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A372917
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a(n) is the number of distinct rectangles with area n whose vertices lie on points of a unit square grid.
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1
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1, 2, 1, 3, 2, 3, 1, 4, 2, 5, 1, 5, 2, 3, 3, 5, 2, 5, 1, 8, 2, 3, 1, 7, 3, 5, 2, 5, 2, 9, 1, 6, 2, 5, 3, 8, 2, 3, 3, 11, 2, 6, 1, 5, 5, 3, 1, 9, 2, 8, 3, 8, 2, 6, 3, 7, 2, 5, 1, 15, 2, 3, 3, 7, 5, 6, 1, 8, 2, 9, 1, 11, 2, 5, 5, 5, 2, 9, 1, 14, 3, 5, 1, 10, 5, 3
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OFFSET
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1,2
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COMMENTS
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A rectangle in the square unit grid has the sides W = w*sqrt(r) and H = h*sqrt(r). The area is therefore n = w*h*r. Let r be a squarefree divisor of n that can be written as the sum of two squares x^2 + y^2. The number of distinct rectangles is then the sum of the number of ways for each value of r to decompose n/r into two factors w and h (with w >= h).
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LINKS
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FORMULA
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a(n) = ceiling(Product_{i=1..omega(n)}(k[i]*e[i] + 1)/2), with k[i] = 2 if p[i] mod 4 = 3 and k[i] = 1 else, where p[i]^e[i] is the prime factorization of n.
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EXAMPLE
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See also the linked illustrations of the terms a(4) = 3, a(8) = 4, a(15) = 3.
n = 4 has the three divisors 1, 2, 4. Since 4 is not squarefree, r can have the values 1 or 2. For r = 1 = 1^2 + 0^2 there are two rectangles (2,2), (4,1). For r = 2 = 1^2 + 1^2 and n/r = 4/2 = 2 = w*h there is the rectangle (2*sqrt(2), 1*sqrt(2)). That's a total of a(4) = 3 distinct rectangles.
n = 8 has the four divisors 1, 2, 4, 8. Since 4 and 8 are not squarefree, r can have the values 1 or 2. For r = 1 = 1^2 + 0^2 there are two rectangles (4,2), (8,1). For r = 2 = 1^2 + 1^2 and n/r = 8/2 = 4 = w*h there are the rectangles (4*sqrt(2), 1*sqrt(2)) and (2*sqrt(2), 2*sqrt(2)). That's a total of a(8) = 4 distinct rectangles.
n = 15 has the four divisors 1, 3, 5, 15. They are all squarefree, but 3 and 15 cannot be written as a sum of two squares, r can only have the values 1 or 5. For r = 1 = 1^2 + 0^2 there are two rectangles (5,3), (15,1). For r = 5 = 2^2 + 1^2 and n/r = 15/5 = 3 = w*h there is the rectangles (3*sqrt(5), 1*sqrt(5)). That's a total of a(15) = 3 distinct rectangles.
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MAPLE
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local f, i, prod;
f:=ifactors(n)[2];
prod:=1;
for i from 1 to numelems(f) do
if f[i][1] mod 4 = 3 then
prod:=prod*(1*f[i][2]+1);
else
prod:=prod*(2*f[i][2]+1);
end if;
end do;
return round(prod/2);
end proc;
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PROG
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(PARI) a(n) = my(f=factor(n)); prod(i=1, #f[, 1], if(f[i, 1]%4==3, 1, 2)*f[i, 2] + 1) \/ 2; \\ Kevin Ryde, Jun 09 2024
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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