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
Keywords
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.
Programs
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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
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
Formula
Extensions
Corrected and extended by Michel Marcus, Dec 22 2021
Comments