

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.


1



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
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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

Table of n, a(n) for n=1..100.
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) = (A1) * (B1) / 4 for t = 0 and a(n) = A * (B1) / 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,changed


AUTHOR

Hein van Winkel, Nov 16 2020


STATUS

approved



