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 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. 9
 15, 30, 45, 60, 65, 75, 90, 105, 120, 130, 135, 150, 165, 175, 180, 195, 210, 225, 240, 255, 260, 270, 285, 300, 315, 325, 330, 345, 350, 360, 369, 375, 390, 405, 420, 435, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS 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). There are two types of solutions, the second one is proposed here, while type 1 is described in A344331. If m is a term and k > 1, k * m is another term. Every term (primitive or not primitive) is the side of an elementary square of type 2 (see A346263). 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. -> Primitive squares 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. 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. 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: (n^4 - (n-1)^4)^2 = (2*n-1)^2 * ((2*n-1)^2 + (2*n*(n-1))^2). 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. 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. -> Non-primitive squares 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. 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, 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). REFERENCES Ivan Yashchenko, Invitation to a Mathematical Festival, pp. 10 and 102, MSRI, Mathematical Circles Library, 2013. LINKS Table of n, a(n) for n=1..76. Index to sequences related to Olympiads. EXAMPLE Primitive square with s = 15: 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 Non-primitive square k*s = 2*15 = 30: a = 3, b = 4, c = 5, s = 30, z = 36, this square is obtained with 4 copies of the primitive square as below. a = 6, b = 8, c = 10, s = 30, z = 9, this square and its tiling are exactly as the primitive square with scale 2. b = 4 (or = 8) a = 3 (or = 6) ________ ________ ________ ______ ______________________________ | | | | | | | | | | | | | | | |______| | |_______ |________|________| | | | | | | | | | | | |______| | | | | | | | |________|________|________| | | | | | |______| | | | | | | | | | | | | | |_____ __|___ ____|_ ______|______| | | | | | | | | | | | | | | | |_____|______|______|______|______|______________________________| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |_________________________________|______________________________| s = 15 s = 30 PROG (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 isdp4(s) = my(k=1, x); while(((x=k^4 - (k-1)^4) <= s), if (x == s, return (1)); k++); return(0); 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)); ); ); } isok2(s) = {if (isokp2(s), return (1)); fordiv(s, d, if ((d>1) || (d

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