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A334763 Ceiling of circumradius of triangle whose sides are consecutive Ulam numbers (A002858).

%I A334763 #10 Jun 13 2020 01:45:05
%S A334763 3,4,5,6,7,9,10,14,15,19,21,24,26,29,31,34,37,40,43,45,48,52,55,58,60,
%T A334763 63,68,72,77,80,84,87,93,99,103,104,107,110,115,118,123,126,131,134,
%U A334763 138,139,142,146,149,153,158,168,176,182,185,190,194,200,204,208
%N A334763 Ceiling of circumradius of triangle whose sides are consecutive Ulam numbers (A002858).
%C A334763 It has been proved that three consecutive Ulam numbers U(n) for n > 1 satisfy the triangle inequality. See Wikipedia link below. Consequently it is possible to create n-gons using n consecutive Ulam numbers. The sequence starts at offset 2 because using the first Ulam number generates a triangle with sides (1,2,3) that is degenerate with infinite circumradius.
%C A334763 Conjecture: Triangles whose sides are consecutive Ulam numbers are acute apart from (1,2,3), (2,3,4), (3,4,6), (4,6,8), (6,8,11) and (16,18,26).
%H A334763 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Circumradius.html">Circumradius</a>.
%H A334763 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/UlamSequence.html">Ulam Sequence</a>.
%H A334763 Wikipedia, <a href="https://en.wikipedia.org/wiki/Ulam_number">Ulam number</a>.
%F A334763 Circumradius of a triangle with sides a, b, c is given by R = a*b*c/(4A) where the Area A is given by Heron's formula A = sqrt(s(s-a)(s-b)(s-c)) and where s = (a+b+c)/2.
%e A334763 a(2)=3 because a triangle with sides 2,3,4 has area = (1/4)*sqrt((2+3+4)(2+3-4)(2-3+4)(-2+3+4)) = 2.904... and circumradius = 2*3*4/(4A) = 2.065...
%t A334763 lst1=ReadList["https://oeis.org/A002858/b002858.txt", {Number, Number}]; lst={}; Do[{a, b, c}={lst1[[n]][[2]], lst1[[n+1]][[2]], lst1[[n+2]][[2]]}; s=(a+b+c)/2; A=Sqrt[s(s-a)(s-b)(s-c)]; R=a*b*c/(4 A); AppendTo[lst, Ceiling@R], {n, 2, 100}]; lst
%Y A334763 Cf. A002858, A331676.
%K A334763 nonn
%O A334763 2,1
%A A334763 _Frank M Jackson_, May 10 2020