A344331 - OEIS

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.

%I #44 Dec 23 2021 05:31:49

%S 10,20,30,40,50,60,68,70,78,80,90,100,110,120,130,136,140,150,156,160,

%T 170,180,190,200,204,210,220,222,230,234,240,250,260,270,272,280,290,

%U 300,310,312,320,330,340,350,360,370,380,390,400,408,410,420,430,440,444,450,460,468,470

%N 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.

%C This sequence is relative to the generalization of the 4th problem proposed for the pupils in grade 6 during the 19th Mathematical Festival at Moscow in 2008 (see A344330).

%C There are two types of solutions, the first one is proposed here, while type 2 is described in A344332.

%C Some notations: s = side of the tiled squares, a = side of small squares, b = side of large squares, and z = number of small squares = number of large squares.

%C -> Primitive squares

%C Side s of primitive squares of type 1 must satisfy the Diophantine equation s^2 = z * (a^2+b^2), with gcd(a, b) = 1, and without using the conditions a^2+b^2 = c^2, when a and b belong to a Pythagorean triple (a, b, c).

%C In this case, the sides of the primitive squares of type 1 are s = a*b * (a^2+b^2) with 1 <= a < b and gcd(a, b) = 1 (A344333), then corresponding z = (a*b)^2 * (a^2+b^2) (A344334).

%C Every primitive square is composed of m = a*b * (a^2+b^2) elementary rectangles of length L = a^2+b^2 and width W = a*b, so with an area A = a*b * (a^2+b^2) = m.

%C In particular: for a = 1, b = n, s = n*(n^2+1) form the subsequence A034262 \ {0, 1} and z = n^2*(n^2+1) form the subsequence A071253 \ {0, 2}).

%C See example with design for a square of side s = 10 with a = 1, b = 2, m = 10, z = 20.

%C -> Non-primitive squares

%C If s is the side of a primitive square of type 1 with z squares of side a and z squares of side b, then every k * s is a non-primitive term that gives one or two distinct tilings of type 1, depending of value of k:

%C - For every k > 1, the square ks X ks can be tiled with k^2*z squares of side a and k^2*z squares of side b (see example).

%C - For every k = r^4, r>1, the square ks X ks also can be tiled with z squares of side ka and z squares of side kb.

%C ---> Consequences:

%C 1) For every pair (a, b), 1 <= a < b, there is a square of side s = a*b * (a^2+b^2) that can be tiled with squares of side a and side b so that the number z of squares of side a and side b is the same, this number z = (a*b)^2 * (a^2+b^2).

%C 2) If q is a term and K > 1, K * q is another term.

%C 3) Every term is even.

%D Ivan Yashchenko, Invitation to a Mathematical Festival, pp. 10 and 102, MSRI, Mathematical Circles Library, 2013.

%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.

%e Primitive square with s = 10:

%e a = 1, b = 2, s = 10, m = 10, z = 20, and

%e Non-primitive square with s = 20:

%e a = 1, b = 2, s = 20, m = 40, z = 80.

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%e with respectively m = 10 (and m = 40) elementary 2 X 5 rectangles as below:

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%e There are these three possibilities:

%e - 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.

%e - 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.

%e - 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,

%e 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.

%o (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); }

%o isok(s) = {if (isokp1(s), return (1)); fordiv(s, d, if ((d>1) || (d<s), if (isokp1(s/d), return (1)))); } \\ _Michel Marcus_, Dec 22 2021

%Y Cf. A344330, A344332, A344333, A344334.

%Y Cf. A034262, A071253.

%K nonn

%O 1,1

%A _Bernard Schott_, May 20 2021