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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
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
Sequence in context: A051908 A056614 A126309 * A048838 A181872 A239264
KEYWORD
nonn
AUTHOR
Hein van Winkel, Nov 16 2020
STATUS
approved

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Last modified March 28 20:05 EDT 2024. Contains 371254 sequences. (Running on oeis4.)