%I #23 Aug 26 2019 08:27:11

%S 0,0,0,0,1,1,1,3,3,6,6,6,10,10,15,15,21,21,21,28,28,36,36,36,45,45,55,

%T 55,66,66,66,78,78,91,91,91,105,105,120,120,120,136,136,153,153,171,

%U 171,171,190,190,210,210,210,231,231,253,253,276,276,276,300,300,325,325

%N a(n) is the number of different sizes of integer-sided rectangles which can be placed inside an n X n square and with length greater than n.

%C Conditions for rectangles L x W which have length L > n: n - L/sqrt(2) > W/sqrt(2) where L/sqrt(2) and W/sqrt(2) are projections on the n X n square's sides.

%C If a rectangle with sides k X m fits in an n X n square then it fits by putting a parallel (without loss of generality, the longest) on the diagonal of the n X n square. Only the sum k + m really matters in order to see if the k X m rectangle fits in the square; it fits if k + m < sqrt(2) * n. - _David A. Corneth_, Aug 24 2019

%F a(n) = binomial(floor(n * sqrt(2)) - n, 2). - _David A. Corneth_, Aug 24 2019

%F a(n) = A327141(n) - A000217(n).

%e For n = 1, 2, 3, 4 we cannot place rectangles with side length L > n.

%e For n = 5 we can place a 6 X 1 rectangle inside a 5 X 5 square, so a(5) = 1.

%e For n = 8 we can place 9 X 1, 9 X 2 and 10 X 1 rectangles inside an 8 X 8 square, so a(8) = 3.

%o (C++)

%o #include <iostream>

%o #include <cmath>

%o #include <vector>

%o using namespace std;

%o int main() {

%o int n;

%o cin>>n;

%o vector <int> v;

%o for (int i=1; i<=n; i++)

%o {int count=0;

%o // ii-length, jj-width of diagonal rectangle

%o for (int ii=i+1; ii<=int(sqrt(2)*i); ii++)

%o for (int jj=1; jj<=i; jj++)

%o if ((double(i)-double(ii)/sqrt(2))-double(jj)/sqrt(2)>0)

%o {count++;}

%o v.push_back(count);

%o }

%o for (int i=0; i<v.size(); i++)

%o cout << v[i] << ", ";

%o return 0;

%o }

%o (PARI) a(n) = {binomial(floor(n * sqrt(2)) - n, 2)} \\ _David A. Corneth_, Aug 24 2019

%Y Cf. A080754, A327141.

%K nonn,easy

%O 1,8

%A _Kirill Ustyantsev_, Aug 23 2019