A251607 Number of non-congruent integer triangles ABC with largest side BC of length n whose opposite vertex A lies inside the closed boundary formed by BC and a locus of points whose real distance from B is x, from C is y and x^3 + y^3 = n^3.

0, 0, 1, 2, 3, 5, 8, 10, 14, 16, 20, 24, 29, 35, 39, 45, 51, 58, 66, 71, 80, 87, 97, 104, 115, 124, 133, 145, 154, 166, 178, 189, 202, 213, 228, 241, 254, 269, 283, 297, 315, 329, 346, 361, 380, 396, 413, 431, 450, 470, 487, 508, 528, 548, 569, 588, 613, 632, 655, 677, 701

Offset: 1

Author

Frank M Jackson, Dec 05 2014

Keywords

• nonn

Comments

Acute integer triangles ABC with longest side BC of length n (A247588) are segregated from obtuse or right integer triangles with the same longest side BC (A236384) by the closed boundary formed by a semicircle and BC as its diameter. The right integer triangles will lie on this boundary and the obtuse integer triangles within this boundary. Define a closed boundary S(q) that is formed by BC and a locus of points whose real distance from B is x, from C is y and x^q + y^q = n^q for integer q > 0. Then S(2) is that closed boundary formed by a semicircle with BC as diameter. Euler proved that there are no integer triangles that lie on S(3) and Wiles for all S(q) where q > 2. This sequence identifies all integer triangles with longest side BC of length n that lie inside S(3).

Examples

			a(5)=3 as there are 3 non-congruent integer triangles with base length of 5 whose apex lies inside S(3). The integer triples are (3,3,5), (2,4,5), (3,4,5). The other triangles from the complete set of non-congruent integer triangles with longest side length 5 (A002620(5+1)) are (4,4,5), (1,5,5), (2,5,5), (3,5,5), (4,5,5), (5,5,5) and lie outside the closed boundary.
		

Crossrefs

Cf. A002620, A236384, A247588.

Programs

GeoGebra

n = Slider(1, 20, 1);
L = Flatten(Sequence(Sequence(((a^2+n^2-(n-a+k)^2)/(2n),((a+(n-a+k)+n)(a+(n-a+k)-n)(a-(n-a+k)+n)(-a+(n-a+k)+n))^(1/2)/(2n)),a,k,(n+k)/2),k,1,n));
p = 3;
C = Curve((a^2+n^2-(n^p-a^p)^(2/p))/(2n),((a+(n^p-a^p)^(1/p)+n)(a+(n^p-a^p)^(1/p)-n)(a-(n^p-a^p)^(1/p)+n)(-a+(n^p-a^p)^(1/p)+n))^(1/2)/(2n),a,0,n);
a_n = CountIf((x(A)^2+y(A)^2)^(p/2)+((n-x(A))^2+y(A)^2)^(p/2)Frank M Jackson, Jan 02 2024

Mathematica

sumtriangles[c_] := (n = 0; Do[If[a^3+b^3c, n++], {b, 1, c}, {a, 1, b}]; n); Table[sumtriangles[m], {m, 1, 200}]