Puzzle Fun, December 1997
%H Rodolfo Kurchan, <a href="http://www.puzzlefun.online/puzzle-fun-20">Solutions to problem 578</a>, Puzzle Fun, December 1998
%e From _Omar E. Pol_, Jul 30 2020: (Start)
%e For n = 1 we draw a 1 X 1 square, so a(1) = 1.
%e 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:
%e _ _
%e |_ |
%e |_|_|
%e .
%e We can see only one 1 X 1 square in the arrangement, so a(2) = 1.
%e 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:
%e . _ _
%e _ _|_ |
%e | |_|_|
%e | |
%e |_ _ _|
%e .
%e Note that on the cell (3,3) we have formed a 1 X 1 square.
%e Finally we draw a 1 X 1 square on the cell (4,4) as shown below:
%e . _ _
%e _ _|_|_| <-- cell (4,4)
%e | |_|_|
%e | |
%e |_ _ _|.
%e 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)
%Y Cf. A336659 (another version), A337515 (with links to list of solutions and illustrations).
%K nonn,hard,more
%O 1,3
%A _Rodolfo Kurchan_, Jul 28 2020
%E a(7)-a(11) from Hermann Jurksch and _Hugo Pfoertner_, Aug 28 2020
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