A344331 Side s of squares of type 1 that can be tiled with squares of two different sizes so that the number of large or small squares is the same.
10, 20, 30, 40, 50, 60, 68, 70, 78, 80, 90, 100, 110, 120, 130, 136, 140, 150, 156, 160, 170, 180, 190, 200, 204, 210, 220, 222, 230, 234, 240, 250, 260, 270, 272, 280, 290, 300, 310, 312, 320, 330, 340, 350, 360, 370, 380, 390, 400, 408, 410, 420, 430, 440, 444, 450, 460, 468, 470
Offset: 1
Keywords
Examples
Primitive square with s = 10: a = 1, b = 2, s = 10, m = 10, z = 20, and Non-primitive square with s = 20: a = 1, b = 2, s = 20, m = 40, z = 80. ___ ___ _ ___ ___ _ ___________________ | | |_| | |_| | |___|___|_|___|___|_| | | | |_| | |_| | |___|___|_|___|___|_| | | | |_| | |_| | |___|___|_|___|___|_| | | | |_| | |_| | |___|___|_|___|___|_| | | | |_| | |_| | |___|___|_|___|___|_|___________________| | | | | | | | | | | | | | | | | | | | | | | | | | | | |___________________|___________________| with respectively m = 10 (and m = 40) elementary 2 X 5 rectangles as below: ___ ___ _ | | |_| |___|___|_| There are these three possibilities: - 10 is a primitive term because the square 10 X 10 can be tiled with 20 squares of size 1 X 1 and 20 squares of size 2 X 2, and no smaller square can be tiled with a same number of squares of size 1 X 1 and of squares of size 2 X 2. - 20 is a non-primitive term because the square 20 X 20 can be tiled with 80 squares of size 1 X 1 and 80 squares of size 2 X 2. - 30 is a primitive term because the square 30 X 30 can be tiled with 90 squares of size 1 X 1 and 90 squares of size 3 X 3, and no smaller square can be tiled with a same number of squares of size 1 X 1 and of squares of size 3 X 3, but also, 30 is a non-primitive term because the square 30 X 30 can be tiled with 180 squares of size 1 X 1 and 180 squares of size 2 X 2.
References
- Ivan Yashchenko, Invitation to a Mathematical Festival, pp. 10 and 102, MSRI, Mathematical Circles Library, 2013.
Programs
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PARI
isokp1(s) = {if (!(s % 2) && ispower(s/2, 4), return (0)); my(m = sqrtnint(s, 3)); vecsearch(setbinop((x, y)->x*y*(x^2+y^2), [1..m]), s); } isok(s) = {if (isokp1(s), return (1)); fordiv(s, d, if ((d>1) || (d
Michel Marcus, Dec 22 2021
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