A338940 a(n) is the number of solutions to the Diophantine equation p * x * (x + n) = y^2 with p = a*b a perfect square and a+b = n.

0, 0, 0, 0, 1, 1, 0, 1, 0, 3, 0, 1, 1, 1, 4, 2, 1, 2, 0, 3, 0, 1, 0, 4, 4, 3, 0, 1, 1, 12, 0, 3, 0, 3, 4, 2, 1, 1, 4, 12, 1, 4, 0, 1, 7, 1, 0, 7, 0, 10, 4, 3, 1, 3, 4, 4, 0, 3, 0, 12, 1, 1, 0, 4, 16, 4, 0, 3, 0, 12, 0, 7, 1, 3, 14, 1, 0, 12, 0, 21, 0, 3, 0, 4, 16, 1, 4, 4, 1, 21, 4, 1, 0, 1, 4, 10, 1, 2, 0, 10

Offset

1,10

Comments

Related to Heron triangles with a partition point on a side of length n where the incircle is tangent. Some partitions correspond to a finite number of Heron triangles. The numbers a(n) in this sequence are the numbers of Heron triangles that match these 'finite' partitions.

Links

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Hein van Winkel, Heron triangles, See the sections SQO and SQE.

Formula

Let n = 2^t * p_1^a_1 * p_2^a_2 *...* p_r^a_r * q_1^b_1 * q_2^b_2 *...* q_s^b_s with t>=0, a_i>=0 for i=1..r, where p_i == 1 (mod 4) for i=1..r and q_j = -1 (mod 4) for j=1..s.

Further let A = (2a_1 + 1) * (2a_2 + 1) *...* (2a_r + 1) and B = A * (2b_1 + 1) * (2b_2 + 1) *...* (2b_s + 1).

Then a(n) = (A-1) * (B-1) / 4 for t = 0 and a(n) = A * (B-1) / 2 for t = 1 AND t = 2 and a(n) = (2*t - 3) * A * (B+1) / 2 for t > 2.

a(n) = A338939(n) * A115878(n).

Example

n = 25 = 5 + 20 = 9 + 16 gives 100 * x * (x + 25) = y^2 or 144 * x * (x + 25) = y^2 or 144 * x * (x + 25) = y^2. And the solutions are (x,y) = (144,1560) or (20,300) or (144,1872) or (20,360).

Crossrefs

Cf. A115878, A338939.

Sequence in context: A051908 A056614 A126309 * A048838 A181872 A239264

Adjacent sequences: A338937 A338938 A338939 * A338941 A338942 A338943

Keyword

nonn

Author

Hein van Winkel, Nov 16 2020

Status

approved