%I #50 Dec 23 2021 12:50:39

%S 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,

%T 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,

%U 10,2,14,4,2,4,8,6,6,2,20,2,3,4,4,10,7,6,12,2,2,10,6

%N 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.

%C 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.

%C 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.

%C 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, ...).

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

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

%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.

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

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

%e 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).

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%e a(1) = 1

%e 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:

%e -> 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.

%e -> 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.

%e -> 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).

%o (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); }

%o isok1(s) = {if (isokp1(s), return (1)); fordiv(s, d, if ((d>1) || (d<s), if (isokp1(s/d), return (1)))); } \\ A344331

%o 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

%o sd(x) = sumdiv(x, d, if (isok3(d), numdiv(x/d)));

%o lista(nn) = my(v1 = select(isok1, [1..nn])); apply(sd, v1); \\ _Michel Marcus_, Dec 22 2021

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

%K nonn

%O 1,2

%A _Bernard Schott_, Jun 14 2021

%E Corrected and extended by _Michel Marcus_, Dec 22 2021