A331040 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.
1, 35, 3, 7, 3, 15, 8, 35, 55, 63, 95, 119, 135, 56, 231, 255, 80, 351, 455, 495, 855, 216, 224, 1071, 360
Offset: 1
Examples
b(1) = a(1)/A331041(1) = 1/12: Triangle (1,1,1) has the least possible radius of incircle = sqrt(1/12). 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. b(3) = a(3)/A331041(3) = 3/4: Triangles are (2,7,7), (3,3,3), and (3,5,7). 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). 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.
Links
- Hugo Pfoertner, Illustration of side lengths for terms a(13)-a(25).
Crossrefs
Programs
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PARI
\\ Only suitable for demonstration of initial terms rh2(a,b,c)={my(s=(a+b+c)/2);(s-a)*(s-b)*(s-c)/s}; lim_a(x)=ceil(4*(x^2+2)); lim_b(x)=ceil(4*(x^4+2*x^2+1)); target=35/4; v=vector(333938); n=0; 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))); 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));
Comments