A344332 - OEIS

A344332

Side s of squares of type 2 that can be tiled with squares of two different sizes so that the number of large or small squares is the same.

%I #39 Jul 13 2021 01:37:39

%S 15,30,45,60,65,75,90,105,120,130,135,150,165,175,180,195,210,225,240,

%T 255,260,270,285,300,315,325,330,345,350,360,369,375,390,405,420,435,

%U 450,455,465,480,495,510,520,525,540,555,570,585,600,615,630,645,650,660,671,675,690,700,705,715,720,735,738,750,765,780,795,810,825,840,845,855,870,875,885,900

%N Side s of squares of type 2 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 second one is proposed here, while type 1 is described in A344331.

%C If m is a term and k > 1, k * m is another term.

%C Every term (primitive or not primitive) is the side of an elementary square of type 2 (see A346263).

%C Some notations: s = side of the tiled square, 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 2 must satisfy the Diophantine equation s^2 = z * (a^2+b^2) with the conditions a^2+b^2 = c^2 and gcd(a, b, c) = 1.

%C In this case, q = a/(c-b) must be odd, and side s = q*c = a*c/(c-b) = (a+b)*c/a with a number of squares z = q^2 = (a/(c-b))^2 = ((b+c)/a)^2.

%C Indeed, these conditions give exactly the following solutions for n >= 2: s = n^4-(n-1)^4 (A005917), a = 2*n-1 (A005408), b = 2*n*(n-1) (A046092), c = 2*n*(n-1)+1 (A001844), z = (2*n-1)^2 (A016754); this results come from the identity:

%C (n^4 - (n-1)^4)^2 = (2*n-1)^2 * ((2*n-1)^2 + (2*n*(n-1))^2).

%C For n >= 2, every primitive square is composed by a square S1 of z = (2*n-1)^2 large squares with side b = 2*n*(n-1), then an edge on two sides of this square S1 of z = (2*n-1)^2 small squares with side a = 2*n-1.

%C See example with design of square of side s = 15 with a = 3, b = 4, c = 5, q = 3, z = 9, obtained with n= 2.

%C -> Non-primitive squares

%C If s is the side of a primitive square of type 2, then every k * s, k > 1 is a non-primitive term that gives two distinct tilings of type 2.

%C The square ks X ks can be tiled with z = q^2 = (2n-1)^2 = (a/(c-b))^2 = ((b+c)/a)^2 squares of side ka and of side kb, but also,

%C The square ks X ks can be tiled with z = k^2*q^2 = ((2n-1)*k)^2 = (k*a/(c-b))^2 = (k*(b+c)/a)^2 squares of side a and of side b (see example).

%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 = 15:

%e a = 3, b = 4, c = 5, s = 15, z = 9; s = 15 is the side of primitive square, with z = 9 squares of size 3 x 3 and 9 squares of size 4 x 4

%e Non-primitive square k*s = 2*15 = 30:

%e a = 3, b = 4, c = 5, s = 30, z = 36, this square is obtained with 4 copies of the primitive square as below.

%e a = 6, b = 8, c = 10, s = 30, z = 9, this square and its tiling are exactly as the primitive square with scale 2.

%e b = 4 (or = 8) a = 3 (or = 6)

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%e s = 15 s = 30

%o (PARI) pts(lim) = my(v=List(), m2, s2, h2, h); for(middle=4, lim-1, m2=middle^2; for(small=1, middle, s2=small^2; if(issquare(h2=m2+s2, &h), if(h>lim, break); listput(v, [small, middle, h])))); vecsort(Vec(v)); \\ A009000

%o isdp4(s) = my(k=1, x); while(((x=k^4 - (k-1)^4) <= s), if (x == s, return (1)); k++); return(0);

%o isokp2(s) = {if (!isdp4(s), return(0)); if (s % 2, my(vp = pts(s)); for (i=1, #vp, my(vpi = vp[i], a = vpi[1], b = vpi[2], c = vpi[3]); if (a*c/(c-b) == s, return(1)); ); ); }

%o isok2(s) = {if (isokp2(s), return (1)); fordiv(s, d, if ((d>1) || (d<s), if (isokp2(s/d), return (1)))); } \\ _Michel Marcus_, Jun 04 2021

%Y Cf. A001844, A005408, A005917, A046092, A046754, A344330, A344331.

%K nonn

%O 1,1

%A _Bernard Schott_, May 20 2021