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

1,2

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).

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..87.

Index to sequences related to Olympiads.

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

Example

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).

Prog

(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) || (d<s), if (isokp1(s/d), return (1)))); } \\ A344331
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

Crossrefs

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

Sequence in context: A091813 A238114 A291300 * A023139 A328484 A319442

Adjacent sequences: A345284 A345285 A345286 * A345288 A345289 A345290

Keyword

nonn

Author

Bernard Schott, Jun 14 2021

Extensions

Corrected and extended by Michel Marcus, Dec 22 2021

Status

approved