%I #28 Jun 17 2024 15:25:58

%S 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,

%T 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,

%U 8,2,9,1,11,2,5,5,5,2,9,1,14,3,5,1,10,5,3

%N a(n) is the number of distinct rectangles with area n whose vertices lie on points of a unit square grid.

%C 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).

%H Felix Huber, <a href="/A372917/b372917.txt">Table of n, a(n) for n = 1..10000</a>

%H Felix Huber, <a href="/A372917/a372917.pdf">Illustrations of terms a(4), a(8), a(15)</a>

%F 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.

%e See also the linked illustrations of the terms a(4) = 3, a(8) = 4, a(15) = 3.

%e 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.

%e 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.

%e 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.

%p A372917:= proc(n)

%p local f,i,prod;

%p f:=ifactors(n)[2];

%p prod:=1;

%p for i from 1 to numelems(f) do

%p if f[i][1] mod 4 = 3 then

%p prod:=prod*(1*f[i][2]+1);

%p else

%p prod:=prod*(2*f[i][2]+1);

%p end if;

%p end do;

%p return round(prod/2);

%p end proc;

%p seq(A372917(n),n=1..86);

%o (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

%Y Cf. A001481, A005117, A038548, A085582, A113751, A122225, A285956, A330805, A354704, A372915.

%K nonn

%O 1,2

%A _Felix Huber_, Jun 08 2024