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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.
4
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
1,2
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).
REFERENCES
Ivan Yashchenko, Invitation to a Mathematical Festival, pp. 10 and 102, MSRI, Mathematical Circles Library, 2013.
FORMULA
a(n) = Sum_{(k>=2) & (A005917(k)|A344332(n))} tau(A344332(n)/A005917(k)).
EXAMPLE
---> 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).
PROG
(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);
CROSSREFS
KEYWORD
nonn
AUTHOR
Bernard Schott, Aug 09 2021
STATUS
approved