%I #39 Mar 30 2023 09:16:03
%S 3,11,195,35,51,627,91,115,51,19,203,2139,11,35,3219,403,451,4515,555,
%T 611,123,731,795,7755,19,1003,9699,1155,1235,11859,1403,1491,14235,67,
%U 1771,16827,219,83,19635,2291,267,22659,2635,2755,25899,3003,3131,29355,3395,3531,33027,3811,3955,36915,4251
%N Squarefree part of numerator of the squared area of the Heronian triangle with sequential odd sides whose shortest leg is 2*n+1.
%C These are the values within the irreducible square roots, which are part of the area resulting from Heronian triangles, which have all sides being sequential odd integers (starting with the second odd number, since the triangle {1,3,5} has no real area). The triangles follow this sequence of sides: {2*n+1, 2*n+3, 2*n+5} and their area is represented by {(r/s)*sqrt(t)}, where r, s, t are integers and a(n) is the number t.
%H Claudio Chaib, <a href="https://community.wolfram.com/groups/-/m/t/1963471">Squared area (squarefree): Triangles (OEIS-A334176 odd/A334177 prime)</a>, Wolfram Community.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Heron%27s_formula">Heron's formula</a>
%F a(n) = A007913(3*(2*n+7)*(2*n-1)). - _Bernard Schott_, Apr 18 2020
%e a(1) = 3 because the first possible Heronian triangle with its sequential odd integer sides is the triangle {3,5,7} and its respective area is {15*sqrt(3)/4}. a(2) = 11 since the second possible triangle is {5,7,9} which has area {21*sqrt(11)/4}. And so on.
%t a[n_]:=Module[{y,z},z=Area@SSSTriangle[2*n+1,2*n+3,2*n+5];({z}/.Coefficient[{z}/.Sqrt[_]->y,y][[1]]->1)[[1]]^2]
%o (PARI) a(n) = core((3*(2*n+7)*(2*n-1))); \\ _Michel Marcus_, Apr 18 2020
%Y Cf. A007913 (squarefree part), A334177.
%K nonn
%O 1,1
%A _Claudio Lobo Chaib Filho_, Apr 17 2020