cp's OEIS Frontend

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.

%I A251607 #22 Jan 21 2024 15:43:00
%S A251607 0,0,1,2,3,5,8,10,14,16,20,24,29,35,39,45,51,58,66,71,80,87,97,104,
%T A251607 115,124,133,145,154,166,178,189,202,213,228,241,254,269,283,297,315,
%U A251607 329,346,361,380,396,413,431,450,470,487,508,528,548,569,588,613,632,655,677,701
%N 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.
%C A251607 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).
%e A251607 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.
%t A251607 sumtriangles[c_] := (n = 0; Do[If[a^3+b^3<c^3&&a+b>c, n++], {b, 1, c}, {a, 1, b}]; n); Table[sumtriangles[m], {m, 1, 200}]
%o A251607 (GeoGebra)
%o A251607 n = Slider(1, 20, 1);
%o A251607 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));
%o A251607 p = 3;
%o A251607 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);
%o A251607 a_n = CountIf((x(A)^2+y(A)^2)^(p/2)+((n-x(A))^2+y(A)^2)^(p/2)<n^p,A,L);
%o A251607 # _Frank M Jackson_, Jan 02 2024
%Y A251607 Cf. A002620, A236384, A247588.
%K A251607 nonn
%O A251607 1,4
%A A251607 _Frank M Jackson_, Dec 05 2014