|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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
|
|
|
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.
|
|
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
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|