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

1,15

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).

Links

Table of n, a(n) for n=1..87.

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.

Example

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.

Prog

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
end

Crossrefs

Sequence in context: A019141 A230686 A341001 * A279372 A086077 A178408

Adjacent sequences: A221833 A221834 A221835 * A221837 A221838 A221839

Keyword

nonn,tabl

Author

Eric M. Schmidt, Jan 26 2013

Status

approved