A346264 a(n) is the number of distinct possible tilings of type 2 for squares with side = A344332(n) and that can be tiled with squares of two different sizes so that the numbers of large or small squares are equal.

1, 2, 2, 3, 1, 2, 4, 2, 4, 2, 3, 4, 2, 1, 6, 4, 4, 4, 5, 2, 3, 6, 2, 6, 4, 2, 4, 2, 2, 8, 1, 3, 8, 4, 6, 2, 8, 2, 2, 6, 4, 4, 4, 6, 9, 2, 4, 7, 8, 2, 8, 2, 4, 6, 1, 6, 4, 3, 2, 2, 10, 3, 2, 6, 4, 12, 2, 8, 4, 8, 2, 4, 4, 2, 2, 12, 4, 2, 4, 6, 7, 8, 8, 2, 6, 4, 5, 12, 2, 12, 2, 3, 3, 4

Offset: 1

Author

Bernard Schott, Aug 09 2021

Keywords

• nonn

Comments

Every side of square of type 2 in A344332 is also the side of an elementary square of type 2. An elementary square of type 2 is the smallest square that can be tiled with squares of two different sides a < b satisfying a^2+b^2 = c^2 and so that the numbers of small and large squares are equal.

Some notation: 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.

a(n) = 1 iff A344332(n) is a term of A005917 that is not a multiple of another term of A005917 (15, 65, 175, 369, 671, 2465, ...).

The first side that is a multiple of two primitive sides is 195 = 13*15 = 3*65 (see 3rd example).

Examples

			---> For a(1), A344332(1) = 15, then, with the formula, we get a(1) = tau(A344332(1)/A005917(2)) = tau(15/15) = tau(1) = 1, and the corresponding tiling of this smallest square 15 X 15 of type 2 consists of z = 9 squares whose sides (a,b) = (3,4) (see below).
            ________ ________ ________ ______
           |        |        |        |      |
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           |        |        |        |______|
           |_______ |________|________|      |
           |        |        |        |      |
           |        |        |        |______|
           |        |        |        |      |
           |________|________|________|      |
           |        |        |        |______|
           |        |        |        |      |
           |        |        |        |      |
           |_____ __|___ ____|_ ______|______|
           |     |      |      |      |      |
           |     |      |      |      |      |
           |_____|______|______|______|______|
                       a(1) = 1
---> For a(2), A344332(2) = 30, then, with the formula, we get a(2) = tau(A344332(2)/A005917(2)) = tau(30/15) = tau(2) = 2, and these 2 distinct tilings are:
1) 30 = 2*A344332(1) = 2*15, z(30) = 2^2 * z(15) = 4*9 = 36 and square 30 X 30 can be tiled with z = 36 squares whose sides (a,b) = (3,4), that is 4 copies of the elementary and primitive square 15 X 15 (as above). Also,
2) 30 = 1*A344332(2) = 1*30, z(30) = 1^2 * z(15) = 1*9 = 9 and the elementary square 30 X 30 can be tiled with z = 9 squares whose sides (a,b) = (6,8) (see link with corresponding drawings).
---> For a(16), A344332(16) = 195, then, with the formula, we get a(16) = tau(A344333(16)/A005917(2)) + tau(A344333(16)/A005917(3)) = tau(195/15) + tau(195/65) = tau(13) + tau(3) = 2+2 = 4, and these 4 distinct tilings are:
1) 195 = 13*A344332(1) = 13*15, z_1(195) = 13^2 * z(15) = 169*9 = 1521 and square 195 X 195 can be tiled with z = 1521 squares whose sides (a,b) = (3,4), that is 169 copies of the elementary and primitive square 15 X 15, as above;
2) 195 = 1*A344332(16) = 1*195, z_2(195) = 1^2 * z(195) = 1*9 = 9 and the elementary square 195 X 195 can be tiled with z = 9 squares whose sides (a,b) = (39,52);
3) 195 = 3*A344332(5) = 3*65, z_3(195) = 3^2 * z(65) = 9*25 = 225 [z(65) = A346263(5) = T(5,1) = 25] and square 195 X 195 can be tiled with z = 225 squares whose sides (a,b) = (5,12), that is 9 copies of the elementary and primitive square 65 X 65;
4) 195 = 1*A344332(16) = 1*195, z_4(195) = 1^2 * z(195) = 1*25 = 25 and the elementary square 195 X 195 can be tiled with z = 25 squares whose sides (a,b) = (15,36).
		

References

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

Links

• Bernard Schott, Tilings of type 2 for square 30 X 30 with a(2) = 2.

Crossrefs

Cf. A005917, A016754, A344332, A346263.

Programs

PARI

\\ using isok2 from A344332; isok7 is for A005917
isok7(k) = my(kk= sqrtnint(k\4, 3)+2); vecsearch(vector(kk, i, (i+1)^4 - i^4), k);
sd(x) = sumdiv(x, d, if (isok7(d), numdiv(x/d)));
lista(nn) = my(v2 = select(isok2, [1..nn])); apply(sd, v2);

Formula

a(n) = Sum_{(k>=2) & (A005917(k)|A344332(n))} tau(A344332(n)/A005917(k)).