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%I #10 Jul 29 2020 13:00:43
%S 1,35,3,7,3,15,8,35,55,63,95,119,135,56,231,255,80,351,455,495,855,
%T 216,224,1071,360
%N Numerator of squared radius of inscribed circle of a triangle with integer sides i <= j <= k, such that the number of triangles with this radius sets a new record. Denominators are A331041.
%C The radius rho of the inscribed circle of a triangle (a,b,c) is rho = sqrt((s-a)*(s-b)*(s-c)/s), with s=(a+b+c)/2. For given integer values of a <= b and a rational target value r2 of the squared incircle radius, c is given by the two positive real roots of the polynomial P(a,b,x,r2) = x^3 - x^2 * (a+b) + x * (4*r2-(b-a)^2) + (a+b)^3 + 4*(a+b)*(r2-a*b). P(a,b,x,r2) = 0 may have 0, 1 or 2 positive integer solutions.
%C The potential ranges of the side lengths of the triangles can be determined in analogy to the ranges for the case of integer radii of the incircles, see A120062 for the relevant formulas and sequences.
%H Hugo Pfoertner, <a href="/A331040/a331040.pdf">Illustration of side lengths for terms a(13)-a(25)</a>.
%e b(1) = a(1)/A331041(1) = 1/12: Triangle (1,1,1) has the least possible radius of incircle = sqrt(1/12).
%e b(2) = a(2)/A331041(2) = 35/52: Triangles (2,18,19) and (3,4,6) are the first occurrence of more than one triangle with the same radius of their incircles. rho = sqrt(35/52) in both cases.
%e b(3) = a(3)/A331041(3) = 3/4: Triangles are (2,7,7), (3,3,3), and (3,5,7).
%e b(4) = a(4)/A331041(4) = 7/4: (3,12,12), (3,22,23), (4,5,6), (5,18,22), (6,11,16) are the A331043(4) = 5 triangles with rho^2 = b(4).
%e b(15) = 231/4 includes the rare case, where two distinct integer solutions for the same pair of sides a and b exist: (20,37,38) and (20,37,39), both with rho^2=231/4 and thus contributing 2 of the A331043(15)=84 triangles with this squared radius of the incircle.
%o (PARI) \\ Only suitable for demonstration of initial terms
%o rh2(a,b,c)={my(s=(a+b+c)/2);(s-a)*(s-b)*(s-c)/s};
%o lim_a(x)=ceil(4*(x^2+2));
%o lim_b(x)=ceil(4*(x^4+2*x^2+1));
%o target=35/4; v=vector(333938); n=0;
%o for(a=1,lim_a(sqrt(target)), for(b=a,lim_b(sqrt(target)), for(c=b,a+b-1, f=rh2(a,b,c);v[n++]=f)));
%o v=vecsort(v); print("A331040 A331041 A331043"); print(numerator(v[1])," ",denominator(v[1])," ",1); m=0; mm=0; for(k=2,#v, if(v[k]>target,break); if(v[k]==v[k-1], m++; if(m>mm&&v[k+1]>v[k], print(numerator(v[k])," ",denominator(v[k])," ",m);mm=m),m=1));
%Y Cf. A057721, A082044, A120062, A120572, A331012.
%Y Cf. A331041 (corresponding denominators), A331042 (floor(4*a(n)/A331042(n)), A331043 (records of numbers of triangles).
%K nonn,frac,more
%O 1,2
%A _Hugo Pfoertner_, Jan 11 2020
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