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%I #21 Sep 15 2021 00:27:19
%S 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,
%T 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,
%U 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
%N 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.
%C 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.
%H Hein van Winkel, <a href="http://heinvanwinkel.nl/wi/pdfs/Heron%20-%20Bhaskara%20-%20I.pdf">Heron triangles</a>, See the sections SQO and SQE.
%F 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.
%F Further let A = (2a_1 + 1) * (2a_2 + 1) *...* (2a_r + 1) and B = A * (2b_1 + 1) * (2b_2 + 1) *...* (2b_s + 1).
%F 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.
%F a(n) = A338939(n) * A115878(n).
%e 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).
%Y Cf. A115878, A338939.
%K nonn
%O 1,10
%A _Hein van Winkel_, Nov 16 2020
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