A374020 Number of solutions to a^2 + d^2 = b^2 + c^2 with 1 <= a < b < c < d <= n.

0, 0, 0, 0, 0, 0, 0, 1, 2, 2, 4, 5, 7, 9, 10, 12, 15, 18, 23, 26, 29, 33, 39, 43, 48, 54, 60, 65, 74, 79, 87, 96, 105, 114, 122, 129, 140, 151, 162, 171, 185, 194, 210, 223, 233, 247, 264, 277, 293, 308, 323, 338, 360, 376, 392, 407, 425, 444, 468, 484

Offset: 1

Author

Thorsten Ehlers, Jun 25 2024

Keywords

• nonn

Examples

			For n = 9 the a(9) = 2 solutions are 1^2 + 8^2 = 4^2 + 7^2 = 65 and 2^2 + 9^2 = 6^2 + 7^2 = 85.
For n = 18 three of the a(18) = 18 solutions sum up to 325: 1^2 + 18^2 = 6^2 + 17^2 = 10^2 + 15^2.
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..1000
• Quora, Can the sum of two squares a^2+b^2 be equal to the sum of two other squares c^2+d^2 where a,b,c,d are distinct positive integers? If so, how do we calculate them?

Programs

C

#include 
#define N    10000ULL
typedef unsigned long long ull_t;
ull_t Sums[2 * N * N];
int main() {
    ull_t sol = 0;
    for (ull_t i = 1; i < N; i++)
        for (ull_t j = i + 1; j <= N; j++)
            sol += Sums[i * i + j * j]++;
    printf("%llu \n", sol);
}

Python

from itertools import count, islice
from collections import Counter
def A374020_gen(): # generator of terms
    c, s = 0, Counter()
    for n in count(1):
        n2 = n**2
        for i in range(1,n):
            c += s[m:=i**2+n2]
            s[m] += 1
        yield c
A374020_list = list(islice(A374020_gen(),20)) # Chai Wah Wu, Jul 18 2024