

A344330


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


10



10, 15, 20, 30, 40, 45, 50, 60, 65, 68, 70, 75, 78, 80, 90, 100, 105, 110, 120, 130, 135, 136, 140, 150, 156, 160, 165, 170, 175, 180, 190, 195, 200, 204, 210, 220, 222, 225, 230, 234, 240, 250, 255, 260, 270, 272, 280, 285, 290, 300, 310, 312, 315, 320, 325, 330, 340, 345, 350, 360, 369, 370
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OFFSET

1,1


COMMENTS

This sequence is a generalization of the 4th problem proposed for the pupils in grade 6 during the 19th Mathematical Festival at Moscow in 2008.
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.
Side s of such tiled squares must satisfy the Diophantine equation s^2 = z * (a^2+b^2).
There are two types of solutions. See A344331 for type 1 and A344332 for type 2.
If q is a term, k * q is another term for k > 1.


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..62.
Index to sequences related to Olympiads.


EXAMPLE

> Example of type 1:
Square 10 x 10 with a = 1, b = 2, s = 10, z = 20.
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  _  _ with 10 elementary 2 x 5 rectangles
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> Example of type 2:
Square 15 x 15 with a = 3, b = 4, s = 15, z = 9.
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Remarks:
 With terms as 10, 20, ... we only obtain sides of squares of type 1:
10 is a term of this type because the square 10 X 10 only can be tiled with 20 squares of size 1 X 1 and 20 squares of size 2 X 2 (see first example),
20 is another term of this type because the square 20 X 20 only can be tiled with 80 squares of size 1 x 1 and 80 squares of size 2 x 2.
 With terms as 15, 65, ... we only obtain sides of squares of type 2:
15 is a term of this type because the square 15 X 15 only can be tiled with 9 squares of size 3 X 3 and 9 squares of size 4 X 4 (see second example),
65 is another term of this type because the square 65 X 65 only can be tiled with 25 squares of size 5 X 5 and 25 squares of size 12 X 12.
 With terms as 30, 60, ... we obtain both sides of squares of type 1 and of type 2:
30 is a term of type 1 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, but,
30 is also a term of type 2 because the square 30 X 30 can be tiled with 9 squares of size 6 X 6 and 9 squares of size 8 X 8.


PROG

(PARI) pts(lim) = my(v=List(), m2, s2, h2, h); for(middle=4, lim1, 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  (k1)^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/(cb) == s, return(1)); ); ); }
isok2(s) = {if (isokp2(s), return (1)); fordiv(s, d, if ((d>1)  (d<s), if (isokp2(s/d), return (1)))); }
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); }
isok1(s) = {if (isokp1(s), return (1)); fordiv(s, d, if ((d>1)  (d<s), if (isokp1(s/d), return (1)))); }
isok(s) = isok1(s)  isok2(s); \\ Michel Marcus, Jun 04 2021


CROSSREFS

Subsequences: A008592 \ {0}, A008597 \ {0}, A034262 \ {0,1}.
Cf. A344331, A344332, A344333, A344334.
Sequence in context: A035166 A129495 A101258 * A091418 A055986 A045161
Adjacent sequences: A344327 A344328 A344329 * A344331 A344332 A344333


KEYWORD

nonn


AUTHOR

Bernard Schott, May 15 2021


EXTENSIONS

Corrected by Michel Marcus, May 18 2021
Incorrect term 145 removed by Michel Marcus, Jun 04 2021


STATUS

approved



