A345287 a(n) is the number of distinct possible tilings of type 1 for squares with side = A344331(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, 3, 3, 2, 6, 1, 2, 1, 4, 5, 4, 2, 9, 3, 2, 4, 6, 2, 5, 2, 10, 2, 6, 2, 6, 4, 1, 2, 2, 12, 3, 6, 7, 3, 6, 3, 13, 2, 3, 6, 6, 6, 5, 15, 2, 4, 10, 8, 4, 2, 12, 2, 6, 2, 10, 4, 4, 2, 2, 15, 3, 6, 7, 10, 2, 14, 4, 2, 4, 8, 6, 6, 2, 20, 2, 3, 4, 4, 10, 7, 6, 12, 2, 2, 10, 6

Offset: 1

Author

Bernard Schott, Jun 14 2021

Keywords

• nonn

Comments

Every side of square of type 1 in A344331 is also the side of an elementary square of type 1. An elementary square of type 1 is the smallest square that can be tiled with squares of two different sides a < b 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 A344331(n) is a term of A344333 that is not a multiple of another term of A344333 (10, 68, 78, 222, ...).

The first side that is a multiple of two primitive sides is 30 = 3*10 = 1*30 (see 2nd example).

Examples

			For a(1), A344331(1) = 10, then, with the formula, we get a(1) = tau(A344331(1)/A344333(1)) = tau(10/10) = tau(1) = 1. This unique corresponding tiling of this square 10 x 10 of type 1 with side s = 10 consists of z = 20 squares whose sides (a,b) = (1,2) (see below).
          ___ ___ _ ___ ___ _
         |   |   |_|   |   |_|
         |___|___|_|___|___|_|
         |   |   |_|   |   |_|
         |___|___|_|___|___|_|
         |   |   |_|   |   |_|
         |___|___|_|___|___|_|
         |   |   |_|   |   |_|
         |___|___|_|___|___|_|
         |   |   |_|   |   |_|
         |___|___|_|___|___|_|
                a(1) = 1
For a(3), A344331(3) = 30, then, with the formula, we get a(3) = tau(A344331(3)/A344333(1)) + tau(A344331(3)/A344333(2)) = tau(30/10) + tau(30/30) = tau(3) + tau(1) = 3. The 3 corresponding tilings of the square 30 x 30 of type 1 with side s = 30 consists of:
-> from 30 = 3*A344333(1) = 3*10, square with side s = 30 can be tiled with z = 180 squares with sides (a,b) = (1,2), indeed with 9 copies of primitive square 10 x 10, as above.
-> from 30 = 1*A344331(3) = 1*30, square with side s = 30 can be tiled with z = 20 squares with sides (a,b) = (3,6), indeed, it is the above square with scale 3.
-> from 30 = 1*A344331(3) = 1*30, square with side s = 30 can also be tiled with z = 90 squares with sides (a,b) = (1,3), indeed that is primitive square 30 x 30 with squares (a,b) = (1,3).
		

References

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

Links

• Index to sequences related to Olympiads.

Crossrefs

Cf. A344330, A344331, A344333, A344334, A345285, A345286, A346264.

Programs

PARI

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) || (dA344331
isok3(s) = {if (!(s % 2) && ispower(s/2, 4), return (0)); my(m = sqrtnint(s, 3)); vecsearch(setbinop((x, y)->if (gcd(x,y)==1, x*y*(x^2+y^2), 0), [1..m]), s); } \\ A344333
sd(x) = sumdiv(x, d, if (isok3(d), numdiv(x/d)));
lista(nn) = my(v1 = select(isok1, [1..nn])); apply(sd, v1); \\ Michel Marcus, Dec 22 2021

Formula

a(n) = Sum_{(k=1..n) & (A344333(k)|A344331(n))} tau(A344331(n)/A344333(k)).

a(n) = Sum_{(d | A344331(n)) & (d in A344333)} tau(A344331(n)/d) where tau is A000005. - Michel Marcus, Dec 22 2021

Extensions

Corrected and extended by Michel Marcus, Dec 22 2021