A221836 Triangle in which m-th term of n-th row is the number of integer Heron triangles with two of the sides having lengths n, m.

0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0

Offset: 1

Author

Eric M. Schmidt, Jan 26 2013

Keywords

• nonn
• tabl

Comments

If the primes divisors of nm congruent to 1 mod 4 have multiplicities e_1, ..., e_r, then a(n, m) <= (3 + (-1)^(nm))/2 * (Product(2*e_j - 1, j = 1..r) - 1).

Examples

			Triangle begins
0;
0, 0;
0, 0, 0;
0, 0, 1, 0;
0, 0, 1, 1, 2;
0, 0, 0, 0, 1, 0;
0, 0, 0, 0, 0, 0, 0;
0, 0, 0, 0, 1, 1, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0, 0;
0, 0, 0, 0, 0, 1, 0, 1, 1, 2.
		

Links

• Sourav Sen Gupta, Nirupam Kar, Subhamoy Maitra, Santanu Sarkar, and Pantelimon Stanica, Counting Heron triangles with Constraints, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 13, Paper A3, 2013.
• Eric Weisstein's World of Mathematics, Heronian Triangle.

Programs

Sage

def A221836(n, m) :
    count = 0
    for k in range(abs(n-m)+1, n+m) :
        s = (n + m + k)/2
        Asq = s * (s-n) * (s-m) * (s-k)
        if Asq.is_integral() and Asq.is_square() : count += 1
    return count